Surface triangulation and interpolation in python 3 - python-3.x

I have 3 lists of equal length of x, y and z coordinates.
With them, I need to triangulate a surface, and retrieve values that lie in a line over that surface. In other words, I need the values that lie on that surface that intersect a given plane.
Problem is, I have no idea where to start.
I have tried scipy interp2d, but it seems I need more z values them what I actually have (like shown in this answer: Python interpolation and extracting value of z for x and y?.
# this is the data I have
x = [0.0, 17.67599999997765, 49.08499999996275, 90.57299999985844, 136.60500000044703]
y = [0.0, 45.22349889159747, 66.50303846438841, 114.04427618243405, 187.7707039612985]
z = [0.0, 1.8700000000000045, 1.9539999999999509, 1.3929999999999154, 1.6299999999999955]
I need a final grid with x y z values that look something like this:
I don't really need too much resolution
My desired final result is to be able to retrieve specific values on top of that surface
Like the point line in this image:
I have also tried looking at geospatial libraries, but I couldn't find a solution either.
Maybe it's possible to interpolate the z values that I need? But I'm not really sure how to do this. I have never used scipy library before, and I'm still struggling to understand it.
I'm using python 3.9


You barely have any data, so if you don't choose your intersecting plane carefully, you'll get no results back (or nonsense back). This includes the case of x=y; you can't do that at all - so the graph you've shown is entirely inapplicable to your data.
import numpy as np
import scipy.interpolate
x = [0.0, 17.6759999999776500, 49.0849999999627500, 90.5729999998584400, 136.6050000004470300]
y = [0.0, 45.2234988915974700, 66.5030384643884100, 114.0442761824340500, 187.7707039612985000]
z = [0.0, 1.8700000000000045, 1.9539999999999509, 1.3929999999999154, 1.6299999999999955]
xyi = np.empty((200, 2))
xyi[:, 0] = np.arange(200)
xyi[:, 1] = xyi[:, 0] * 1.374
zi = scipy.interpolate.griddata(
points=(x, y), values=z,
xi=xyi,
method='cubic',
)
good_vals = ~np.isnan(zi)
xyz = np.empty((np.count_nonzero(good_vals), 3))
xyz[:, :2] = xyi[good_vals, :]
xyz[:, 2] = zi[good_vals]
print(xyz)
[[0.00000000e+00 0.00000000e+00 0.00000000e+00]
[1.00000000e+00 1.37400000e+00 4.68988354e-02]
[2.00000000e+00 2.74800000e+00 9.44855957e-02]
[3.00000000e+00 4.12200000e+00 1.42698116e-01]
[4.00000000e+00 5.49600000e+00 1.91474231e-01]
[5.00000000e+00 6.87000000e+00 2.40751776e-01]
[6.00000000e+00 8.24400000e+00 2.90468585e-01]
[7.00000000e+00 9.61800000e+00 3.40562494e-01]
[8.00000000e+00 1.09920000e+01 3.90971337e-01]
[9.00000000e+00 1.23660000e+01 4.41632950e-01]
[1.00000000e+01 1.37400000e+01 4.92485168e-01]
[1.10000000e+01 1.51140000e+01 5.43465824e-01]
[1.20000000e+01 1.64880000e+01 5.94512755e-01]
[1.30000000e+01 1.78620000e+01 6.45563795e-01]
[1.40000000e+01 1.92360000e+01 6.96556779e-01]
[1.50000000e+01 2.06100000e+01 7.47429542e-01]
[1.60000000e+01 2.19840000e+01 7.98119919e-01]
[1.70000000e+01 2.33580000e+01 8.48565744e-01]
[1.80000000e+01 2.47320000e+01 8.98704854e-01]
[1.90000000e+01 2.61060000e+01 9.48475082e-01]
[2.00000000e+01 2.74800000e+01 9.97814263e-01]
[2.10000000e+01 2.88540000e+01 1.04666023e+00]
[2.20000000e+01 3.02280000e+01 1.09495083e+00]
[2.30000000e+01 3.16020000e+01 1.14262388e+00]
[2.40000000e+01 3.29760000e+01 1.18961722e+00]
[2.50000000e+01 3.43500000e+01 1.23586870e+00]
[2.60000000e+01 3.57240000e+01 1.28131613e+00]
[2.70000000e+01 3.70980000e+01 1.32589737e+00]
[2.80000000e+01 3.84720000e+01 1.36955024e+00]
[2.90000000e+01 3.98460000e+01 1.41221257e+00]
[3.00000000e+01 4.12200000e+01 1.45382221e+00]
[3.10000000e+01 4.25940000e+01 1.49431699e+00]
[3.20000000e+01 4.39680000e+01 1.53363474e+00]
[3.30000000e+01 4.53420000e+01 1.57171329e+00]
[3.40000000e+01 4.67160000e+01 1.60849049e+00]
[3.50000000e+01 4.80900000e+01 1.64390417e+00]
[3.60000000e+01 4.94640000e+01 1.67789216e+00]
[3.70000000e+01 5.08380000e+01 1.71039230e+00]
[3.80000000e+01 5.22120000e+01 1.74134241e+00]
[3.90000000e+01 5.35860000e+01 1.77068035e+00]
[4.00000000e+01 5.49600000e+01 1.79834394e+00]
[4.10000000e+01 5.63340000e+01 1.82427101e+00]
[4.20000000e+01 5.77080000e+01 1.84839941e+00]
[4.30000000e+01 5.90820000e+01 1.87066696e+00]
[4.40000000e+01 6.04560000e+01 1.89101151e+00]
[4.50000000e+01 6.18300000e+01 1.90937088e+00]
[4.60000000e+01 6.32040000e+01 1.92570604e+00]
[4.70000000e+01 6.45780000e+01 1.94056076e+00]
[4.80000000e+01 6.59520000e+01 1.95421968e+00]
[4.90000000e+01 6.73260000e+01 1.96719695e+00]
[5.00000000e+01 6.87000000e+01 1.97928362e+00]
[5.10000000e+01 7.00740000e+01 1.98942207e+00]
[5.20000000e+01 7.14480000e+01 1.99700320e+00]
[5.30000000e+01 7.28220000e+01 2.00180807e+00]
[5.40000000e+01 7.41960000e+01 2.00361776e+00]
[5.50000000e+01 7.55700000e+01 2.00221333e+00]
[5.60000000e+01 7.69440000e+01 1.99737586e+00]
[5.70000000e+01 7.83180000e+01 1.98888640e+00]
[5.80000000e+01 7.96920000e+01 1.97652604e+00]
[5.90000000e+01 8.10660000e+01 1.96007583e+00]
[6.00000000e+01 8.24400000e+01 1.93931685e+00]
[6.10000000e+01 8.38140000e+01 1.91403017e+00]
[6.20000000e+01 8.51880000e+01 1.88497097e+00]
[6.30000000e+01 8.65620000e+01 1.85740456e+00]
[6.40000000e+01 8.79360000e+01 1.83466390e+00]
[6.50000000e+01 8.93100000e+01 1.81970145e+00]
[6.60000000e+01 9.06840000e+01 1.81546971e+00]
[6.70000000e+01 9.20580000e+01 1.82177337e+00]
[6.80000000e+01 9.34320000e+01 1.82842946e+00]
[6.90000000e+01 9.48060000e+01 1.83439551e+00]
[7.00000000e+01 9.61800000e+01 1.83969621e+00]
[7.10000000e+01 9.75540000e+01 1.84435625e+00]
[7.20000000e+01 9.89280000e+01 1.84840034e+00]
[7.30000000e+01 1.00302000e+02 1.85185316e+00]
[7.40000000e+01 1.01676000e+02 1.85473940e+00]
[7.50000000e+01 1.03050000e+02 1.85708377e+00]
[7.60000000e+01 1.04424000e+02 1.85891096e+00]
[7.70000000e+01 1.05798000e+02 1.86024566e+00]
[7.80000000e+01 1.07172000e+02 1.86111256e+00]
[7.90000000e+01 1.08546000e+02 1.86153636e+00]
[8.00000000e+01 1.09920000e+02 1.86154176e+00]
[8.10000000e+01 1.11294000e+02 1.86115344e+00]
[8.20000000e+01 1.12668000e+02 1.86039610e+00]
[8.30000000e+01 1.14042000e+02 1.85929444e+00]
[8.40000000e+01 1.15416000e+02 1.85787402e+00]
[8.50000000e+01 1.16790000e+02 1.85624017e+00]
[8.60000000e+01 1.18164000e+02 1.85445481e+00]
[8.70000000e+01 1.19538000e+02 1.85252055e+00]
[8.80000000e+01 1.20912000e+02 1.85043999e+00]
[8.90000000e+01 1.22286000e+02 1.84821574e+00]
[9.00000000e+01 1.23660000e+02 1.84585039e+00]
[9.10000000e+01 1.25034000e+02 1.84334656e+00]
[9.20000000e+01 1.26408000e+02 1.84070685e+00]
[9.30000000e+01 1.27782000e+02 1.83793385e+00]
[9.40000000e+01 1.29156000e+02 1.83503019e+00]
[9.50000000e+01 1.30530000e+02 1.83199845e+00]
[9.60000000e+01 1.31904000e+02 1.82884125e+00]
[9.70000000e+01 1.33278000e+02 1.82556119e+00]
[9.80000000e+01 1.34652000e+02 1.82216087e+00]
[9.90000000e+01 1.36026000e+02 1.81864290e+00]
[1.00000000e+02 1.37400000e+02 1.81500988e+00]
[1.01000000e+02 1.38774000e+02 1.81126441e+00]
[1.02000000e+02 1.40148000e+02 1.80740911e+00]
[1.03000000e+02 1.41522000e+02 1.80344657e+00]
[1.04000000e+02 1.42896000e+02 1.79937940e+00]
[1.05000000e+02 1.44270000e+02 1.79521020e+00]
[1.06000000e+02 1.45644000e+02 1.79094157e+00]
[1.07000000e+02 1.47018000e+02 1.78657613e+00]
[1.08000000e+02 1.48392000e+02 1.78211648e+00]
[1.09000000e+02 1.49766000e+02 1.77756521e+00]
[1.10000000e+02 1.51140000e+02 1.77292494e+00]
[1.11000000e+02 1.52514000e+02 1.76819827e+00]
[1.12000000e+02 1.53888000e+02 1.76338780e+00]
[1.13000000e+02 1.55262000e+02 1.75849613e+00]
[1.14000000e+02 1.56636000e+02 1.75352588e+00]
[1.15000000e+02 1.58010000e+02 1.74847964e+00]
[1.16000000e+02 1.59384000e+02 1.74336002e+00]
[1.17000000e+02 1.60758000e+02 1.73816962e+00]
[1.18000000e+02 1.62132000e+02 1.73291105e+00]
[1.19000000e+02 1.63506000e+02 1.72758692e+00]
[1.20000000e+02 1.64880000e+02 1.72219982e+00]
[1.21000000e+02 1.66254000e+02 1.71675236e+00]
[1.22000000e+02 1.67628000e+02 1.71124714e+00]
[1.23000000e+02 1.69002000e+02 1.70568677e+00]
[1.24000000e+02 1.70376000e+02 1.70007386e+00]
[1.25000000e+02 1.71750000e+02 1.69441100e+00]
[1.26000000e+02 1.73124000e+02 1.68870081e+00]
[1.27000000e+02 1.74498000e+02 1.68294588e+00]
[1.28000000e+02 1.75872000e+02 1.67714882e+00]
[1.29000000e+02 1.77246000e+02 1.67131224e+00]
[1.30000000e+02 1.78620000e+02 1.66543873e+00]
[1.31000000e+02 1.79994000e+02 1.65953091e+00]
[1.32000000e+02 1.81368000e+02 1.65359138e+00]
[1.33000000e+02 1.82742000e+02 1.64762273e+00]
[1.34000000e+02 1.84116000e+02 1.64162758e+00]
[1.35000000e+02 1.85490000e+02 1.63560853e+00]
[1.36000000e+02 1.86864000e+02 1.62956819e+00]]

Related

Pan Tompkins Lowpass filter overflow

The Pan Tompkins algorithm1 for removing noise from an ECG/EKG is cited often. They use a low pass filter, followed by a high pass filter. The output of the high pass filter looks great. But (depending on starting conditions) the output of the low pass filter will continuously increase or decrease. Given enough time, your numbers will eventually get to a size that the programming language cannot handle and rollover. If I run this on an Arduino (which uses a variant of C), it rolls over on the order of 10 seconds. Not ideal. Is there a way to get rid of this bias? I've tried messing with initial conditions, but I'm fresh out of ideas. The advantage of this algorithm is that it's not very computationally intensive and will run comfortably on a modest microprocessor.
1 Pan, Jiapu; Tompkins, Willis J. (March 1985). "A Real-Time QRS Detection Algorithm". IEEE Transactions on Biomedical Engineering. BME-32 (3): 230–236.
Python code to illustrate problem. Uses numpy and matplotlib:
import numpy as np
import matplotlib.pyplot as plt
#low-pass filter
def lpf(x):
y = x.copy()
for n in range(len(x)):
if(n < 12):
continue
y[n,1] = 2*y[n-1,1] - y[n-2,1] + x[n,1] - 2*x[n-6,1] + x[n-12,1]
return y
#high-pass filter
def hpf(x):
y = x.copy()
for n in range(len(x)):
if(n < 32):
continue
y[n,1] = y[n-1,1] - x[n,1]/32 + x[n-16,1] - x[n-17,1] + x[n-32,1]/32
return y
ecg = np.loadtxt('ecg_data.csv', delimiter=',',skiprows=1)
plt.plot(ecg[:,0], ecg[:,1])
plt.title('Raw Data')
plt.grid(True)
plt.savefig('raw.png')
plt.show()
#Application of lpf
f1 = lpf(ecg)
plt.plot(f1[:,0], f1[:,1])
plt.title('After Pan-Tompkins LPF')
plt.xlabel('time')
plt.ylabel('mV')
plt.grid(True)
plt.savefig('lpf.png')
plt.show()
#Application of hpf
f2 = hpf(f1[16:,:])
print(f2[-300:-200,1])
plt.plot(f2[:-100,0], f2[:-100,1])
plt.title('After Pan-Tompkins LPF+HPF')
plt.xlabel('time')
plt.ylabel('mV')
plt.grid(True)
plt.savefig('hpf.png')
plt.show()
raw data in CSV format:
timestamp,ecg_measurement
96813044,2.2336266040
96816964,2.1798632144
96820892,2.1505377292
96824812,2.1603128910
96828732,2.1554253101
96832660,2.1163244247
96836580,2.0576734542
96840500,2.0381231307
96844420,2.0527858734
96848340,2.0674486160
96852252,2.0283479690
96856152,1.9648094177
96860056,1.9208210945
96863976,1.9159335136
96867912,1.9208210945
96871828,1.8768328666
96875756,1.7986314296
96879680,1.7448680400
96883584,1.7155425548
96887508,1.7057673931
96891436,1.6520038604
96895348,1.5591397285
96899280,1.4809384346
96903196,1.4467253684
96907112,1.4369501113
96911032,1.3978494453
96914956,1.3440860509
96918860,1.2952101230
96922788,1.3000977039
96926684,1.3343108892
96930604,1.3440860509
96934516,1.3489736318
96938444,1.3294233083
96942364,1.3782991170
96946284,1.4222873687
96950200,1.4516129493
96954120,1.4369501113
96958036,1.4320625305
96961960,1.4565005302
96965872,1.4907135963
96969780,1.5053763389
96973696,1.4613881111
96977628,1.4125122070
96981548,1.4076246261
96985476,1.4467253684
96989408,1.4809384346
96993324,1.4760508537
96997236,1.4711632728
97001160,1.4907135963
97005084,1.5444769859
97008996,1.5982404708
97012908,1.5835777282
97016828,1.5591397285
97020756,1.5786901473
97024676,1.6324535369
97028604,1.6911046504
97032516,1.6959922313
97036444,1.6764417648
97040364,1.6813293457
97044296,1.7155425548
97048216,1.7448680400
97052120,1.7253177165
97056048,1.6911046504
97059968,1.6911046504
97063880,1.7302052974
97067796,1.7741935253
97071724,1.7693059444
97075644,1.7350928783
97079564,1.7595307826
97083480,1.8719452857
97087396,2.0381231307
97091316,2.2482893466
97095244,2.4828934669
97099156,2.7468230724
97103088,2.9960899353
97106996,3.0987291336
97110912,2.9178886413
97114836,2.5171065330
97118756,2.0185728073
97122668,1.5053763389
97126584,1.1094819307
97130492,0.8015640258
97134396,0.5767350673
97138308,0.4545454502
97142212,0.4349951267
97146124,0.4692081928
97150020,0.4887585639
97153924,0.4594330310
97157828,0.4105571746
97161740,0.3861192512
97165660,0.3763440847
97169580,0.3714565038
97173492,0.3225806236
97177404,0.2639296054
97181316,0.2394916772
97185236,0.2297165155
97189148,0.2443792819
97193060,0.2248289346
97196972,0.1857282543
97200900,0.1808406734
97204812,0.2199413537
97208732,0.2492668628
97212652,0.2443792819
97216572,0.2199413537
97220484,0.2248289346
97224404,0.2834799575
97228316,0.3274682044
97232228,0.3665689229
97236132,0.3861192512
97240036,0.4398827075
97243936,0.5083088874
97247836,0.6109481811
97251748,0.7086998939
97255660,0.7771260738
97259568,0.8553275108
97263476,0.9775171279
97267392,1.1094819307
97271308,1.1974585056
97275228,1.2512218952
97279148,1.2952101230
97283056,1.3734115362
97286992,1.4760508537
97290900,1.5493645668
97294820,1.5738025665
97298740,1.5982404708
97302652,1.6471162796
97306584,1.7106549739
97310500,1.7546432018
97314420,1.7546432018
97318340,1.7644183635
97322272,1.8084066390
97326168,1.8621701240
97330072,1.8963831901
97333988,1.8817204475
97337912,1.8572825431
97341840,1.8670577049
97345748,1.8866080284
97349668,1.8768328666
97353580,1.8230694770
97357500,1.7595307826
97361424,1.7302052974
97365332,1.7350928783
97369252,1.6959922313
97373168,1.6226783752
97377092,1.5298142433
97381012,1.4613881111
97384940,1.4320625305
97388860,1.4076246261
97392780,1.3440860509
97396676,1.2658846378
97400604,1.2121212482
97404532,1.1974585056
97408444,1.1779080629
97412356,1.1192570924
97416264,1.0361680984
97420164,0.9628542900
97424068,0.9286412239
97427988,0.9042033195
97431892,0.8406646728
97435804,0.7575757503
97439708,0.6940371513
97443628,0.6793744087
97447540,0.6793744087
97451452,0.6549364089
97455356,0.6060606002
97459240,0.5767350673
97463140,0.6011730194
97467044,0.6451612472
97470964,0.6842619895
97474884,0.6891495704
97478796,0.7184750556
97482700,0.8064516067
97486612,0.8846529960
97490516,0.9335288047
97494428,0.9530791282
97498340,0.9481915473
97502256,0.9726295471
97506156,0.9921798706
97510060,0.9726295471
97513980,0.8846529960
97517884,0.7869012832
97521796,0.7086998939
97525692,0.6549364089
97529604,0.5913978099
97533516,0.4887585639
97537428,0.3567937374
97541348,0.2639296054
97545260,0.2003910064
97549148,0.1417399787
97553060,0.0928641223
97556980,0.0537634420
97560892,0.0342130994
97564804,0.0146627559
97568708,0.0244379281
97572628,0.0048875851
97576500,0.0000000000
97580324,0.0000000000
97584172,0.0097751703
97588060,0.0244379281
97591980,0.0195503416
97595900,0.0146627559
97599812,0.0488758563
97603724,0.1319648027
97607628,0.2248289346
97611548,0.3030303001
97615444,0.3665689229
97619364,0.4496578693
97623276,0.5718474864
97627176,0.7038123130
97631076,0.8064516067
97634988,0.8699902534
97638900,0.9384163856
97642816,1.0361680984
97646720,1.1485825777
97650644,1.2365591526
97654572,1.2658846378
97658492,1.2805473804
97662404,1.3294233083
97666324,1.3782991170
97670244,1.3831867027
97674148,1.3489736318
97678068,1.3049852848
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100179188,1.9501466751
100183132,2.1016616821
100187084,2.3216030597
100191036,2.5904202461
100194996,2.8787879943
100198956,3.1769306659
100202916,3.4555230140
100206876,3.5826001167
100210836,3.4115347862
100214804,3.0205278396
100218748,2.5317692756
100222708,2.1016616821
100226676,1.7937438488
100230640,1.5933528900
100234592,1.4858260154
100238548,1.5053763389
100242500,1.6422286987
100246456,1.8377322196
100250424,1.9990224838
100254380,2.0967741012
100258340,2.1652004718
100262284,2.2434017658
100266236,2.3411533832
100270196,2.4242424964
100274140,2.4731183052
100278112,2.4975562095
100282068,2.5562071800
100286020,2.6441838741
100289992,2.7028348445
100293948,2.7077224254
100297908,2.7126100063
100301868,2.7468230724
100305820,2.8054740905
100309764,2.8250244140
100313724,2.7908113002
100317676,2.7370479106
100321636,2.7223851680
100325600,2.7419354915
100329556,2.7517106533
100333516,2.7126100063
100337468,2.6735093593
100341428,2.6686217784
100345392,2.7028348445
100349348,2.7272727489
100353316,2.6881721019
100357260,2.6441838741
100361212,2.6588466167
100365180,2.6832845211
100369140,2.7077224254
100373100,2.6783969402
100377052,2.6148581504
100381012,2.6001954078
100384960,2.6246333122
100388916,2.6490714550
100392860,2.6197457313
100396828,2.5659823417
100400788,2.5562071800
100404740,2.5806450843
100408692,2.6099705696
100412644,2.5904202461
100416588,2.5366568565
100420548,2.5268816947
100424508,2.5610947608
100428460,2.5953078269
100432412,2.5757575035
100436372,2.5171065330
100440324,2.4926686286
100444276,2.5219941139
100448228,2.5366568565
100452196,2.5073313713
100456148,2.4389052391
100460108,2.3949170112
100464068,2.3753666877
100468028,2.3655912876
100471988,2.3069403171

The trick seems to be the initial conditions. Load the first 13 values of input and output of low pass filter to zero and the bias goes away.
#low-pass filter
def lpf(x):
y = x.copy()
for n in range(13):
y[n,1] = 0
x[n,1] = 0
for n in range(len(x)):
if(n < 12):
continue
y[n,1] = 2*y[n-1,1] - y[n-2,1] + x[n,1] - 2*x[n-6,1] + x[n-12,1]
return y

How to find the shortest distance between two line segments capturing the sign values with python

I have a pandas dataframe of the form:
benchmark_x benchmark_y ref_point_x ref_point_y
0 525039.140 175445.518 525039.145 175445.539
1 525039.022 175445.542 525039.032 175445.568
2 525038.944 175445.558 525038.954 175445.588
3 525038.855 175445.576 525038.859 175445.576
4 525038.797 175445.587 525038.794 175445.559
5 525038.689 175445.609 525038.679 175445.551
6 525038.551 175445.637 525038.544 175445.577
7 525038.473 175445.653 525038.459 175445.594
8 525038.385 175445.670 525038.374 175445.610
9 525038.306 175445.686 525038.289 175445.626
I am trying to find the shortest distance from the line to the benchmark such that if the line is above the benchmark the distance is positive and if it is below the benchmark the distance is negative. See image below:
I used the KDTree from scipy like so:
from scipy.spatial import KDTree
tree=KDTree(df[["benchmark_x", "benchmark_y"]])
test = df.apply(lambda row: tree.query(row[["ref_point_x", "ref_point_y"]]), axis=1)
test=test.apply(pd.Series, index=["distance", "index"])
This seems to work except that it fails to capture the negative values as a result that the line is below the benchmark.

# recreating your example
columns = "benchmark_x benchmark_y ref_point_x ref_point_y".split(" ")
data = """525039.140 175445.518 525039.145 175445.539
525039.022 175445.542 525039.032 175445.568
525038.944 175445.558 525038.954 175445.588
525038.855 175445.576 525038.859 175445.576
525038.797 175445.587 525038.794 175445.559
525038.689 175445.609 525038.679 175445.551
525038.551 175445.637 525038.544 175445.577
525038.473 175445.653 525038.459 175445.594
525038.385 175445.670 525038.374 175445.610
525038.306 175445.686 525038.289 175445.626"""
data = [float(x) for x in data.replace("\n"," ").split(" ") if len(x)>0]
arr = np.array(data).reshape(-1,4)
df = pd.DataFrame(arr, columns=columns)
# adding your two new columns to the df
from scipy.spatial import KDTree
tree=KDTree(df[["benchmark_x", "benchmark_y"]])
df["distance"], df["index"] = tree.query(df[["ref_point_x", "ref_point_y"]])
Now to compare if one line is above the other or not, we have to evaluate y at the same x position. Therefore we need to interpolate the y points for the x positions of the other line.
df = df.sort_values("ref_point_x") # sorting is required for interpolation
xy_refpoint = df[["ref_point_x", "ref_point_y"]].values
df["ref_point_y_at_benchmark_x"] = np.interp(df["benchmark_x"], xy_refpoint[:,0], xy_refpoint[:,1])
And finally your criterium can be evaluated and applied:
df["distance"] = np.where(df["ref_point_y_at_benchmark_x"] < df["benchmark_y"], -df["distance"], df["distance"])
# or change the < to <,>,<=,>= as you wish

Obtaining hyperpolarization depth from electrophysiological graph

I am working on electrophysiological data which is in .abf format.
I want to obtain the hyperpolarization depth as indicated above in the figure. This is what I have done so far;
import matplotlib.pyplot as plt
import pyabf
import pandas as pd
abf = pyabf.ABF("test.abf")
abf.setSweep(10) # I can access a given sweep. Here sweep 10
df = pd.DataFrame({'time': abf.sweepX, 'current':abf.sweepY})
df1 = df.loc[15650:15800]
df1.plot(x='time', y='current')
I am thinking to apply change in derivative to find the first point of interest (x1,y1) and then lower point (x2,y2), but it looks complex. I would appreciate if someone give some hint or procedure.
The dataset as follow,
time current
0.7825 -63.323975
0.78255 -63.171387
0.7826 -62.89673
0.78265 -62.713623
0.7827 -62.469482
0.78275 -62.37793
0.7828 -62.10327
0.78285 -61.950684
0.7829 -61.76758
0.78295 -61.584473
0.783 -61.401367
0.78305 -61.24878
0.7831 -61.035156
0.78315 -60.85205
0.7832 -60.72998
0.78325 -60.516357
0.7833 -60.455322
0.78335 -60.2417
0.7834 -60.08911
0.78345 -59.96704
0.7835 -59.814453
0.78355 -59.661865
0.7836 -59.509277
0.78365 -59.417725
0.7837 -59.23462
0.78375 -59.11255
0.7838 -58.95996
0.78385 -58.86841
0.7839 -58.685303
0.78395 -58.59375
0.784 -58.441162
0.78405 -58.34961
0.7841 -58.19702
0.78415 -58.044434
0.7842 -57.922363
0.78425 -57.769775
0.7843 -57.678223
0.78435 -57.434082
0.7844 -57.34253
0.78445 -56.9458
0.7845 -56.274414
0.78455 -54.96216
0.7846 -53.253174
0.78465 -51.208496
0.7847 -48.950195
0.78475 -46.325684
0.7848 -43.09082
0.78485 -38.42163
0.7849 -31.036377
0.78495 -22.033691
0.785 -13.397217
0.78505 -6.072998
0.7851 -0.61035156
0.78515 2.7160645
0.7852 3.9367676
0.78525 3.4179688
0.7853 1.3427734
0.78535 -1.4953613
0.7854 -5.0964355
0.78545 -9.185791
0.7855 -13.641357
0.78555 -18.249512
0.7856 -23.132324
0.78565 -27.98462
0.7857 -32.714844
0.78575 -37.261963
0.7858 -41.47339
0.78585 -45.22705
0.7859 -48.553467
0.78595 -51.54419
0.786 -53.985596
0.78605 -56.18286
0.7861 -58.013916
0.78615 -59.539795
0.7862 -60.760498
0.78625 -61.88965
0.7863 -62.652588
0.78635 -63.323975
0.7864 -63.934326
0.78645 -64.2395
0.7865 -64.60571
0.78655 -64.78882
0.7866 -65.00244
0.78665 -64.971924
0.7867 -65.093994
0.78675 -65.03296
0.7868 -64.971924
0.78685 -64.819336
0.7869 -64.78882
0.78695 -64.66675
0.787 -64.48364
0.78705 -64.42261
0.7871 -64.2395
0.78715 -64.11743
0.7872 -63.964844
0.78725 -63.842773
0.7873 -63.659668
0.78735 -63.568115
0.7874 -63.446045
0.78745 -63.26294
0.7875 -63.171387
0.78755 -62.98828
0.7876 -62.89673
0.78765 -62.74414
0.7877 -62.713623
0.78775 -62.530518
0.7878 -62.438965
0.78785 -62.37793
0.7879 -62.25586
0.78795 -62.164307
0.788 -62.042236
0.78805 -62.01172
0.7881 -61.88965
0.78815 -61.88965
0.7882 -61.73706
0.78825 -61.706543
0.7883 -61.645508
0.78835 -61.61499
0.7884 -61.523438
0.78845 -61.462402
0.7885 -61.431885
0.78855 -61.340332
0.7886 -61.37085
0.78865 -61.279297
0.7887 -61.279297
0.78875 -61.157227
0.7888 -61.187744
0.78885 -61.09619
0.7889 -61.157227
0.78895 -61.12671
0.789 -61.09619
0.78905 -61.12671
0.7891 -61.00464
0.78915 -61.00464
0.7892 -60.97412
0.78925 -60.97412
0.7893 -60.943604
0.78935 -61.00464
0.7894 -60.913086
0.78945 -60.97412
0.7895 -60.943604
0.78955 -60.913086
0.7896 -60.943604
0.78965 -60.85205
0.7897 -60.85205
0.78975 -60.821533
0.7898 -60.88257
0.78985 -60.88257
0.7899 -60.913086
0.78995 -60.88257
0.79 -60.913086

We can plot the difference in current between consecutive points (which essentially is to a constant factor the derivative, since times are evenly spaced). First chart shows the actual diffs. Based on this we can set some threshold, such as 0.3, and apply it to filter the main DataFrame. The filtered values are shown in orange on the second chart:
fig, ax = plt.subplots(2, figsize=(8,8))
# plot derivative
df['current'].diff().plot(ax=ax[0])
# current
threshold = 0.4
df['filtered'] = df.loc[df['current'].diff().abs() > threshold]
df.plot(ax=ax[1])
# add spans
x = df['filtered'].dropna()
ax[1].axhspan(x.iloc[0], x.iloc[-1], alpha=0.3, edgecolor='skyblue', facecolor="none", hatch='////')
ax[1].axvspan(x.index.min(), x.index.max(), alpha=0.3, edgecolor='orange', facecolor="none", hatch='\\\\')
Output:
If you're interested in range values, you can dropna values in the filtered subset and find min and max from the index:
print('min', df['filtered'].dropna().index.min())
print('max', df['filtered'].dropna().index.max())
Output:
min 0.78445
max 0.7865
For the value of the gap you can use:
abs(df['filtered'].dropna().iloc[-1] - df['filtered'].dropna().iloc[0])
Output:
7.6599100000000035
Note: We can alternatively also get left edges of these spans as points where diff in the point is lower than the threshold and diff in the next point is higher than the threshold, and similarly for the right edges. This would also work in case we have multiple peaks:
threshold = 0.3
x = df['current'].diff().abs()
spanA = df.loc[(x < threshold) & (x.shift(-1) >= threshold)]
spanB = df.loc[(x >= threshold) & (x.shift(-1) < threshold)]
print(spanA)
current
time
0.7844 -57.34253
print(spanB)
current
time
0.7865 -64.60571

How to record the value of a variable within odeint?

I would like to know if there is a way to record the value of a specific variable within the function of integration, without having to print it within the definition of the function, which in many cases, due to the algorithm of prediction-correction, lead to more or less values than the final vector returned by the function?
Example let's try with this code:
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
def essai(y, t):
a = y[0]
c1 = a
a = c1 / a**2
return [a]
# Solving
essai0 = [10]
t = np.linspace(0, 2000, 10)
y = odeint(essai, essai0, t)
a = y[:, 0]
# Graphs
fig, ax = plt.subplots()
ax.plot(t, a, 'k--', label='a')
legend = ax.legend(loc='lower right', shadow=True, fontsize='x-large')
legend.get_frame().set_facecolor('#FFFCCC') # 00FFCC
plt.xlabel('x')
plt.ylabel('y')
plt.title('y vs x')
plt.show()
I would like to record the values of c1 which depends on a. What should I do?
If I print, it I get (because of pred-corr algorithm):
10.0
10.001203411814794
10.00120326701222
10.002406534059283
10.00240638930896
10.031168251789499
10.03116843523562
10.059847893733858
10.059848247411573
10.088446178306066
10.088446526968276
10.178981333917179
10.1789826635142
10.26872274187664
10.268720875457465
10.251795853148066
10.251794757670828
10.324093402400061
10.324093338929458
10.395889284010963
10.395889126663482
10.467192620394076
10.467192470562162
10.60836217080531
10.608361512785885
10.747675991273601
10.747676529983982
10.885208084361661
10.88520861500753
11.021024408838219
11.021024559158226
11.15518691385528
11.15518704871583
11.389028983440005
11.389029612664437
11.618166387462095
11.618166372845774
11.842871925632974
11.842870666797078
12.063390475531826
12.0633901508557
12.279950446401756
12.279950250452782
12.492757035192547
12.492756877414479
12.790475076345272
12.79047467718475
13.081418818481728
13.081418595295522
13.366029970579808
13.366030900758636
13.644707388512776
13.644707798536366
13.917805722870085
13.917805853240296
14.185647189512732
14.185647276304193
14.448524340486092
14.44852440612534
14.849045554474056
14.849045812160185
15.239043242348172
15.239044113472564
15.619306858637934
15.619307570817467
15.990530200625596
15.990530706701604
16.353328829257094
16.35332918566708
16.70825155213741
16.708251810028536
17.055790075751844
17.055790265472186
17.52054793291328
17.520548366986496
17.97329155702487
17.97329263337524
18.414908470097206
18.41490919183692
18.84617978510828
18.846180323693773
19.26780035288661
19.26780072790131
19.68039039537204
19.680390669145883
20.084506483562638
20.084506685872917
20.63204921728682
20.632049705019547
21.165431430483114
21.16543268212929
21.685699626883885
21.685700483180575
22.193774842932424
22.193775478119036
22.69047628806277
22.69047673120133
23.176535191516802
23.1765355148269
23.652607704971896
23.652607943862492
24.296731084127696
24.296731656936466
24.92421316694978
24.924214631653445
25.536282592848192
25.536283593100098
26.134020839947766
26.134021582629195
26.718389929663125
26.718390447872228
27.290248649274574
27.290249027491374
27.8503676838429
27.85036796338048
28.60821935477876
28.608220025227006
29.346505899333515
29.346507613905608
30.066670806260635
30.066671977520553
30.769984796557875
30.769985666417984
31.457578314647648
31.457578921761066
32.13046057231114
32.13046101551341
32.78953730742519
32.789537635058444
33.68118868621462
33.68118947182226
34.54983545122736
34.549837459883506
35.39717380841791
35.397175180698845
36.22469707822626
36.224698097642104
37.033733817898586
37.03373452954837
37.82547018189015
37.82547070150822
38.60097077071101
38.60097115490064
39.65004988104156
39.650050802111195
40.67207751401193
40.67207986867377
41.669047220267416
41.66904882908885
42.64271422854618
42.64271542393563
43.594640193459966
43.59464102811222
44.52621945824691
44.52622006777859
45.43870353935591
45.438703990091476
46.67300975177773
46.673010832232926
47.87550305124021
47.87550581301012
49.04852683447106
49.04852872160157
50.194144483083306
50.19414588551954
51.3141919066777
51.31419288605143
52.41030839692969
52.41030911225109
53.48396538985435
53.483965918885744
54.9362075454971
54.93620881348237
56.35103457439806
56.35103781516747
57.73120149400896
57.731203708595864
59.07913425147381
59.07913589751868
60.39699143853227
60.39699258818307
61.68670054765226
61.68670138744394
62.94999176730058
62.949992388453296
64.65865496068966
64.65865644932029
Which is much more values than I may expect with t = np.linspace(0, 2000, 10) which divide the intervale of time in tenth of 200.
I have thought to this problem for a long time without find a really good way to do it and I would be delighted to know how to bypass this problem.

There is no relation between the evaluation points of the ODE function in the internal solver steps and the requested sample points of the solution for the output. Moreover, the evaluation points can deviate from the solution trajectory with some error of an order lower than the order of the integration method.
The easiest way to do what you want in a structured fashion is to define the c1 function as a separate function and then to call it on the results
def c1_func(y): return y[0]
def essai(y, t):
a = y[0]
c1 = c1_func(y)
a = c1 / a**2
return [a]
...
y = odeint(...
c1_val = c1_func(y.T)
plt.plot(x, c1_val)
or so.

How to fit multiple spectra to a spectrum by extracting the coefficients from a least-squares linear fit?

I want to fit two different spectra to my original spectrum. The two different spectra have x and y values of:
x_1 = 1700.42
1700.9
1701.38
1701.86
1702.34
1702.83
1703.31
1703.79
1704.27
1704.75
1705.24
1705.72
1706.2
1706.68
1707.17
1707.65
1708.13
1708.61
1709.09
1709.58
1710.06
1710.54
1711.02
1711.5
1711.99
1712.47
1712.95
1713.43
1713.91
1714.4
1714.88
1715.36
1715.84
1716.33
1716.81
1717.29
1717.77
1718.25
1718.74
1719.22
1719.7
1720.18
1720.66
1721.15
1721.63
1722.11
1722.59
1723.08
1723.56
1724.04
1724.52
1725
1725.49
1725.97
1726.45
1726.93
1727.41
1727.9
1728.38
1728.86
1729.34
1729.82
1730.31
1730.79
1731.27
1731.75
1732.24
1732.72
1733.2
1733.68
1734.16
1734.65
1735.13
1735.61
1736.09
1736.57
1737.06
1737.54
1738.02
1738.5
1738.98
1739.47
1739.95
1740.43
1740.91
1741.4
1741.88
1742.36
1742.84
1743.32
1743.81
1744.29
1744.77
1745.25
1745.73
1746.22
1746.7
1747.18
1747.66
1748.14
1748.63
1749.11
1749.59
1750.07
1750.56
y_1 = 0.00285
0.00289
0.00290
0.00292
0.00297
0.00304
0.00310
0.00314
0.00319
0.00323
0.00327
0.00333
0.00340
0.00344
0.00347
0.00352
0.00358
0.00364
0.00369
0.00374
0.00382
0.00388
0.00392
0.00397
0.00403
0.00408
0.00414
0.00420
0.00428
0.00436
0.00444
0.00451
0.00456
0.00461
0.00468
0.00474
0.00480
0.00486
0.00493
0.00501
0.00509
0.00517
0.00524
0.00530
0.00535
0.00543
0.00551
0.00558
0.00564
0.00571
0.00578
0.00587
0.00594
0.00599
0.00607
0.00615
0.00623
0.00631
0.00636
0.00645
0.00657
0.00666
0.00673
0.00682
0.00688
0.00695
0.00704
0.00713
0.00722
0.00732
0.00741
0.00750
0.00758
0.00768
0.00777
0.00783
0.00788
0.00792
0.00795
0.00799
0.00803
0.00806
0.00807
0.00804
0.00800
0.00795
0.00787
0.00779
0.00767
0.00753
0.00737
0.00719
0.00699
0.00677
0.00652
0.00626
0.00599
0.00572
0.00546
0.00519
0.00492
0.00465
0.00437
0.00413
0.00391
and x_2 = 1700.42 1700.9 1701.38 1701.86 1702.34 1702.83 1703.31 1703.79 1704.27 1704.75 1705.24 1705.72 1706.2 1706.68 1707.17 1707.65 1708.13 1708.61 1709.09 1709.58 1710.06 1710.54 1711.02 1711.5 1711.99 1712.47 1712.95 1713.43 1713.91 1714.4 1714.88 1715.36 1715.84 1716.33 1716.81 1717.29 1717.77 1718.25 1718.74 1719.22 1719.7 1720.18 1720.66 1721.15 1721.63 1722.11 1722.59 1723.08 1723.56 1724.04 1724.52 1725 1725.49 1725.97 1726.45 1726.93 1727.41 1727.9 1728.38 1728.86 1729.34 1729.82 1730.31 1730.79 1731.27 1731.75 1732.24 1732.72 1733.2 1733.68 1734.16 1734.65 1735.13 1735.61 1736.09 1736.57 1737.06 1737.54 1738.02 1738.5 1738.98 1739.47 1739.95 1740.43 1740.91 1741.4 1741.88 1742.36 1742.84 1743.32 1743.81 1744.29 1744.77 1745.25 1745.73 1746.22 1746.7 1747.18 1747.66 1748.14 1748.63 1749.11 1749.59 1750.07 1750.56
y_2 = 0.00182478
0.00198449
0.0021542
0.00230491
0.00248363
0.00269334
0.00289705
0.00308676
0.00330747
0.00358919
0.0038779
0.00415561
0.00444332
0.00474103
0.00507474
0.00542346
0.00576517
0.00613688
0.00651859
0.0068873
0.00727502
0.00767773
0.00808544
0.00851815
0.00894486
0.00935658
0.00979429
0.010245
0.0106727
0.0110844
0.0115191
0.0119878
0.0124556
0.0128823
0.013274
0.0137237
0.0142374
0.0147181
0.0151798
0.0156495
0.0160963
0.016534
0.0169657
0.0173574
0.0177211
0.0180818
0.0184125
0.0187012
0.0189339
0.0191077
0.0192454
0.0193291
0.0193638
0.0193495
0.0192672
0.0191119
0.0188696
0.0185614
0.0181941
0.0176948
0.0170465
0.0162762
0.0153449
0.0142406
0.0129863
0.0115801
0.0100468
0.00844248
0.00692419
0.0055719
0.00435861
0.00340132
0.00270704
0.00213775
0.00168046
0.00134117
0.00109188
9.16595E-4
7.80307E-4
6.65019E-4
5.62731E-4
4.75443E-4
4.42155E-4
4.49867E-4
4.29579E-4
3.9929E-4
3.83002E-4
3.51714E-4
3.38426E-4
3.40138E-4
3.2985E-4
3.27562E-4
3.24274E-4
3.06986E-4
2.92698E-4
3.0041E-4
3.12121E-4
2.84833E-4
2.47545E-4
2.41257E-4
2.34969E-4
2.27681E-4
2.47393E-4
2.60105E-4
2.25817E-4
My original data:
x_orig = 1700.42
1700.9
1701.38
1701.86
1702.34
1702.83
1703.31
1703.79
1704.27
1704.75
1705.24
1705.72
1706.2
1706.68
1707.17
1707.65
1708.13
1708.61
1709.09
1709.58
1710.06
1710.54
1711.02
1711.5
1711.99
1712.47
1712.95
1713.43
1713.91
1714.4
1714.88
1715.36
1715.84
1716.33
1716.81
1717.29
1717.77
1718.25
1718.74
1719.22
1719.7
1720.18
1720.66
1721.15
1721.63
1722.11
1722.59
1723.08
1723.56
1724.04
1724.52
1725
1725.49
1725.97
1726.45
1726.93
1727.41
1727.9
1728.38
1728.86
1729.34
1729.82
1730.31
1730.79
1731.27
1731.75
1732.24
1732.72
1733.2
1733.68
1734.16
1734.65
1735.13
1735.61
1736.09
1736.57
1737.06
1737.54
1738.02
1738.5
1738.98
1739.47
1739.95
1740.43
1740.91
1741.4
1741.88
1742.36
1742.84
1743.32
1743.81
1744.29
1744.77
1745.25
1745.73
1746.22
1746.7
1747.18
1747.66
1748.14
1748.63
1749.11
1749.59
1750.07
1750.56
y_orig = 0.011507
0.0121121
0.0127542
0.0132673
0.0137554
0.0143684
0.0148995
0.0154036
0.0159997
0.0165907
0.0172408
0.0178499
0.018388
0.019089
0.0197701
0.0203572
0.0210393
0.0216564
0.0222324
0.0228305
0.0233166
0.0238667
0.0244387
0.0248918
0.0254159
0.025865
0.026158
0.0265131
0.0267652
0.0269333
0.0271824
0.0273214
0.0274515
0.0274626
0.0271257
0.0269957
0.0270148
0.0267899
0.026651
0.026427
0.0260381
0.0257212
0.0252253
0.0247254
0.0243314
0.0237925
0.0233076
0.0227997
0.0221607
0.0216288
0.0210079
0.020299
0.019702
0.0189881
0.0182382
0.0175053
0.0165944
0.0157524
0.0149355
0.0139746
0.0131167
0.0122307
0.0112948
0.0105009
0.00964397
0.00886105
0.00821613
0.0074542
0.00685928
0.00640136
0.00589444
0.00568351
0.00555559
0.00529467
0.00514074
0.00495682
0.0047789
0.00469697
0.00453005
0.00441613
0.0042912
0.00408328
0.00409536
0.00412444
0.00400951
0.00397959
0.00389367
0.00375074
0.00372082
0.0036819
0.00365497
0.00363905
0.00353413
0.00348721
0.00346528
0.00336936
0.00334044
0.00331251
0.00322459
0.00316767
0.00308874
0.00304882
0.0030859
0.00301798
0.00287005
How do I fit the two spectra to the original spectrum by extracting the coefficients from a least-squares linear fit? I use scipy.optimize.curve_fit to fit using gaussians, but now I need to just fit data.

When I scatterplot the data, it looks like three separate, somewhat asymmetrical peaks - I would think this requires one peak fit for each set of data. Please see my image and code below.
import numpy, matplotlib
import matplotlib.pyplot as plt
x_1 = numpy.array([1700.42, 1700.9, 1701.38, 1701.86, 1702.34, 1702.83, 1703.31, 1703.79, 1704.27, 1704.75, 1705.24, 1705.72, 1706.2, 1706.68, 1707.17, 1707.65, 1708.13, 1708.61, 1709.09, 1709.58, 1710.06, 1710.54, 1711.02, 1711.5, 1711.99, 1712.47, 1712.95, 1713.43, 1713.91, 1714.4, 1714.88, 1715.36, 1715.84, 1716.33, 1716.81, 1717.29, 1717.77, 1718.25, 1718.74, 1719.22, 1719.7, 1720.18, 1720.66, 1721.15, 1721.63, 1722.11, 1722.59, 1723.08, 1723.56, 1724.04, 1724.52, 1725, 1725.49, 1725.97, 1726.45, 1726.93, 1727.41, 1727.9, 1728.38, 1728.86, 1729.34, 1729.82, 1730.31, 1730.79, 1731.27, 1731.75, 1732.24, 1732.72, 1733.2, 1733.68, 1734.16, 1734.65, 1735.13, 1735.61, 1736.09, 1736.57, 1737.06, 1737.54, 1738.02, 1738.5, 1738.98, 1739.47, 1739.95, 1740.43, 1740.91, 1741.4, 1741.88, 1742.36, 1742.84, 1743.32, 1743.81, 1744.29, 1744.77, 1745.25, 1745.73, 1746.22, 1746.7, 1747.18, 1747.66, 1748.14, 1748.63, 1749.11, 1749.59, 1750.07, 1750.56])
y_1 = numpy.array([0.00285, 0.00289, 0.00290, 0.00292, 0.00297, 0.00304, 0.00310, 0.00314, 0.00319, 0.00323, 0.00327, 0.00333, 0.00340, 0.00344, 0.00347, 0.00352, 0.00358, 0.00364, 0.00369, 0.00374, 0.00382, 0.00388, 0.00392, 0.00397, 0.00403, 0.00408, 0.00414, 0.00420, 0.00428, 0.00436, 0.00444, 0.00451, 0.00456, 0.00461, 0.00468, 0.00474, 0.00480, 0.00486, 0.00493, 0.00501, 0.00509, 0.00517, 0.00524, 0.00530, 0.00535, 0.00543, 0.00551, 0.00558, 0.00564, 0.00571, 0.00578, 0.00587, 0.00594, 0.00599, 0.00607, 0.00615, 0.00623, 0.00631, 0.00636, 0.00645, 0.00657, 0.00666, 0.00673, 0.00682, 0.00688, 0.00695, 0.00704, 0.00713, 0.00722, 0.00732, 0.00741, 0.00750, 0.00758, 0.00768, 0.00777, 0.00783, 0.00788, 0.00792, 0.00795, 0.00799, 0.00803, 0.00806, 0.00807, 0.00804, 0.00800, 0.00795, 0.00787, 0.00779, 0.00767, 0.00753, 0.00737, 0.00719, 0.00699, 0.00677, 0.00652, 0.00626, 0.00599, 0.00572, 0.00546, 0.00519, 0.00492, 0.00465, 0.00437, 0.00413, 0.00391])
x_2 = numpy.array([1700.42, 1700.9, 1701.38, 1701.86, 1702.34, 1702.83, 1703.31, 1703.79, 1704.27, 1704.75, 1705.24, 1705.72, 1706.2, 1706.68, 1707.17, 1707.65, 1708.13, 1708.61, 1709.09, 1709.58, 1710.06, 1710.54, 1711.02, 1711.5, 1711.99, 1712.47, 1712.95, 1713.43, 1713.91, 1714.4, 1714.88, 1715.36, 1715.84, 1716.33, 1716.81, 1717.29, 1717.77, 1718.25, 1718.74, 1719.22, 1719.7, 1720.18, 1720.66, 1721.15, 1721.63, 1722.11, 1722.59, 1723.08, 1723.56, 1724.04, 1724.52, 1725, 1725.49, 1725.97, 1726.45, 1726.93, 1727.41, 1727.9, 1728.38, 1728.86, 1729.34, 1729.82, 1730.31, 1730.79, 1731.27, 1731.75, 1732.24, 1732.72, 1733.2, 1733.68, 1734.16, 1734.65, 1735.13, 1735.61, 1736.09, 1736.57, 1737.06, 1737.54, 1738.02, 1738.5, 1738.98, 1739.47, 1739.95, 1740.43, 1740.91, 1741.4, 1741.88, 1742.36, 1742.84, 1743.32, 1743.81, 1744.29, 1744.77, 1745.25, 1745.73, 1746.22, 1746.7, 1747.18, 1747.66, 1748.14, 1748.63, 1749.11, 1749.59, 1750.07, 1750.56])
y_2 = numpy.array([0.00182478, 0.00198449, 0.0021542, 0.00230491, 0.00248363, 0.00269334, 0.00289705, 0.00308676, 0.00330747, 0.00358919, 0.0038779, 0.00415561, 0.00444332, 0.00474103, 0.00507474, 0.00542346, 0.00576517, 0.00613688, 0.00651859, 0.0068873, 0.00727502, 0.00767773, 0.00808544, 0.00851815, 0.00894486, 0.00935658, 0.00979429, 0.010245, 0.0106727, 0.0110844, 0.0115191, 0.0119878, 0.0124556, 0.0128823, 0.013274, 0.0137237, 0.0142374, 0.0147181, 0.0151798, 0.0156495, 0.0160963, 0.016534, 0.0169657, 0.0173574, 0.0177211, 0.0180818, 0.0184125, 0.0187012, 0.0189339, 0.0191077, 0.0192454, 0.0193291, 0.0193638, 0.0193495, 0.0192672, 0.0191119, 0.0188696, 0.0185614, 0.0181941, 0.0176948, 0.0170465, 0.0162762, 0.0153449, 0.0142406, 0.0129863, 0.0115801, 0.0100468, 0.00844248, 0.00692419, 0.0055719, 0.00435861, 0.00340132, 0.00270704, 0.00213775, 0.00168046, 0.00134117, 0.00109188, 9.16595E-4, 7.80307E-4, 6.65019E-4, 5.62731E-4, 4.75443E-4, 4.42155E-4, 4.49867E-4, 4.29579E-4, 3.9929E-4, 3.83002E-4, 3.51714E-4, 3.38426E-4, 3.40138E-4, 3.2985E-4, 3.27562E-4, 3.24274E-4, 3.06986E-4, 2.92698E-4, 3.0041E-4, 3.12121E-4, 2.84833E-4, 2.47545E-4, 2.41257E-4, 2.34969E-4, 2.27681E-4, 2.47393E-4, 2.60105E-4, 2.25817E-4])
x_orig = numpy.array([1700.42, 1700.9, 1701.38, 1701.86, 1702.34, 1702.83, 1703.31, 1703.79, 1704.27, 1704.75, 1705.24, 1705.72, 1706.2, 1706.68, 1707.17, 1707.65, 1708.13, 1708.61, 1709.09, 1709.58, 1710.06, 1710.54, 1711.02, 1711.5, 1711.99, 1712.47, 1712.95, 1713.43, 1713.91, 1714.4, 1714.88, 1715.36, 1715.84, 1716.33, 1716.81, 1717.29, 1717.77, 1718.25, 1718.74, 1719.22, 1719.7, 1720.18, 1720.66, 1721.15, 1721.63, 1722.11, 1722.59, 1723.08, 1723.56, 1724.04, 1724.52, 1725, 1725.49, 1725.97, 1726.45, 1726.93, 1727.41, 1727.9, 1728.38, 1728.86, 1729.34, 1729.82, 1730.31, 1730.79, 1731.27, 1731.75, 1732.24, 1732.72, 1733.2, 1733.68, 1734.16, 1734.65, 1735.13, 1735.61, 1736.09, 1736.57, 1737.06, 1737.54, 1738.02, 1738.5, 1738.98, 1739.47, 1739.95, 1740.43, 1740.91, 1741.4, 1741.88, 1742.36, 1742.84, 1743.32, 1743.81, 1744.29, 1744.77, 1745.25, 1745.73, 1746.22, 1746.7, 1747.18, 1747.66, 1748.14, 1748.63, 1749.11, 1749.59, 1750.07, 1750.56])
y_orig = numpy.array([0.011507, 0.0121121, 0.0127542, 0.0132673, 0.0137554, 0.0143684, 0.0148995, 0.0154036, 0.0159997, 0.0165907, 0.0172408, 0.0178499, 0.018388, 0.019089, 0.0197701, 0.0203572, 0.0210393, 0.0216564, 0.0222324, 0.0228305, 0.0233166, 0.0238667, 0.0244387, 0.0248918, 0.0254159, 0.025865, 0.026158, 0.0265131, 0.0267652, 0.0269333, 0.0271824, 0.0273214, 0.0274515, 0.0274626, 0.0271257, 0.0269957, 0.0270148, 0.0267899, 0.026651, 0.026427, 0.0260381, 0.0257212, 0.0252253, 0.0247254, 0.0243314, 0.0237925, 0.0233076, 0.0227997, 0.0221607, 0.0216288, 0.0210079, 0.020299, 0.019702, 0.0189881, 0.0182382, 0.0175053, 0.0165944, 0.0157524, 0.0149355, 0.0139746, 0.0131167, 0.0122307, 0.0112948, 0.0105009, 0.00964397, 0.00886105, 0.00821613, 0.0074542, 0.00685928, 0.00640136, 0.00589444, 0.00568351, 0.00555559, 0.00529467, 0.00514074, 0.00495682, 0.0047789, 0.00469697, 0.00453005, 0.00441613, 0.0042912, 0.00408328, 0.00409536, 0.00412444, 0.00400951, 0.00397959, 0.00389367, 0.00375074, 0.00372082, 0.0036819, 0.00365497, 0.00363905, 0.00353413, 0.00348721, 0.00346528, 0.00336936, 0.00334044, 0.00331251, 0.00322459, 0.00316767, 0.00308874, 0.00304882, 0.0030859, 0.00301798, 0.00287005])
plt.plot(x_1, y_1, 'o')
plt.plot(x_2, y_2, 'o')
plt.plot(x_orig, y_orig, 'o')
plt.xlabel('X Data') # X axis data label
plt.ylabel('Y Data') # Y axis data label
plt.show()

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