Overwrite GPS coordinates in Image Exif using Python 3.6 - python-3.x

I am trying to transform image geotags so that images and ground control points lie in the same coordinate system inside my software (Pix4D mapper).
The answer here says:
Exif data is standardized, and GPS data must be encoded using
geographical coordinates (minutes, seconds, etc) described above
instead of a fraction. Unless it's encoded in that format in the exif
tag, it won't stick.
Here is my code:
import os, piexif, pyproj
from PIL import Image
img = Image.open(os.path.join(dirPath,fn))
exif_dict = piexif.load(img.info['exif'])
breite = exif_dict['GPS'][piexif.GPSIFD.GPSLatitude]
lange = exif_dict['GPS'][piexif.GPSIFD.GPSLongitude]
breite = breite[0][0] / breite[0][1] + breite[1][0] / (breite[1][1] * 60) + breite[2][0] / (breite[2][1] * 3600)
lange = lange[0][0] / lange[0][1] + lange[1][0] / (lange[1][1] * 60) + lange[2][0] / (lange[2][1] * 3600)
print(breite) #48.81368778730952
print(lange) #9.954511162420633
x, y = pyproj.transform(wgs84, gk3, lange, breite) #from WGS84 to GaussKrüger zone 3
print(x) #3570178.732528623
print(y) #5408908.20172699
exif_dict['GPS'][piexif.GPSIFD.GPSLatitude] = [ ( (int)(round(y,6) * 1000000), 1000000 ), (0, 1), (0, 1) ]
exif_bytes = piexif.dump(exif_dict) #error here
img.save(os.path.join(outPath,fn), "jpeg", exif=exif_bytes)
I am getting struct.error: argument out of range in the dump method. The original GPSInfo tag looks like: {0: b'\x02\x03\x00\x00', 1: 'N', 2: ((48, 1), (48, 1), (3449322402, 70000000)), 3: 'E', 4: ((9, 1), (57, 1), (1136812930, 70000000)), 5: b'\x00', 6: (3659, 10)}
I am guessing I have to offset the values and encode them properly before writing, but have no idea what is to be done.

It looks like you are already using PIL and Python 3.x, not sure if you want to continue using piexif but either way, you may find it easier to convert the degrees, minutes, and seconds into decimal first. It looks like you are trying to do that already but putting it in a separate function may be clearer and account for direction reference.
Here's an example:
def get_decimal_from_dms(dms, ref):
degrees = dms[0][0] / dms[0][1]
minutes = dms[1][0] / dms[1][1] / 60.0
seconds = dms[2][0] / dms[2][1] / 3600.0
if ref in ['S', 'W']:
degrees = -degrees
minutes = -minutes
seconds = -seconds
return round(degrees + minutes + seconds, 5)
def get_coordinates(geotags):
lat = get_decimal_from_dms(geotags['GPSLatitude'], geotags['GPSLatitudeRef'])
lon = get_decimal_from_dms(geotags['GPSLongitude'], geotags['GPSLongitudeRef'])
return (lat,lon)
The geotags in this example is a dictionary with the GPSTAGS as keys instead of the numeric codes for readability. You can find more detail and the complete example from this blog post: Getting Started with Geocoding Exif Image Metadata in Python 3

After much hemming & hawing I reached the pages of py3exiv2 image metadata manipulation library. One will find exhaustive lists of the metadata tags as one reads through but here is the list of EXIF tags just to save few clicks.
It runs smoothly on Linux and provides many opportunities to edit image-headers. The documentation is also quite clear. I recommend this as a solution and am interested to know if it solves everyone else's problems as well.

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
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96883584,1.7155425548
96887508,1.7057673931
96891436,1.6520038604
96895348,1.5591397285
96899280,1.4809384346
96903196,1.4467253684
96907112,1.4369501113
96911032,1.3978494453
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97951988,1.3929618644
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98022592,1.6129032135
98026516,1.6715541839
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98038268,1.6422286987
98042192,1.6715541839
98046124,1.7204301357
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98434076,2.0527858734
98437988,2.0381231307
98441900,2.0625610351
98445820,2.1065492630
98449740,2.1309874057
98453660,2.1065492630
98457572,2.0869989395
98461492,2.0918865203
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98473236,2.1749756336
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98481052,2.1652004718
98484972,2.1945259571
98488900,2.2287390232
98492820,2.2091886997
98496732,2.1700880527
98500644,2.1652004718
98504556,2.2091886997
98508476,2.2531769275
98512404,2.2336266040
98516324,2.1994135379
98520244,2.2043011188
98524152,2.2531769275
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98531988,2.2727272510
98535908,2.2238514423
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98602528,1.8572825431
98606456,1.8621701240
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98645640,1.9941349029
98649556,2.1456501483
98653476,2.3313782215
98657404,2.5708699226
98661316,2.8885631561
98665236,3.2306940555
98669148,3.4799609184
98673064,3.4604105949
98676972,3.1769306659
98680884,2.7614858150
98684796,2.2678396701
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98692628,1.4418377876
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98908216,1.8328446388
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98994396,0.9237536430
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99143156,1.2903225421
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99154924,1.2707722187
99158844,1.2072336673
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99174508,1.1974585056
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99201916,1.1192570924
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99225412,1.2267839908
99229332,1.2707722187
99233244,1.2609970569
99237152,1.2365591526
99241068,1.2463343143
99244988,1.2805473804
99248900,1.2952101230
99252820,1.2805473804
99256732,1.2316715717
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99284212,1.3782991170
99288116,1.4271749496
99292040,1.4369501113
99295964,1.4076246261
99299892,1.4076246261
99303816,1.4662756919
99307740,1.5395894050
99311652,1.5640274047
99315564,1.5444769859
99319484,1.5444769859
99323412,1.5786901473
99327332,1.6275659561
99331252,1.6520038604
99335156,1.6422286987
99339076,1.6275659561
99343004,1.6422286987
99346924,1.6666666030
99350844,1.6568914413
99354764,1.6031280517
99358676,1.5542521476
99362604,1.5542521476
99366532,1.5835777282
99370460,1.5982404708
99374372,1.5835777282
99378300,1.5640274047
99382204,1.5835777282
99386132,1.6373411178
99390056,1.6715541839
99393980,1.6520038604
99397892,1.6275659561
99401812,1.6422286987
99405736,1.6862169265
99409664,1.7106549739
99413580,1.6911046504
99417500,1.6568914413
99421432,1.6715541839
99425348,1.7204301357
99429256,1.8084066390
99433164,1.9208210945
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99440980,2.3655912876
99444912,2.7321603298
99448828,3.0596284866
99452752,3.2453567981
99456680,3.1867058277
99460600,2.9374389648
99464516,2.5610947608
99468428,2.1163244247
99472356,1.6813293457
99476284,1.3343108892
99480200,1.1436949968
99484112,1.1339198350
99488036,1.2365591526
99491956,1.3440860509
99495864,1.4320625305
99499780,1.5298142433
99503708,1.6422286987
99507636,1.7350928783
99511556,1.7644183635
99515480,1.7399804592
99519396,1.7350928783
99523320,1.7448680400
99527220,1.7350928783
99531140,1.6862169265
99535064,1.5933528900
99538980,1.5102639198
99542892,1.4711632728
99546820,1.4467253684
99550748,1.3978494453
99554668,1.3049852848
99558588,1.2072336673
99562504,1.1485825777
99566428,1.1192570924
99570348,1.0752688646
99574256,1.0068426132
99578176,0.9384163856
99582084,0.9188660621
99585988,0.9188660621
99589900,0.9188660621
99593812,0.8895405769
99597716,0.8748778343
99601636,0.8651026725
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99609436,0.9481915473
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99617268,0.9237536430
99621180,0.9335288047
99625080,1.0019550323
99628980,1.0752688646
99632888,1.0801564407
99636792,1.0703812837
99640704,1.0899316024
99644616,1.1436949968
99648536,1.2170088291
99652444,1.2170088291
99656356,1.2023460865
99660268,1.2072336673
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99668084,1.3000977039
99671980,1.3147605657
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99687652,1.4027370452
99691568,1.4222873687
99695484,1.3978494453
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99703328,1.4173997879
99707248,1.4565005302
99711156,1.4760508537
99715064,1.4271749496
99718988,1.3929618644
99722908,1.3929618644
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99734668,1.3147605657
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99742492,1.2414467334
99746420,1.2658846378
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99758168,1.2121212482
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99777780,1.3929618644
99781704,1.4662756919
99785620,1.5640274047
99789532,1.6568914413
99793460,1.6959922313
99797392,1.7057673931
99801312,1.7399804592
99805228,1.7937438488
99809148,1.8377322196
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99816996,1.8230694770
99820920,1.8475073814
99824840,1.9061583518
99828756,1.9501466751
99832680,1.9599218368
99836608,1.9501466751
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99844452,2.0087976455
99848364,2.0527858734
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99856184,2.0283479690
99860092,2.0185728073
99864012,2.0576734542
99867932,2.0967741012
99871836,2.0869989395
99875740,2.0478982925
99879652,2.0234603881
99883564,2.0527858734
99887484,2.1065492630
99891404,2.1163244247
99895332,2.0772237777
99899236,2.0527858734
99903156,2.0821113586
99907076,2.1065492630
99910996,2.1016616821
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99946220,2.1309874057
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99977564,2.1945259571
99981492,2.2043011188
99985404,2.2482893466
99989332,2.2922775745
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100016924,2.1798632144
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100024844,2.1798632144
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100032756,2.1798632144
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100048604,2.1603128910
100052556,2.1407625675
100056516,2.0967741012
100060468,2.0918865203
100064420,2.1163244247
100068384,2.1407625675
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100076292,2.0478982925
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100084196,2.0478982925
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100107928,1.9208210945
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100123776,1.8084066390
100127712,1.8475073814
100131672,1.8523949623
100135636,1.8181818962
100139604,1.8035190582
100143560,1.8377322196
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100151488,1.8719452857
100155448,1.8523949623
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100175232,1.9061583518
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100183132,2.1016616821
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100191036,2.5904202461
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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
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100242500,1.6422286987
100246456,1.8377322196
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100254380,2.0967741012
100258340,2.1652004718
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100274140,2.4731183052
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100289992,2.7028348445
100293948,2.7077224254
100297908,2.7126100063
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100305820,2.8054740905
100309764,2.8250244140
100313724,2.7908113002
100317676,2.7370479106
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100325600,2.7419354915
100329556,2.7517106533
100333516,2.7126100063
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100341428,2.6686217784
100345392,2.7028348445
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100365180,2.6832845211
100369140,2.7077224254
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100377052,2.6148581504
100381012,2.6001954078
100384960,2.6246333122
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100396828,2.5659823417
100400788,2.5562071800
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100408692,2.6099705696
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100416588,2.5366568565
100420548,2.5268816947
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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 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.

Plot x and y if z == [value]

Only just started using python this week, so I'm a total beginner. Imagine I have a massive dataset with data like so:
close high low open time symbol
0.04951 0.04951 0.04951 0.04951 7/16/2010 BTC
0.08584 0.08585 0.05941 0.04951 7/17/2010 BTC
0.0808 0.09307 0.07723 0.08584 7/18/2010 ETH
How, using matplotlib, can I plot close with time, only if symbol = BTC? I was thinking something like
bitgroup = df.groupby('symbol')
if bitgroup == 'BTC':
df(['close','time']).plot()
plt.show()
Building on this, I'd then like to use these new groups to create new columns, such as returns, (calculated using (p1-p0)/p0) doing something like this:
def createnewcolumn()
for i in bitgroup
df[returns] = (bitgroup['close'].ix[i] - bitgroup['close'].ix[i-1]) / bitgroup['close'].ix[i-1]
createnewcolumn()
Any help would be greatly appreciated in turning this pseudocode into real code!
df.symbol == 'BTC'
returns a list of [0, 1, 1, 0, 0, 0 ... ] for each row, and then you can use that as a mask on the original data -
df[df.symbol == 'BTC']

R simplify heatmap to pdf

I want to plot a simplified heatmap that is not so difficult to edit with the scalar vector graphics program I am using (inkscape). The original heatmap as produced below contains lots of rectangles, and I wonder if they could be merged together in the different sectors to simplify the output pdf file:
nentries=100000
ci=rainbow(nentries)
set.seed=1
mean=10
## Generate some data (4 factors)
i = data.frame(
a=round(abs(rnorm(nentries,mean-2))),
b=round(abs(rnorm(nentries,mean-1))),
c=round(abs(rnorm(nentries,mean+1))),
d=round(abs(rnorm(nentries,mean+2)))
)
minvalue = 10
# Discretise values to 1 or 0
m0 = matrix(as.numeric(i>minvalue),nrow=nrow(i))
# Remove rows with all zeros
m = m0[rowSums(m0)>0,]
# Reorder with 1,1,1,1 on top
ms =m[order(as.vector(m %*% matrix(2^((ncol(m)-1):0),ncol=1)), decreasing=TRUE),]
rowci = rainbow(nrow(ms))
colci = rainbow(ncol(ms))
colnames(ms)=LETTERS[1:4]
limits=c(which(!duplicated(ms)),nrow(ms))
l=length(limits)
toname=round((limits[-l]+ limits[-1])/2)
freq=(limits[-1]-limits[-l])/nrow(ms)
rn=rep("", nrow(ms))
for(i in toname) rn[i]=paste(colnames(ms)[which(ms[i,]==1)],collapse="")
rn[toname]=paste(rn[toname], ": ", sprintf( "%.5f", freq ), "%")
heatmap(ms,
Rowv=NA,
labRow=rn,
keep.dendro = FALSE,
col=c("black","red"),
RowSideColors=rowci,
ColSideColors=colci,
)
dev.copy2pdf(file="/tmp/file.pdf")
Why don't you try RSvgDevice? Using it you could save your image as svg file, which is much convenient to Inkscape than pdf
I use the Cairo package for producing svg. It's incredibly easy. Here is a much simpler plot than the one you have in your example:
require(Cairo)
CairoSVG(file = "tmp.svg", width = 6, height = 6)
plot(1:10)
dev.off()
Upon opening in Inkscape, you can ungroup the elements and edit as you like.
Example (point moved, swirl added):
I don't think we (the internet) are being clear enough on this one.
Let me just start off with a successful export example
png("heatmap.png") #Ruby dev's think of this as kind of like opening a `File.open("asdfsd") do |f|` block
heatmap(sample_matrix, Rowv=NA, Colv=NA, col=terrain.colors(256), scale="column", margins=c(5,10))
dev.off()
The dev.off() bit, in my mind, reminds me of an end call to a ruby block or method, in that, the last line of the "nested" or enclosed (between png() and dev.off()) code's output is what gets dumped into the png file.
For example, if you ran this code:
png("heatmap4.png")
heatmap(sample_matrix, Rowv=NA, Colv=NA, col=terrain.colors(32), scale="column", margins=c(5,15))
heatmap(sample_matrix, Rowv=NA, Colv=NA, col=greenred(32), scale="column", margins=c(5,15))
dev.off()
it would output the 2nd (greenred color scheme, I just tested it) heatmap to the heatmap4.png file, just like how a ruby method returns its last line by default

How to increase resolution of gif image?

How to increase resolution of gif image generated by rgl package of R (plot3d and movie3d functions) - either externally or through R ?
R Code :
MyX<-rnorm(10,5,1)
MyY<-rnorm(10,5,1)
MyZ<-rnorm(10,5,1)
plot3d(MyX, MyY, MyZ, xlab="X", ylab="Y", zlab="Z", type="s", box=T, axes=F)
text3d(MyX, MyY, MyZ, text=c(1:10), cex=5, adj=1)
movie3d(spin3d(axis = c(0, 0, 1), rpm = 4), duration=15, movie="TestMovie",
type="gif", dir=("~/Desktop"))
Output :
Update
Adding this line at the beginning of code solved the problem
r3dDefaults$windowRect <- c(0, 100, 1400, 1400)
I don't think you can do much about the resolution of the gif itself. I think you have to make the image much larger as an alternative, and then when you display it smaller it looks better. This is untested as a recent upgrade broke a thing or two for me, but this did work under 2.15:
par3d(windowRect = c(0, 0, 500, 500)) # make the window large
par3d(zoom = 1.1) # larger values make the image smaller
# you can test your settings interactively at this point
M <- par3d("userMatrix") # save your settings to pass to the movie
movie3d(par3dinterp(userMatrix=list(M,
rotate3d(M, pi, 1, 0, 0),
rotate3d(M, pi, 0, 1, 0) ) ),
duration = 5, fps = 50,
movie = "MyMovie")
HTH. If it doesn't quite work for you, check out the functions used and tune up the settings.

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