scipy.optimize.minimize() not converging giving success=False - python-3.x

I recently tried to apply backpropagation algorithm in python, I tried fmin_tnc,bfgs but none of them actually worked, so please help me to figure out the problem.
def sigmoid(Z):
return 1/(1+np.exp(-Z))
def costFunction(nnparams,X,y,input_layer_size=400,hidden_layer_size=25,num_labels=10,lamda=1):
#input_layer_size=400; hidden_layer_size=25; num_labels=10; lamda=1;
Theta1=np.reshape(nnparams[0:hidden_layer_size*(input_layer_size+1)],(hidden_layer_size,(input_layer_size+1)))
Theta2=np.reshape(nnparams[(hidden_layer_size*(input_layer_size+1)):],(num_labels,hidden_layer_size+1))
m=X.shape[0]
J=0;
y=y.reshape(m,1)
Theta1_grad=np.zeros(Theta1.shape)
Theta2_grad=np.zeros(Theta2.shape)
X=np.concatenate([np.ones([m,1]),X],1)
a2=sigmoid(Theta1.dot(X.T));
a2=np.concatenate([np.ones([1,a2.shape[1]]),a2])
h=sigmoid(Theta2.dot(a2))
c=np.array(range(1,11))
y=y==c;
for i in range(y.shape[0]):
J=J+(-1/m)*np.sum(y[i,:]*np.log(h[:,i]) + (1-y[i,:])*np.log(1-h[:,i]) );
DEL2=np.zeros(Theta2.shape); DEL1=np.zeros(Theta1.shape);
for i in range(m):
z2=Theta1.dot(X[i,:].T);
a2=sigmoid(z2).reshape(-1,1);
a2=np.concatenate([np.ones([1,a2.shape[1]]),a2])
z3=Theta2.dot(a2);
# print('z3 shape',z3.shape)
a3=sigmoid(z3).reshape(-1,1);
# print('a3 shape = ',a3.shape)
delta3=(a3-y[i,:].T.reshape(-1,1));
# print('y shape ',y[i,:].T.shape)
delta2=((Theta2.T.dot(delta3)) * (a2 * (1-a2)));
# print('shapes = ',delta3.shape,a3.shape)
DEL2 = DEL2 + delta3.dot(a2.T);
DEL1 = DEL1 + (delta2[1,:])*(X[i,:]);
Theta1_grad=np.zeros(np.shape(Theta1));
Theta2_grad=np.zeros(np.shape(Theta2));
Theta1_grad[:,0]=(DEL1[:,0] * (1/m));
Theta1_grad[:,1:]=(DEL1[:,1:] * (1/m)) + (lamda/m)*(Theta1[:,1:]);
Theta2_grad[:,0]=(DEL2[:,0] * (1/m));
Theta2_grad[:,1:]=(DEL2[:,1:]*(1/m)) + (lamda/m)*(Theta2[:,1:]);
grad=np.concatenate([Theta1_grad.reshape(-1,1),Theta2_grad.reshape(-1,1)]);
return J,grad
This is how I called the function (op is scipy.optimize)
r2=op.minimize(fun=costFunction, x0=nnparams, args=(X, dataY.flatten()),
method='TNC', jac=True, options={'maxiter': 400})
r2 is like this
fun: 3.1045444063663266
jac: array([[-6.73218494e-04],
[-8.93179045e-05],
[-1.13786179e-04],
...,
[ 1.19577741e-03],
[ 5.79555099e-05],
[ 3.85717533e-03]])
message: 'Linear search failed'
nfev: 140
nit: 5
status: 4
success: False
x: array([-0.97996948, -0.44658952, -0.5689309 , ..., 0.03420931,
-0.58005183, -0.74322735])
Please help me to find correct way of minimizing this function, Thanks in advance

Finally Solved it, The problem was I used np.randn() to generate random Theta values which gives random values in a standard normal distribution, therefore as too many values were within the same range,therefore this lead to symmetricity in the theta values. Due to this symmetricity problem the optimization terminates in the middle of the process.
Simple solution was to use np.rand() (which provide uniform random distribution) instead of np.randn()

Related

(Wald)Test for comparing two nested feols/plm models

I want to test if
model_1 <- feols(auth ~ dummy_past1 + dummy_past2 + dummy_past3
| region + date_f, # Fixed Effects
data=df)
is just as good as
model_2 <- feols(auth ~ i(dummy_past1,specific_regions) + i(dummy_past2,specific_regions) + i(dummy_past3, specific_regions) # interaction(dummies * specific regions)
| region + date_f, # Fixed Effects
data=df)
If this was a normal linear regression I would conduct a lrtest (likelihood ratio test). As I heard that a waldtest can also be applied to panel data (?) I conducted a waldtest
waldtest(model_1, model_2). Yet, when doing that I get the warning: "Error in solve.default(vc[ovar, ovar]) : system is computationally singular: reciprocal condition number = 2.10749e-44"
Does someone know if doing a waldtest is the correct approach here, or if there are any other tests that I could do on plm or feols regressions? Or it would also be very helpful if you have ideas on how I could get the waldtest working.

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
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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

PuLP solvers do not respond to options fed to them

So I've got a fairly large optimization problem and I'm trying to solve it within a sensible amount of time.
Ive set it up as:
import pulp as pl
my_problem = LpProblem("My problem",LpMinimize)
# write to problem file
my_problem.writeLP("MyProblem.lp")
And then alternatively
solver = CPLEX_CMD(timeLimit=1, gapRel=0.1)
status = my_problem .solve(solver)
solver = pl.apis.CPLEX_CMD(timeLimit=1, gapRel=0.1)
status = my_problem .solve(solver)
path_to_cplex = r'C:\Program Files\IBM\ILOG\CPLEX_Studio1210\cplex\bin\x64_win64\cplex.exe' # and yes this is the actual path on my machine
solver = pl.apis.cplex_api.CPLEX_CMD(timeLimit=1, gapRel=0.1, path=path_to_cplex)
status = my_problem .solve(solver)
solver = pl.apis.cplex_api.CPLEX_CMD(timeLimit=1, gapRel=0.1, path=path_to_cplex)
status = my_problem .solve(solver)
It runs in each case.
However, the solver does not repond to the timeLimit or gapRel instructions.
If I use timelimit it does warn this is depreciated for timeLimit. Same for fracgap: it tells me I should use relGap. So somehow I am talking to the solver.
However, nor matter what values i pick for timeLimit and relGap, it always returns the exact same answer and takes the exact same amount of time (several minutes).
Also, I have tried alternative solvers, and I cannot get any one of them to accept their variants of time limits or optimization gaps.
In each case, the problem solves and returns an status: optimal message. But it just ignores the time limit and gap instructions.
Any ideas?
out of the zoo example:
import pulp
import cplex
bus_problem = pulp.LpProblem("bus", pulp.LpMinimize)
nbBus40 = pulp.LpVariable('nbBus40', lowBound=0, cat='Integer')
nbBus30 = pulp.LpVariable('nbBus30', lowBound=0, cat='Integer')
# Objective function
bus_problem += 500 * nbBus40 + 400 * nbBus30, "cost"
# Constraints
bus_problem += 40 * nbBus40 + 30 * nbBus30 >= 300
solver = pulp.CPLEX_CMD(options=['set timelimit 40'])
bus_problem.solve(solver)
print(pulp.LpStatus[bus_problem.status])
for variable in bus_problem.variables():
print ("{} = {}".format(variable.name, variable.varValue))
Correct way to pass solver option as dictionary
pulp.CPLEX_CMD(options={'timelimit': 40})
#Alex Fleisher has it correct with pulp.CPLEX_CMD(options=['set timelimit 40']). This also works for CBC using the following syntax:
prob.solve(COIN_CMD(options=['sec 60','Presolve More','Multiple 15', 'Node DownFewest','HEUR on', 'Round On','PreProcess Aggregate','PassP 10','PassF 40','Strong 10','Cuts On', 'Gomory On', 'CutD -1', 'Branch On', 'Idiot -1', 'sprint -1','Reduce On','Two On'],msg=True)).
It is important to understand that the parameters, and associated options, are specific to a solver. PuLP seems to be calling CBC via the command line so an investigation of those things is required. Hope that helps

How to use custom mean, median, mode functions with array of 2500 in python?

So I am trying to solve mean, median and mode challenge on Hackerrank. I defined 3 functions to calculate mean, median and mode for a given array with length between 10 and 2500, inclusive.
I get an error with an array of 2500 integers, not sure why. I looked into python documentation and found no mentions of max length for lists. I know I can use statistics module but trying the hard way and being stubborn I guess. Any help and criticism is appreciated regarding my code. Please be honest and brutal if need be. Thanks
N = int(input())
var_list = [int(x) for x in input().split()]
def mean(sample_list):
mean = sum(sample_list)/N
print(mean)
return
def median(sample_list):
sorted_list = sorted(sample_list)
if N%2 != 0:
median = sorted_list[(N//2)]
else:
median = (sorted_list[N//2] + sorted_list[(N//2)-1])/2
print(median)
return
def mode(sample_list):
sorted_list = sorted(sample_list)
mode = min(sorted_list)
max_count = sorted_list.count(mode)
for i in sorted_list:
if (i <= mode) and (sorted_list.count(i) >= max_count):
mode = i
print(mode)
return
mean(var_list)
median(var_list)
mode(var_list)
Compiler Message
Wrong Answer
Input (stdin)
2500
19325 74348 68955 98497 26622 32516 97390 64601 64410 10205 5173 25044 23966 60492 71098 13852 27371 40577 74997 42548 95799 26783 51505 25284 49987 99134 33865 25198 24497 19837 53534 44961 93979 76075 57999 93564 71865 90141 5736 54600 58914 72031 78758 30015 21729 57992 35083 33079 6932 96145 73623 55226 18447 15526 41033 46267 52486 64081 3705 51675 97470 64777 31060 90341 55108 77695 16588 64492 21642 56200 48312 5279 15252 20428 57224 38086 19494 57178 49084 37239 32317 68884 98127 79085 77820 2664 37698 84039 63449 63987 20771 3946 862 1311 77463 19216 57974 73012 78016 9412 90919 40744 24322 68755 59072 57407 4026 15452 82125 91125 99024 49150 90465 62477 30556 39943 44421 68568 31056 66870 63203 43521 78523 58464 38319 30682 77207 86684 44876 81896 58623 24624 14808 73395 92533 4398 8767 72743 1999 6507 49353 81676 71188 78019 88429 68320 59395 95307 95770 32034 57015 26439 2878 40394 33748 41552 64939 49762 71841 40393 38293 48853 81628 52111 49934 74061 98537 83075 83920 42792 96943 3357 83393{-truncated-}
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Expected Output
49921.5
49253.5
2184
Your issue seems to be that you are actually using standard list operations rather than calculating things on the fly, while looping through the data once (for the average). sum(sample_list) will almost surely give you something which exceeds the double-limit, i.a.w. it becomes really big.
Further reading
Calculating the mean, variance, skewness, and kurtosis on the fly
How do I determine the standard deviation (stddev) of a set of values?
Rolling variance algorithm
What is a good solution for calculating an average where the sum of all values exceeds a double's limits?
How do I determine the standard deviation (stddev) of a set of values?
How to efficiently compute average on the fly (moving average)?
I figured out that you forgot to change the max_count variable inside the if block. Probably that causes the wrong result. I tested the debugged version on my computer and they seem to work well when I compare their result with the scipy's built-in functions. The correct mode function should be
def mode(sample_list):
N = len(sample_list)
sorted_list = sorted(sample_list)
mode = min(sorted_list)
max_count = sorted_list.count(mode)
for i in sorted_list:
if (sorted_list.count(i) >= max_count):
mode = i
max_count = sorted_list.count(i)
print(mode)
I was busy with some stuff and now came back to completing this. I am happy to say that I have matured enough as a coder and solved this issue.
Here is the solution:
# Enter your code here. Read input from STDIN. Print output to STDOUT
# Input an array of numbers, convert it to integer array
n = int(input())
my_array = list(map(int, input().split()))
my_array.sort()
# Find mean
array_mean = sum(my_array) / n
print(array_mean)
# Find median
if (n%2) != 0:
array_median = my_array[n//2]
else:
array_median = (my_array[n//2 - 1] + my_array[n//2]) / 2
print(array_median)
# Find mode(I could do this using multimode method of statistics module for python 3.8)
def sort_second(array):
return array[1]
modes = [[i, my_array.count(i)] for i in my_array]
modes.sort(key = sort_second, reverse=True)
array_mode = modes[0][0]
print(array_mode)

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.

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