joris267
joris267

Reputation: 33

Python 2.7 memory leak with scipy.minimze

During a fit procedure, my RAM memory slowly but steadily (about 2.8 mb every couple of seconds) increases until I get a memory error or I terminate the program. This happens when I try to fit some 80 measurements by fitting a model to them. This fitting is done by using scipy.minimze to minimize Chi_squared.

So far I've tried:

From these test, it seems that my RAM usage goes up while the total space my variables take up is constant. Where my memory goes I would really like to know.

The code below reproduces the problem for me:

import numpy as np
import scipy
import scipy.optimize as op
import scipy.stats
import scipy.integrate



def fit_model(model_pmt, x_list, y_list, PMT_parra, PMT_bounds=None, tolerance=10**-1, PMT_start_gues=None):
    result = op.minimize(chi_squared, PMT_start_gues, args=(x_list, y_list, model_pmt, PMT_parra[0], PMT_parra[1], PMT_parra[2]),
                     bounds=PMT_bounds, method='SLSQP', options={"ftol": tolerance})
    print result



def chi_squared(fit_parm, x, y_val, model, *non_fit_parm):
    parm = np.concatenate((fit_parm, non_fit_parm))
    y_mod = model(x, *parm)
    X2 = sum(pow(y_val - y_mod, 2))
    return X2



def basic_model(cb_list, max_intesity, sigma_e, noise, N, centre1, centre2, sigma_eb, min_dist=10**-5):
        """
        plateau function consisting of two gaussian CDF functions.
        """
        def get_distance(x, r):
            dist = abs(x - r)
            if dist < min_dist:
                dist = min_dist
            return dist

        def amount_of_material(x):
            A = scipy.stats.norm.cdf((x - centre1) / sigma_e)
            B = (1 - scipy.stats.norm.cdf((x - centre2) / sigma_e))
            cube =  A * B
            return cube

        def amount_of_field_INTEGRAL(x, cb):
        """Integral that is part of my sum"""
            result = scipy.integrate.quad(lambda r: scipy.stats.norm.pdf((r - cb) / sigma_b) / pow(get_distance(x, r), N),
                                          start, end, epsabs=10 ** -1)[0]
            return result



        # Set some constants, not important
        sigma_b = (sigma_eb**2-sigma_e**2)**0.5
        start, end = centre1 - 3 * sigma_e, centre2 + 3 * sigma_e
        integration_range = np.linspace(start, end, int(end - start) / 20)
        intensity_list = []

        # Doing a riemann sum, this is what takes the most time.
        for i, cb_point in enumerate(cb_list):
            intensity = sum([amount_of_material(x) * amount_of_field_INTEGRAL(x, cb_point) for x in integration_range])
            intensity *= (integration_range[1] - integration_range[0])
            intensity_list.append(intensity)


        model_values = np.array(intensity_list) / max(intensity_list)* max_intesity + noise
        return model_values


def get_dummy_data():
"""Can be ignored, produces something resembling my data with noise"""
    # X is just a range
    x_list = np.linspace(0, 300, 300)

    # Y is some sort of step function with noise
    A = scipy.stats.norm.cdf((x_list - 100) / 15.8)
    B = (1 - scipy.stats.norm.cdf((x_list - 200) / 15.8))
    y_list = A * B * .8 + .1 + np.random.normal(0, 0.05, 300)

    return x_list, y_list


if __name__=="__main__":
    # Set some variables
    start_pmt = [0.7, 8, 0.15, 0.6]
    pmt_bounds = [(.5, 1.3), (4, 15), (0.05, 0.3), (0.5, 3)]
    pmt_par = [110, 160, 15]
    x_list, y_list = get_dummy_data()

    fit_model(basic_model, x_list, y_list,  pmt_par, PMT_start_gues=start_pmt, PMT_bounds=pmt_bounds, tolerance=0.1)

Thanks for trying to help!

Upvotes: 1

Views: 463

Answers (1)

MB-F
MB-F

Reputation: 23647

I narrowed down the problem by successively removing layer after layer of indirection. (@joris267 This is something you really should have done yourself before asking.) The minimal remaining code to reproduce the problem looks like this:

import scipy.integrate

if __name__=="__main__":    
    while True:
        scipy.integrate.quad(lambda r: 0, 1, 100)

Conclusion:

  1. Yes, there is e memory leak.
  2. No, the leak is not in scipy.minimize but in scipy.quad.

However, this is a known issue with scipy 0.19.0. Upgrade to 0.19.1 should supposedly fix the problem, but I don't know for sure because I'm still with 0.19.0 myself :)

Update:

After upgrading scipy to 0.19.1 (and numpy to 1.13.3 for compatibility) the leak disapeared on my system.

Upvotes: 4

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