Reputation: 551
I'm trying to implement a simple Monte Carlo in Python (to which I'm fairly new). Coming from C I'm probably following the wrongest path since my code is far too slow for what I'm asking: I have a potential hard sphere-like (see V_pot(r)
in the code) for 60 3d particles and periodic boundary conditions (PBC), so I defined the following functions
import timeit
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
from numpy import inf
#
L, kb, d, eps, DIM = 100, 1, 1, 1, 3
r_c, T = L/2, eps/(.5*kb)
beta = 1/(kb*T)
#
def dist(A, B):
d = A - B
d -= L*np.around(d/L)
return np.sqrt(np.sum(d**2))
#
def V_pot(r):
V = -eps*(d**6/r**6 - d**6/r_c**6)
if r > r_c:
V = 0
elif r < d:
V = inf
return V
#
def ener(config):
V_jk_val, j = 0, N
#
while (j > 0):
j -= 1
i = 0
while (i < j):
V_jk_val += V_pot(dist(config[j,:], config[i,:]))
i += 1
#
return V_jk_val
#
def acc(en_n, en_o):
d_en = en_n-en_o
if (d_en <= 0):
acc_val = 1
else:
acc_val = np.exp(-beta*(d_en))
return acc_val
#
then, starting from the configuration (where every line of the array represents the coordinates of a 3D particle)
config = np.array([[16.24155657, 57.41672173, 94.39565792],
[76.38121764, 55.88334066, 5.72255163],
[38.41393783, 58.09432145, 6.26448054],
[86.44286438, 61.37100899, 91.97737383],
[37.7315366 , 44.52697269, 23.86320444],
[ 0.59231801, 39.20183376, 89.63974115],
[38.00998141, 3.84363202, 52.74021401],
[99.53480756, 69.97688928, 21.43528924],
[49.62030291, 93.60889503, 15.73723259],
[54.49195524, 0.6431965 , 25.37401196],
[33.82527814, 25.37776021, 67.4320553 ],
[64.61952893, 46.8407798 , 4.93960443],
[60.47322732, 16.48140136, 33.26481306],
[19.71667792, 46.56999616, 35.61044526],
[ 5.33252557, 4.44393836, 60.55759256],
[44.95897856, 7.81728046, 10.26000715],
[86.5548395 , 49.74079452, 4.80480133],
[52.47965686, 42.831448 , 22.03890639],
[ 2.88752006, 59.84605062, 22.75760029],
[ 9.49231045, 42.08653603, 40.63380097],
[13.90093641, 74.40377984, 32.62917915],
[97.44839233, 90.47695772, 91.60794836],
[51.29501624, 27.03796277, 57.09525454],
[10.30180295, 21.977336 , 69.54173272],
[59.61327648, 14.29582325, 11.70942289],
[89.52722796, 26.87758644, 76.34934637],
[82.03736088, 78.5665713 , 23.23587395],
[79.77571695, 66.140968 , 53.6784269 ],
[82.86070472, 40.82189833, 51.48739072],
[99.05647523, 98.63386809, 6.33888993],
[31.02997123, 66.99709163, 95.88332332],
[97.71654767, 59.24793618, 5.20183793],
[ 6.79964473, 45.01258652, 48.69477807],
[93.34977049, 55.20537774, 82.35693526],
[17.35577815, 20.45936211, 29.27981422],
[55.51942207, 52.22875901, 3.6616131 ],
[61.45612224, 36.50170405, 62.89796773],
[23.55822368, 7.09069623, 37.38274914],
[39.57082799, 58.95457592, 48.0304924 ],
[93.94997617, 64.34383203, 77.63346308],
[17.47989107, 90.01113402, 81.00648645],
[86.79068539, 66.35768515, 56.64402907],
[98.71924121, 38.33749023, 73.4715132 ],
[ 0.42356139, 78.32172925, 15.19883322],
[77.75572529, 2.60088767, 56.4683935 ],
[49.76486142, 3.01800153, 93.48019286],
[42.54483899, 4.27174457, 4.38942325],
[66.75777178, 41.1220603 , 19.64484167],
[19.69520773, 41.09230171, 2.51986091],
[73.20493772, 73.16590392, 99.19174281],
[94.16756184, 72.77653334, 10.32128552],
[29.95281655, 27.58596604, 85.12791195],
[ 2.44803886, 32.82333962, 41.6654683 ],
[23.9665915 , 49.94906612, 37.42701059],
[30.40282934, 39.63854309, 47.16572743],
[56.04809276, 30.19705527, 29.15729635],
[ 2.50566522, 70.37965564, 16.78016719],
[28.39713572, 4.04948368, 27.72615789],
[26.11873563, 41.49557167, 14.38703697],
[81.91731981, 12.10514972, 12.03083427]])
I make the 5000 time steps of the simulation with the following code
N = 60
TIME_MC = 5000
DELTA_LIST = [d]
#d/6, d/3, d, 2*d, 3*d
np.random.seed(19680801)
en_mc_delta = np.zeros((TIME_MC, len(DELTA_LIST)))
start = timeit.default_timer()
config_tmp = config
#
for iD, Delta in enumerate(DELTA_LIST):
t=0
while (t < TIME_MC):
for k in range(N):
RND = np.random.rand()
config_tmp[k,:] = config[k,:] + Delta*(np.random.random_sample((1,3))-.5)
en_o, en_n = ener(config), ener(config_tmp)
ACC = acc(en_n, en_o)
if (RND < ACC):
config[k,:] = config_tmp[k,:]
en_o = en_n
en_mc_delta[t][iD] = en_o
t += 1
stop = timeit.default_timer()
print('Time: ', stop-start)
following the rule of the Metropolis algorithm for the acceptance of the proposed move extracted with config_tmp[k,:] = config[k,:] + Delta*(np.random.random_sample((1,3))-.5)
.
I made some attempts to check where the code get stuck and I found that the function ener
(also because of the function dist
) is extremely slow: it takes something like ~0.02s
to calculate the energy of a configuration, which means something around ~6000s
to run the complete simulation (60 particles, 5000 proposed moves).
The outer for it's just to calculate the results for different values of Delta
.
Running this code with TIME_MC=60
can make you an idea of how much slow is this code (~218s
) which takes just some seconds if implemented in C. I read some other question about how to speed up Python codes but I can't understand how to do it here.
I'm now almost sure that the problem is in the function dist
, since just to calculate PBC distance between two 3D vectors it takes around ~0.0012s
which gives crazy long times when you calculate it 5000*60 times.
Upvotes: 1
Views: 711
Reputation: 676
Note that this is a partial answer continued from comments on the original question.
Here's an example of how "unrolling" numpy's function can improve performance when replaced with a more direct calculation of the distance. Note that this was not verified to be equivalent, especially concerning the rounding. The principle still applies, I think.
import random
import time
import numpy as np
L = 100
inv_L = 0.01
vec_length = 10
repetitions = 100000
def dist_np(A, B):
d = A - B
d -= L*np.around(d/L)
return np.sqrt(np.sum(d**2))
def dist_direct(A, B):
sum = 0
for i in range(0, len(A)):
diff = (A[0,i] - B[0,i])
diff -= L * int(diff * inv_L)
sum += diff * diff
return np.sqrt(sum)
vec1 = np.zeros((1,vec_length))
vec2 = np.zeros((1,vec_length))
for i in range(0, vec_length):
vec1[0,i] = random.random()
vec2[0,i] = random.random()
print("with numpy method:")
start = time.time()
for i in range(0, repetitions):
dist_np(vec1, vec2)
print("done in {}".format(time.time() - start))
print("with direct method:")
start = time.time()
for i in range(0, repetitions):
dist_direct(vec1, vec2)
print("done in {}".format(time.time() - start))
Output:
with numpy method:
done in 6.332799911499023
with direct method:
done in 1.0938000679016113
Play around with the average vector length and the repetitions to see where the sweet spot is. I expect the performance gain is not constant when varying these meta-parameters.
Upvotes: 1