问题
It's a call to the community to see if anyone has an idea to improve the speed of this MSD calculation implementation. It is largely based on the implementation from this blog post : http://damcb.com/mean-square-disp.html
For now the current implementation takes about 9s for a 2D trajectory of 5 000 points. It's really way too much if you need to compute a lot of trajectories...
I didn't try to parallelize it (with multiprocess or joblib) but I have the feeling that creating new processes will be too heavy for this kind of algorithm.
Here is the code :
import os
import matplotlib
import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
# Parameters
N = 5000
max_time = 100
dt = max_time / N
# Generate 2D brownian motion
t = np.linspace(0, max_time, N)
xy = np.cumsum(np.random.choice([-1, 0, 1], size=(N, 2)), axis=0)
traj = pd.DataFrame({'t': t, 'x': xy[:,0], 'y': xy[:,1]})
print(traj.head())
# Draw motion
ax = traj.plot(x='x', y='y', alpha=0.6, legend=False)
# Set limits
ax.set_xlim(traj['x'].min(), traj['x'].max())
ax.set_ylim(traj['y'].min(), traj['y'].max())
And the output :
t x y
0 0.000000 -1 -1
1 0.020004 -1 0
2 0.040008 -1 -1
3 0.060012 -2 -2
4 0.080016 -2 -2
def compute_msd(trajectory, t_step, coords=['x', 'y']):
tau = trajectory['t'].copy()
shifts = np.floor(tau / t_step).astype(np.int)
msds = np.zeros(shifts.size)
msds_std = np.zeros(shifts.size)
for i, shift in enumerate(shifts):
diffs = trajectory[coords] - trajectory[coords].shift(-shift)
sqdist = np.square(diffs).sum(axis=1)
msds[i] = sqdist.mean()
msds_std[i] = sqdist.std()
msds = pd.DataFrame({'msds': msds, 'tau': tau, 'msds_std': msds_std})
return msds
# Compute MSD
msd = compute_msd(traj, t_step=dt, coords=['x', 'y'])
print(msd.head())
# Plot MSD
ax = msd.plot(x="tau", y="msds", logx=True, logy=True, legend=False)
ax.fill_between(msd['tau'], msd['msds'] - msd['msds_std'], msd['msds'] + msd['msds_std'], alpha=0.2)
And the output :
msds msds_std tau
0 0.000000 0.000000 0.000000
1 1.316463 0.668169 0.020004
2 2.607243 2.078604 0.040008
3 3.891935 3.368651 0.060012
4 5.200761 4.685497 0.080016
And some profiling :
%timeit msd = compute_msd(traj, t_step=dt, coords=['x', 'y'])
Give this :
1 loops, best of 3: 8.53 s per loop
Any idea ?
回答1:
The MSD calculations mentioned so far are all O(N**2) where N is the number of time steps. Using the FFT this can be reduced to O(N*log(N)). See this question and answer for an explanation and an implementation in python.
EDIT: A small benchmark (I also added this benchmark to this answer): Generate a trajectory with
r = np.cumsum(np.random.choice([-1., 0., 1.], size=(N, 3)), axis=0)
For N=100.000, we get
$ %timeit msd_straight_forward(r)
1 loops, best of 3: 2min 1s per loop
$ %timeit msd_fft(r)
10 loops, best of 3: 253 ms per loop
回答2:
It did some profiling line by line and it appears that pandas is making this slow. This pure numpy version is about 14x faster:
def compute_msd_np(xy, t, t_step):
shifts = np.floor(t / t_step).astype(np.int)
msds = np.zeros(shifts.size)
msds_std = np.zeros(shifts.size)
for i, shift in enumerate(shifts):
diffs = xy[:-shift if shift else None] - xy[shift:]
sqdist = np.square(diffs).sum(axis=1)
msds[i] = sqdist.mean()
msds_std[i] = sqdist.std(ddof=1)
msds = pd.DataFrame({'msds': msds, 'tau': t, 'msds_std': msds_std})
return msds
回答3:
Adding to moarningsun answer above:
- you can speed up using numexpr
if you plot the MSD in log scale anyway, you don't need to compute it for every time
import numpy as np import numexpr def logSpaced(L, pointsPerDecade=15): """Generate an array of log spaced integers smaller than L""" nbdecades = np.log10(L) return np.unique(np.logspace( start=0, stop=nbdecades, num=nbdecades * pointsPerDecade, base=10, endpoint=False ).astype(int)) def compute_msd(xy, pointsPerDecade=15): dts = logSpaced(len(xy), pointsPerDecade) msd = np.zeros(len(idts)) msd_std = np.zeros(len(idts)) for i, dt in enumerate(dts): sqdist = numexpr.evaluate( '(a-b)**2', {'a': xy[:-dt], 'b':xy[dt:]} ).sum(axis=-1) msd[i] = sqdist.mean() msd_std[i] = sqdist.std(ddof=1) msds = pd.DataFrame({'msds': msd, 'tau': dt, 'msds_std': msd_std}) return msds
回答4:
With the comments I designed this function :
def get_msd(traj, dt, with_nan=True):
shifts = np.arange(1, len(traj), dtype='int')
msd = np.empty((len(shifts), 2), dtype='float')
msd[:] = np.nan
msd[:, 1] = shifts * dt
for i, shift in enumerate(shifts):
diffs = traj[:-shift] - traj[shift:]
if with_nan:
diffs = diffs[~np.isnan(diffs).any(axis=1)]
diffs = np.square(diffs).sum(axis=1)
if len(diffs) > 0:
msd[i, 0] = np.mean(diffs)
msd = pd.DataFrame(msd)
msd.columns = ["msd", "delay"]
msd.set_index('delay', drop=True, inplace=True)
msd.dropna(inplace=True)
return msd
With the following features :
- It takes
numpyarray as trajectory input. - It returns a
pandas.DataFramewith almost no overlay. with_nanallows to deal with trajectory containingNaNvalues but it adds a big overhead (more than 100%) so I put it as a function parameter.- It can deal with multi dimensional trajectories (1D, 2D, 3D, etc)
Some profiling :
$ print(traj.shape)
(2108, 2)
$ %timeit get_msd(traj, with_nan=True, dt=0.1)
10 loops, best of 3: 143 ms per loop
$ %timeit get_msd(traj, with_nan=False, dt=0.1)
10 loops, best of 3: 68 ms per loop
回答5:
Maybe is not the topic, however MSD has to be calculated not be the mean like in line 37:
msds[i] = sqdist.mean()
Taking as mean=N
You must divide by:
msds[i] = sqdist/N-1 // for lag1
Then:
msds[i] = sqdist/N-2 // for lag2 .... msds[i] = sqdist/N-n // for lag n
And so on.
As a result you don't get standard deviation, just the MSD for a single trajectory
来源：https://stackoverflow.com/questions/32988269/speedup-msd-calculation-in-python