Speedup MSD calculation in Python

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 numpy array as trajectory input.
• It returns a pandas.DataFrame with almost no overlay.
• with_nan allows to deal with trajectory containing NaN values 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