Pandas expanding/rolling window correlation calculation with p-value
Approach #1
corr2_coeff_rowwise
lists how to do element-wise correlation between rows. We could strip it down to a case for element-wise correlation between two columns. So, we would end up with a loop that uses corr2_coeff_rowwise
. Then, we would try to vectorize it and see that there are pieces in it that could be vectorized :
- Getting average values with
mean
. This could be vectorized with use of uniform filter. - Next up was getting the differences between those average values against sliding elements from input arrays. To port to a vectorized one, we would make use of
broadcasting
.
Rest stays the same to get the first off the two outputs from pearsonr
.
To get the second output, we go back to the source code
. This should be straight-forward given the first coefficient output.
So, with those in mind, we would end up with something like this -
import scipy.special as special
from scipy.ndimage import uniform_filter
def sliding_corr1(a,b,W):
# a,b are input arrays; W is window length
am = uniform_filter(a.astype(float),W)
bm = uniform_filter(b.astype(float),W)
amc = am[W//2:-W//2+1]
bmc = bm[W//2:-W//2+1]
da = a[:,None]-amc
db = b[:,None]-bmc
# Get sliding mask of valid windows
m,n = da.shape
mask1 = np.arange(m)[:,None] >= np.arange(n)
mask2 = np.arange(m)[:,None] < np.arange(n)+W
mask = mask1 & mask2
dam = (da*mask)
dbm = (db*mask)
ssAs = np.einsum('ij,ij->j',dam,dam)
ssBs = np.einsum('ij,ij->j',dbm,dbm)
D = np.einsum('ij,ij->j',dam,dbm)
coeff = D/np.sqrt(ssAs*ssBs)
n = W
ab = n/2 - 1
pval = 2*special.btdtr(ab, ab, 0.5*(1 - abs(np.float64(coeff))))
return coeff,pval
Thus, to get the final output off the inputs from the pandas series -
out = sliding_corr1(df['x'].to_numpy(copy=False),df['y'].to_numpy(copy=False),50)
Approach #2
A lot similar to Approach #1
, but we will use numba
to improve on the memory efficiency to replace step #2 from previous approach.
from numba import njit
import math
@njit(parallel=True)
def sliding_corr2_coeff(a,b,amc,bmc):
L = len(a)-W+1
out00 = np.empty(L)
for i in range(L):
out_a = 0
out_b = 0
out_D = 0
for j in range(W):
d_a = a[i+j]-amc[i]
d_b = b[i+j]-bmc[i]
out_D += d_a*d_b
out_a += d_a**2
out_b += d_b**2
out00[i] = out_D/math.sqrt(out_a*out_b)
return out00
def sliding_corr2(a,b,W):
am = uniform_filter(a.astype(float),W)
bm = uniform_filter(b.astype(float),W)
amc = am[W//2:-W//2+1]
bmc = bm[W//2:-W//2+1]
coeff = sliding_corr2_coeff(a,b,amc,bmc)
ab = W/2 - 1
pval = 2*special.btdtr(ab, ab, 0.5*(1 - abs(np.float64(coeff))))
return coeff,pval
Approach #3
Very similar to previous one, except that we are pushing all of the coefficient work to numba
-
@njit(parallel=True)
def sliding_corr3_coeff(a,b,W):
L = len(a)-W+1
out00 = np.empty(L)
for i in range(L):
a_mean = 0.0
b_mean = 0.0
for j in range(W):
a_mean += a[i+j]
b_mean += b[i+j]
a_mean /= W
b_mean /= W
out_a = 0
out_b = 0
out_D = 0
for j in range(W):
d_a = a[i+j]-a_mean
d_b = b[i+j]-b_mean
out_D += d_a*d_b
out_a += d_a*d_a
out_b += d_b*d_b
out00[i] = out_D/math.sqrt(out_a*out_b)
return out00
def sliding_corr3(a,b,W):
coeff = sliding_corr3_coeff(a,b,W)
ab = W/2 - 1
pval = 2*special.btdtr(ab, ab, 0.5*(1 - np.abs(coeff)))
return coeff,pval
Timings -
In [181]: df = pd.DataFrame({'x': np.random.rand(10000), 'y': np.random.rand(10000)})
In [182]: %timeit sliding_corr2(df['x'].to_numpy(copy=False),df['y'].to_numpy(copy=False),50)
100 loops, best of 3: 5.05 ms per loop
In [183]: %timeit sliding_corr3(df['x'].to_numpy(copy=False),df['y'].to_numpy(copy=False),50)
100 loops, best of 3: 5.51 ms per loop
Note :
sliding_corr1
seems to be taking long on this dataset and most probably because of the memory-requirement from its step#2.The bottleneck after using the numba funcs, then transfers to the p-val computation with
special.btdtr
.
I could not think of a clever way to do this in pandas using rolling
directly, but note that you can calculate the p-value given the correlation coefficient.
Pearson's correlation coefficient follows Student's t-distribution and you can get the p-value by plugging it to the cdf defined by the incomplete beta function, scipy.special.betainc
. It sounds complicated but can be done in a few lines of code. Below is a function that computes the p-value given the correlation coefficient corr
and the sample size n
. It is actually based on scipy's implementation you have been using.
import pandas as pd
from scipy.special import betainc
def pvalue(corr, n=50):
df = n - 2
t_squared = corr**2 * (df / ((1.0 - corr) * (1.0 + corr)))
prob = betainc(0.5*df, 0.5, df/(df+t_squared))
return prob
You can then apply this function to the correlation values you already have.
rolling_corr = df['x'].rolling(50).corr(df['y'])
pvalue(rolling_corr)
It might not be the perfect vectorised numpy solution but should be tens of times faster than calculating the correlations over and over again.