resuming this question: Compute the pairwise distance in scipy with missing values
test case: I want to compute the pairwise distance of series with different length taht are grouped together and I have to do it in the most efficient possible way (using euclidean distance).
one way that makes it work could be this:
import pandas as pd
import numpy as np
from scipy.spatial.distance import pdist
a = pd.DataFrame(np.random.rand(10, 4), columns=['a','b','c','d'])
a.loc[0, 'a'] = np.nan
a.loc[1, 'a'] = np.nan
a.loc[0, 'c'] = np.nan
a.loc[1, 'c'] = np.nan
def dropna_on_the_fly(x, y):
return np.sqrt(np.nansum(((x-y)**2)))
pdist(starting_set, dropna_on_the_fly)
but I feel this could be very inefficient as built in methods of the pdist function are internally optimized whereas the function is simply passed over.
I have a hunch that a vectorized solution in numpy for which I broadcast the subtraction and then I proceed with the np.nansum for na resistant sum but I am unsure on how to proceed.
Inspired by this post, there would be two solutions.
Approach #1 : The vectorized solution would be -
ar = a.values
r,c = np.triu_indices(ar.shape[0],1)
out = np.sqrt(np.nansum((ar[r] - ar[c])**2,1))
Approach #2 : The memory-efficient and more performant one for large arrays would be -
ar = a.values
b = np.where(np.isnan(ar),0,ar)
mask = ~np.isnan(ar)
n = b.shape[0]
N = n*(n-1)//2
idx = np.concatenate(( [0], np.arange(n-1,0,-1).cumsum() ))
start, stop = idx[:-1], idx[1:]
out = np.empty((N),dtype=b.dtype)
for j,i in enumerate(range(n-1)):
dif = b[i,None] - b[i+1:]
mask_j = (mask[i] & mask[i+1:])
masked_vals = mask_j * dif
out[start[j]:stop[j]] = np.einsum('ij,ij->i',masked_vals, masked_vals)
# or simply : ((mask_j * dif)**2).sum(1)
out = np.sqrt(out)