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How can I vectorize this for loop in numpy?

The code is below:

import numpy as np
X = np.array(range(15)).reshape(5,3)  # X's element value is meaningless
flag = np.random.randn(5,4)
y = np.array([0, 1, 2, 3, 0])  # Y's element value in range(flag.shape[1]) and Y.shape[0] equals X.shape[0]
dW = np.zeros((3, 4))  # dW.shape equals (X.shape[1], flag.shape[1])
for i in xrange(5):
    for j in xrange(4):
        if flag[i,j] > 0:
            dW[:,j] += X[i,:].T
            dW[:,y[i]] -= X[i,:].T

To compute dW more efficiently, how to vectorize this for loop?

like image 750
Century Xu Avatar asked Feb 15 '16 15:02

Century Xu


2 Answers

Here's how I'd do it:

# has shape (x.shape[1],) + flag.shape
masked = np.where(flag > 0, X.T[...,np.newaxis], 0)

# sum over the i index
dW = masked.sum(axis=1)

# sum over the j index
np.subtract.at(dW, np.s_[:,y], masked.sum(axis=2))

# dW[:,y] -= masked.sum(axis=2) does not work here

See the documentation of ufunc.at for an explanation of that last comment

like image 170
Eric Avatar answered Oct 01 '22 17:10

Eric


Here's a vectorized approach based upon np.add.reduceat -

# --------------------- Setup output array ----------------------------------
dWOut = np.zeros((X.shape[1], flag.shape[1]))

# ------ STAGE #1 : Vectorize calculations for "dW[:,j] += X[i,:].T" --------
# Get indices where flag's transposed version has > 0
idx1 = np.argwhere(flag.T > 0)

# Row-extended version of X using idx1's col2 that corresponds to i-iterator
X_ext1 = X[idx1[:,1]]

# Get the indices at which we need to columns change
shift_idx1 = np.append(0,np.where(np.diff(idx1[:,0])>0)[0]+1)

# Use the changing indices as boundaries for add.reduceat to add 
# groups of rows from extended version of X
dWOut[:,np.unique(idx1[:,0])] += np.add.reduceat(X_ext1,shift_idx1,axis=0).T

# ------ STAGE #2 : Vectorize calculations for "dW[:,y[i]] -= X[i,:].T" -------
# Repeat same philsophy for this second stage, except we need to index into y.
# So, that would involve sorting and also the iterator involved is just "i".
idx2 = idx1[idx1[:,1].argsort()]
cols_idx1 = y[idx2[:,1]]
X_ext2 = X[idx2[:,1]]
sort_idx = (y[idx2[:,1]]).argsort()
X_ext2 = X_ext2[sort_idx]
shift_idx2 = np.append(0,np.where(np.diff(cols_idx1[sort_idx])>0)[0]+1)
dWOut[:,np.unique(cols_idx1)] -= np.add.reduceat(X_ext2,shift_idx2,axis=0).T
like image 27
Divakar Avatar answered Oct 01 '22 16:10

Divakar