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If I have an array, A, with shape (n, m, o) and an array, B, with shape (n, m), is there a way to divide each array at A[n, m] by the scalar at B[n, m] without a list comprehension?

>>> A.shape
(4,173,1469)
>>> B.shape
(4,173)
>>> # Better way to do:
>>> np.array([[A[i, j] / B[i, j] for j in range(len(B[i]))] for i in range(len(B))])

The problem with a list comprehension is that it is slow, it doesn't return an array (so you have to np.array(_) it, which makes it even slower), it is hard to read, and the whole point of numpy was to move loops from Python to C++ or Fortran.

If A was of shape (n) and B was a scalar (of shape ( )), then this would be trivial: A / B, but this property does not scale with dimensions

>>> A / B
ValueError: operands could not be broadcast together with shapes (4,173,1469) (4,173)

I am looking for a fast way to do this (preferably not by tiling B to an array of shape (n, m, o), and preferably using native numpy tools).

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2 Answers 2

up vote 1 down vote accepted

You are absolutely right, there is a better way, I think you are getting the spirit of numpy. The solution in your case is that you have to add a new dimension to B that consists of one entry in that dimension: so if your A is of shape (n,m,o) your B has to be of shape (n,m,1) and then you can use the native broadcasting to get your operation "A/B" done. You can just add that dimension to be by adding a "newaxis" to B there.

import numpy as np
A = np.ones(10,5,3)
B = np.ones(10,5)
Result = A/B[:,:,np.newaxis]

B[:,:,np.newaxis] --> this will turn B into an array of shape of (10,5,1)

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You might want to try B[..., None] for a more elegant solution. –  Ophion Feb 2 '14 at 22:34
up vote 0 down vote

From here, the rules of broadcasting are:

When operating on two arrays, NumPy compares their shapes element-wise. It starts with the trailing dimensions, and works its way forward. Two dimensions are compatible when

they are equal, or
one of them is 1

Your dimensions are n,m,o and n,m so not compatible.

The / division operator will work using broadcasting if you use:

  1. o,n,m divided by n,m
  2. n,m,o divided by n,m,1
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