How to repeat a numPy array value

Question

I was wondering if there was a vertorized way to use two arrays and based on the value of one continuously copy the value in a second array until a new value in the first array is found and then repeat the whole process.

here is an example

   a = np.array([FALSE,TRUE,FALSE,FALSE, FALSE, TRUE, FALSE  FALSE})
   b = np.array([0,10,0,0,0,20,0,0,])
   output = array([0,10,10,10,10,20,20,20])

I can use a loop which is much too slow for this.

for i in range(len(b)):   
   if b[i-1] and not b[i]:
       b[i] = b[i-1]

Updated:

I tried this code below and increased the speed by 5x but I figure there should be a faster more elegant way of doing this.

import numpy as np

nz= np.concatenate((np.nonzero(b)[0], [len(b)]))
np.repeat(b[np.nonzero(b)], np.diff(nz))

What is the output until the first TRUE is discovered in a? — Divakar
– Divakar, Commented Nov 30, 2018 at 5:49
The question doesn't appear to include any attempt at all to solve the problem. StackOverflow expects you to try to solve your own problem first, as your attempts help us to better understand what you want. Please edit the question to show what you've tried, and show a specific roadblock you're running into with Minimal, Complete, and Verifiable example. For more information, please see How to Ask. — Stephan T.
– Stephan T., Commented Nov 30, 2018 at 6:15

Mateen Ulhaq · Accepted Answer · 2018-12-01 05:42:45Z

1

Your example can be improved slightly:

b = np.array([0, 10, 0, 0, 0, 20, 0, 0])
nz = np.squeeze(np.nonzero(b))
diffs = np.empty_like(nz)
diffs[:-1] = np.diff(nz)
diffs[-1] = b.size - nz[-1]
result = np.repeat(b[nz], diffs)

...I'm not sure I understand the application though.

It might be wise to perftools your code and figure out what exactly is causing the bottleneck in your application. You might also wish to state the sizes of your arrays. (For instance, an itertools solution might be faster for fewer elements.)

answered Dec 1, 2018 at 5:42

Mateen Ulhaq

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1 Comment

John Holmes Over a year ago

thanks Mateen. I will post another question to descibe what I'm trying to achieve overall but basically i have variables which are arrays that I need to iterate through row by row and is very slow so i'm trying to vectorize the function. I've up voted yours as it works and you have provided an alternative with itertools.

Tls Chris · Accepted Answer · 2018-12-01 09:49:07Z

0

This generalises to n dimensional arrays. It iterates through n-1d slices of the array.

In [26]: a=np.array([
    ...:     [ 0, 0,10, 0, 0, 0,20,30, 0, 0],
    ...:     [ 4, 0, 0, 0, 0, 0, 0, 0, 0, 0],  
    ...:     [ 0, 5, 0, 6, 0, 7, 0, 8, 0, 9]])

In [27]: bfwd=0
    ...: for ix in range(a.shape[-1]):
    ...:     sel=np.s_[... ,ix]
    ...:     a[sel]=bfwd*(a[sel]==0)+a[sel]  # If a is zero use bfwd else use a
    ...:     bfwd=a[sel]

In [28]: a
Out[28]: 
array([[ 0,  0, 10, 10, 10, 10, 20, 30, 30, 30],
       [ 4,  4,  4,  4,  4,  4,  4,  4,  4,  4],
       [ 0,  5,  5,  6,  6,  7,  7,  8,  8,  9]])

HTH

answered Dec 1, 2018 at 9:49

Tls Chris

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