Modify numpy array based on previous value

Question

I have a np array, let's say

[1, 1, 1, 1, 0, 1, 0, 1, 0]

Now I want to modify the number at index i s.t. it is 0, if one of the previous n elements was a 1. But this calculation should not be static w.r.t to the original array, but dynamic, i.e. considering already changed values. Let's say n=2, then I want to get

[1, 0, 0, 1, 0, 0, 0, 1, 0]

How do I do that efficiently without a for loop? Simple indexing will run over the whole array in one iteration, ignoring newly changed values, meaning the result would be

[1, 0, 0, 0, 0, 0, 0, 0, 0]

right?

Update: In a loop I would do this:

for i in range(a.shape[0]):
    FoundPrevious = False
    for j in range(n):
        if i - j < 0:
            continue
        if a[i - j] == 1:
            FoundPrevious = True
            break
    if FoundPrevious:
        a[i] = 0

Can you show how you would do it in a loop. That will make it easier to convert. — Kenan
– Kenan, Commented Aug 31, 2018 at 19:08
A list solution is probably the best you'll get. The problem is inherently iterative, and it's faster to iterate on a list than an array. — hpaulj
– hpaulj, Commented Aug 31, 2018 at 19:12
Shouldn't the correct output be [1, 0, 0, 1, 0, 0, 0, 1, 0]? — cs95
– cs95, Commented Aug 31, 2018 at 19:18

cs95 · Accepted Answer · 2018-08-31 19:23:00Z

4

The problem is inherently iterative, but you can simplify things by doing a single pass over your list.

arr = [1, 1, 1, 1, 0, 1, 0, 1, 0]
n = 2
newarr = []

for i in arr:
    newarr.append(0 if sum(newarr[-n:]) else i)

print (newarr)
[1, 0, 0, 1, 0, 0, 0, 1, 0]

answered Aug 31, 2018 at 19:23

cs95

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2 Comments

BENY Over a year ago

When I saw the pic, I thought it was scott , LOL :-)

cs95 Over a year ago

@Wen Almost, scott is from TAMU :p

Daniel R. Livingston · Accepted Answer · 2018-08-31 19:25:48Z

1

As other answerers and commenters have noted, the intrinsic interative nature of the problem makes a solution without a for-loop non-obvious.

Another approach to simplifying this problem is:

import numpy as np

a = np.array([1, 1, 1, 1, 0, 1, 0, 1, 0])
n = 2

for i in range(a.shape[0]):
    i0 = i - 2 if i >= 2 else 0
    if 1 in a[i0:i]:
        a[i] = 0

print(a)
# [1 0 0 1 0 0 0 1 0]

answered Aug 31, 2018 at 19:25

Daniel R. Livingston

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