CODEWITHSUNDEEP
python
John C
John C

Reputation: 79

Shift a list in Python

How would you shift left a list by x in Python, and fill the empty values with zeros?

shift left by 1

input: [1, 2, 3, 4] output: [2, 3, 4, 0]

shift left by 2

input [1, 2, 3, 4] output [3, 4, 0, 0]

Upvotes: 4

Views: 2059

Answers (3)

Ahx
Ahx

Reputation: 7995

You can combine numpy's append and roll methods:

import numpy as np


def shift_left(x_, r):
    return np.append(np.roll(x_, -r)[:-r], [0 for _ in range(0, r)])


print(shift_left([1, 2, 3, 4], 1))
print(shift_left([1, 2, 3, 4], 2))

Result:

[2 3 4 0]
[3 4 0 0]

Explanation

When you use roll on a list:

print(np.roll([1, 2, 3, 4], -2))

Result:

[3 4 1 2]

You move the each element to the left by r times (r= -2). But we don't want the last r elements so:

print(np.roll([1, 2, 3, 4], -2)[:-2])

Result:

[3 4]

We want the last r values to be 0. So we can append r 0 to the end of the array.

print(np.append(np.roll([1, 2, 3, 4], -2)[:-2], [0 for _ in range(0, 2)]))

Result:

[3 4 0 0]

Upvotes: 1

João Haas
João Haas

Reputation: 2139

As far as I'm concerned, there's no 'easy' way, since Python's lists are not constrained by a size, but you can easily implement an algorithm which handles this:

def shift_left(arr, n):
    return arr[n:] + [0 for _ in range(n)]

or a bit more consise:

def shift_left(arr, n):
    return arr[n:] + [0] * n

Upvotes: 3

Hamza
Hamza

Reputation: 6035

You can concat two lists as:

arr[shift:]+[0]*shift

Or if you are a fan of chaining like me:

arr[shift:].__add__([0]*shift)

Upvotes: 1

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