how to convert 1d list to 2d list with some some number of columns?

Question

I want to convert a 1d list to 2d in python. I found a way to do it with NumPy, but it gives the following error when the length of the list is not divisible by the number of columns:

ValueError: cannot reshape array of size 9 into shape (4)

In the case that the length is not divisible, I want to add none to fill the spaces. Like so:

a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
b = some_func(a)
print(b)

and the output should be:

[[1, 2, 3, 4], 
[5, 6, 7, 8], 
[9, None, None, None]]

Please help me. My current code is:

a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a = numpy.reshape(a, (-1, 4))
print(a)

puchal · Accepted Answer · 2021-01-27 11:52:19Z

2

First you could fill your list with None values so it matches the size

reshape_size = 4
a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a += [None]*(reshape_size - len(a) % reshape_size) if len(a)%4 != 0 else []
a = numpy.reshape(a, (-1, reshape_size))
print(a)

answered Jan 27, 2021 at 11:52

puchal

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agayev169 · Accepted Answer · 2021-01-27 11:41:16Z

1

This should work.

from math import ceil

def my_reshape(arr, cols):
    rows = ceil(len(arr) / cols)
    res = []
    for row in range(rows):
        current_row = []
        for col in range(cols):
            arr_idx = row * cols + col
            if arr_idx < len(arr):
                current_row.append(arr[arr_idx])
            else:
                current_row.append(None)
        res.append(current_row)
    return res

answered Jan 27, 2021 at 11:41

agayev169

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Comments

Dogunbound hounds · Accepted Answer · 2021-10-12 13:12:24Z

1

With a little bit of ternary love, you can do it in a one-liner with pure python:

arr2d = [[arr1d[i+j*width] if i+j*width < len(arr1d) else 0 for i in range(width)] for j in range(height)]

answered Oct 12, 2021 at 13:12

Dogunbound hounds

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