select 0 in the first n numpy array dimensions

Question

I have a numpy array of shape (1,2,3,4,5,6,...) (arbitrary length) and I would like to select the first entry in the first n columns, i.e.,

def select_first_n_0(arr, n):
    if n == 1:
        return arr[0]
    elif n == 2:
        return arr[0][0]
    elif n == 2:
        return arr[0][0][0]
    # ...

Is there a more comprehensive expression for this?

Create an indexing tuple, e.g. idx=(0,0,0).

hpaulj
– hpaulj

2021-02-20 16:32:08 +00:00
Commented Feb 20, 2021 at 16:32 — hpaulj
– hpaulj, Commented Feb 20, 2021 at 16:32

Rostyslav · Accepted Answer · 2021-02-20 11:32:54Z

1

You can use recursion

def select_first_n_0(arr, n):
   if n == 1:
      return arr[0]
   return select_first_n_0(arr[0], n-1)

answered Feb 20, 2021 at 11:32

Rostyslav

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Nico Schlömer · Accepted Answer · 2021-02-23 17:53:11Z

0

After @hjpaul:

def select_first_n_0(arr, n):
    return arr[tuple([0] * n)]

answered Feb 23, 2021 at 17:53

Nico Schlömer

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