Inserting rows and columns into a numpy array

Question

I would like to insert multiple rows and columns into a NumPy array.

If I have a square array of length n_a , eg: n_a = 3

a = np.array([[1, 2, 3],
              [4, 5, 6],
              [7, 8, 9]])

and I would like to get a new array with size n_b , which contains array a and zeros (or any other 1D array of length n_b ) on certain rows and columns with indices, eg

index = [1, 3] 

so n_b = n_a + len(index) . Then the new array is:

b = np.array([[1, 0, 2, 0, 3],
              [0, 0, 0, 0, 0],
              [4, 0, 5, 0, 6],
              [0, 0, 0, 0, 0],
              [7, 0, 8, 0, 9]])

My question is, how to do this efficiently, with the assumption that by bigger arrays n_a is much larger than len(index) .

EDIT

The results for:

import numpy as np
import random

n_a = 5000
n_index = 100

a=np.random.rand(n_a, n_a)
index = random.sample(range(n_a), n_index)

Warren Weckesser's solution: 0.208 s

wim's solution: 0.980 s

Ashwini Chaudhary's solution: 0.955 s

Thank you to all!

Answer 1

Here's one way to do it. It has some overlap with @wim's answer, but it uses index broadcasting to copy a into b with a single assignment.

import numpy as np

a = np.array([[1, 2, 3],
              [4, 5, 6],
              [7, 8, 9]])

index = [1, 3]
n_b = a.shape[0] + len(index)

not_index = np.array([k for k in range(n_b) if k not in index])

b = np.zeros((n_b, n_b), dtype=a.dtype)
b[not_index.reshape(-1,1), not_index] = a

Answer 2

You can do this by applying two numpy.insert calls on a :

>>> a = np.array([[1, 2, 3],
              [4, 5, 6],
              [7, 8, 9]])
>>> indices = np.array([1, 3])
>>> i = indices - np.arange(len(indices))
>>> np.insert(np.insert(a, i, 0, axis=1), i, 0, axis=0)
array([[1, 0, 2, 0, 3],
       [0, 0, 0, 0, 0],
       [4, 0, 5, 0, 6],
       [0, 0, 0, 0, 0],
       [7, 0, 8, 0, 9]])

Answer 3

Since fancy indexing returns a copy instead of a view, I can only think how to do it in a two-step process. Maybe a numpy wizard knows a better way...

Here you go:

import numpy as np

a = np.array([[1, 2, 3],
              [4, 5, 6],
              [7, 8, 9]])

index = [1, 3]
n = a.shape[0]
N = n + len(index)

non_index = [x for x in xrange(N) if x not in index]

b = np.zeros((N,n), a.dtype)
b[non_index] = a

a = np.zeros((N,N), a.dtype)
a[:, non_index] = b

Answer 4

Why can't you just Slice/splice ? This has zero loops or for statements.

xlen = a.shape[1]
ylen = a.shape[0]
b = np.zeros((ylen * 2 - ylen % 2, xlen * 2 - xlen % 2))  #accomodates both odd and even shapes
b[0::2,0::2] = a