Prepend element to numpy array

Question

I have the following numpy array

import numpy as np

X = np.array([[5.], [4.], [3.], [2.], [1.]])

I want to insert [6.] at the beginning. I've tried:

X = X.insert(X, 0)

how do I insert into X?

Answer 1

I know this is a fairly old one, but a short solution is using numpy slicing tricks:

np.r_[[[6.]], X]

If you need to do it in a second dimension you can use np.c_.

I think this is the least cluttered version I can think of

Answer 2

You can try the following

X = np.append(arr = np.array([[6]]), values = X, axis= 0)

Instead of inserting 6 to the existing X, let append 6 by X.

So, first argument arr is numpy array of scalar 6, second argument is your array to be added, and third is the place where we want to add

Answer 3

I just wrote some code that does this operation ~100,000 times, so I needed to figure out the fastest way to do this. I'm not an expert in code efficiency by any means, but I could figure some things out by using the %%timeit magic function in a jupyter notebook.

My findings:

np.concatenate(([number],array)) requires the least time. Let's call it 1x time.

np.asarray([number] + list(array)) comes in at ~2x.

np.r_[number,array] is ~4x.

np.insert(array,0,number) appears to be the worst option here at 8x.

I have no idea how this changes with the size of array (I used a shape (15,) array) and most of the options I suggested only work if you want to put the number at the beginning. However, since that's what the question is asking about, I figure this is a good place to make these comparisons.

Answer 4

numpy has an insert function that's accesible via np.insert with documentation.

You'll want to use it in this case like so:

X = np.insert(X, 0, 6., axis=0)

the first argument X specifies the object to be inserted into.

The second argument 0 specifies where.

The third argument 6. specifies what is to be inserted.

The fourth argument axis=0 specifies that the insertion should happen at position 0 for every column. We could've chosen rows but your X is a columns vector, so I figured we'd stay consistent.