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?

Camera

Answer 1

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.

Taite

More about axis=0 is here medium.com/@panjeh/… – Weft

Answer 2

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.

Sterner

Awesome! Not only the first option is faster, but can be easily used to insert an entire sequence ... for example, in my case, I wanted to insert several zeros at the begging; aa =np.concatenate(([0]*99, aa)) ... as I was manipulating a signal prior to smoothing, then finding the peaks. – Gabbert

Answer 3

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

Stentor

Answer 4

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

Varicotomy

BTW: you can append this way as well by simply reversing the order of course. – Varicotomy

Answer 5

An addition to the comment of jbf81tb, as of 2023, all variants but the one with conversion to list, are more or less equal:

arr = np.random.rand(1000000)
%timeit np.concatenate(([-1],arr))
1.33 ms ± 23.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

arr = np.random.rand(1000000)
%timeit np.asarray([-1] + list(arr))
63.7 ms ± 691 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

arr = np.random.rand(1000000)
%timeit np.r_[-1,arr]
1.41 ms ± 35.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

arr = np.random.rand(1000000)
%timeit np.insert(arr,0,-1)
1.39 ms ± 24.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Larondalarosa