How to Index multiple items of array with intervals in Python

Question

Suppose that I have a list:

import numpy as np
a = [2, 4, 6, 8, ..., 1000] # total 500 elements
b = np.array(a)             # numpy version

I want to get 1st to 100th, 201st to 300th, 401st to 500th elements and make them into a new array.

To this end, I've tried the following codes:

a_sub = a[0:100] + a[200:300] + a[400:500]
b_sub = np.concatenate((b[0:100], b[200:300], b[400:500]))

But I want to do it with a simple oneline-indexing

Say:

a_sub = a[(0:100, 200:300, 400:500)]
a_sub = a[[0:100, 200:300, 400:500]]
b_sub = b[[0:100, 200:300, 400:500]]
b_sub = b[[0:100, 200:300, 400:500]]

But the above are all invalid and I couldn't find such a oneliner indexing.

I don't know if this is possible . Or do you want this a_sub = b[range(0,100)+range(200,300)+range(400,500)] — Akshay Hazari
– Akshay Hazari, Commented Nov 21, 2015 at 6:22
I mean you can't index a list like that in a line specifically a_sub = a[(0:100, 200:300, 400:500)] — Akshay Hazari
– Akshay Hazari, Commented Nov 21, 2015 at 6:24

Divakar · Accepted Answer · 2015-11-21 07:40:55Z

3

You could use reshaping with np.reshape and slicing, like so -

np.array(a).reshape(-1,100)[::2].ravel()

If a is a NumPy array, you could do like so -

a.reshape(-1,100)[::2].ravel()

edited Nov 21, 2015 at 7:40

answered Nov 21, 2015 at 7:34

Divakar

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1 Comment

Nestor Demeure Over a year ago

Note that this solution only works for interval of constant size and spacing (one interval of size 100 every 100 items) as in the OP's example.

shx2 · Accepted Answer · 2015-11-21 07:29:25Z

2

You can convert the slices to a mask-array (by slicing an ones-array), and union the mask-arrays using the | (or) operator.

ones = np.ones(b.shape, dtype = bool)
mask = ones[ 0:100] | ones[200:300] | ones[400:500]
b_sub = b[mask]

Note that if your slices overlap, or appear in a non-increasing order, this results with a different array than your original code (items will not repeat, and will always appear in the same order as in the original array).

answered Nov 21, 2015 at 7:29

shx2

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Comments

ali_m · Accepted Answer · 2015-11-21 19:30:11Z

2

You could also use np.split:

a = range(2, 1002, 2)
edges = [100, 200, 300, 400]
subarrays = np.split(a, edges)
b = np.hstack(subarrays[i] for i in [0, 2, 4])

answered Nov 21, 2015 at 19:30

ali_m

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Comments

Andrii Rusanov · Accepted Answer · 2015-11-21 07:36:55Z

1

well, it is pure python, but maybe it may solve your question

a = [2, 4, 6, 8, ..., 1000]
slices = ((0, 100), (200, 300), (400, 500))

def new_from_slices(list_, slices):
    return list(itertools.chain(*[list_[s[0]:s[1]] for s in slices]))
new_from_slices(a, slices)

answered Nov 21, 2015 at 7:36

Andrii Rusanov

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B. M. · Accepted Answer · 2015-11-21 15:59:05Z

0

two others one-liner :

[x for i,x in enumerate(a) if i//100%2==0] #python
b[np.arange(len(b))//100%2==0] # numpy

edited Nov 21, 2015 at 15:59

answered Nov 21, 2015 at 13:10

B. M.

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MaxNoe · Accepted Answer · 2015-11-21 20:29:19Z

0

or using hstack (+ adds the values element wise)

a = np.arange(1000)
limits = [(0, 100), (200, 300), (400, 500)]
b = np.hstack(a[low:high] for low, high in limits)

answered Nov 21, 2015 at 20:29

MaxNoe

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