10

Suppose you have the following numpy array,

>>> x = numpy.array([0,1,2,3,4,5,6,7,8,9,10])

and you want to extract a new numpy array consisting of only the first three (3) and last four (4) elements, i.e.,

>>> y = x[something]
>>> print y
[0 1 2 7 8 9 10]

Is this possible? I know that to extract the first three numbers you simply do x[:3] and to extract the last four you do x[-4:], but is there a simple way of extracting all that in a simple slice? I know this can be done by, e.g., appending both calls,

 >>> y = numpy.append(x[:3],x[-4:])

but I was wondering if there is some simple little trick to do it in a more direct, pythonic way, without having to reference x again (i.e., I first thought that maybe x[-4:3] could work but I realized immediately that it didn't made sense).

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  • The question is similar to this one, but the proposed solutions there won't work for newer numpy versions so it's probably not relevant anymore. Still might be interesting: stackoverflow.com/questions/13525266/… Commented Feb 9, 2015 at 15:04

4 Answers 4

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4

I think you should use index arrays.

indices = list(range(3))+list(range(-4,0))
y = x[indices]

You can probably drop list casts (not sure because python 3 changed the behaviour a bit). Or you could use numpy range genreators.

Edit: not sure why the downvote, cause it works:

import numpy
x = numpy.array([0,1,2,3,4,5,6,7,8,9,10])
indices = list(range(3))+list(range(-4,0))
y = x[indices]
print(y)
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2 Comments

Yeah, actually is the same thing that occurred to me after watching @mgilson answer (see the comments). Maybe if you add what we commented in that post regarding the use of list around the range I could accept this as the best answer :).
I was typing my answer before seeing comments, I wanted to make it with "dynamic" ranges, and link to the docs.
4

A simple slice probably won't work. You can use numpy's extended slicing:

>>> import numpy as np
>>> a = np.arange(10)
>>> something = [0, 1, 2, -4, -3, -2, -1]
>>> a[something] 
array([0, 1, 2, 6, 7, 8, 9])

Notice I passed in a list of indices that I wanted to take from the original array ...

Frankly though, your solution with np.append is likely just as good...

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4 Comments

It could seem like the same, but now that I see your example, I think that an elegant way of doing the same could be: something = range(0,3) + range(-4,0,1). So a general way of slicing the first n elements and the last m elements would be: something = range(0,n) + range(-m,0,1)!
Be careful with adding range together, in Python 3 they're range objects, not lists, and so you'll get a TypeError.
Hi @Ffisegydd; thanks for the input. This means that a more version independant way of doing this would be something = list(range(0,n)) + list(range(-m,0,1))?
Yeah that should be fine :)
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np.arange(11)[np.r_[0:3,7:11]]
# array([ 0,  1,  2,  7,  8,  9, 10])

np.r_ is a function, actually an indexable object, in numpy/lib/index_tricks. It takes multiple slices, or indexes, and concatenates them into one indexing array. 

np.r_[0:3,7:11]
# array([ 0,  1,  2,  7,  8,  9, 10])

Or borrowing from luk32s answer, a negative slice works:

a[np.r_[:3,-4:0]]

An alternative to splicing the 2 parts is to delete the middle. s_ lets you use slice notation inside the function.

np.delete(np.arange(11),np.s_[3:7])
# array([ 0,  1,  2,  7,  8,  9, 10])

Look at the code for the .lib.index_tricks, delete, and insert functions for more ideas on how to construct index arrays from multiple pieces.

Sometimes a boolean index is convenient. Make it all True or False, and flip selected elements.

i = np.ones(a.shape,dtype='bool')
i[3:7] = False
a[i]
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1

These days one can use the built-in function np.setdiff1d() like so:

import numpy as np


x = np.asarray([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
x_cut_out = np.asarray([3, 4, 5, 6])

x_new = np.setdiff1d(
    ar1=x,
    ar2=x_cut_out
)

This returns the unique values in x that are not in x_cut_out.

>>> print(x_new)
[ 0  1  2  7  8  9 10]
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