对二维 numpy 数组进行子集化

Question

I have looked into documentations and also other questions here, but it seems I have not got the hang of subsetting in numpy arrays yet.

I have a numpy array, and for the sake of argument, let it be defined as follows:

import numpy as np
a = np.arange(100)
a.shape = (10,10)
# array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
#        [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
#        [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
#        [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
#        [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
#        [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
#        [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
#        [70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
#        [80, 81, 82, 83, 84, 85, 86, 87, 88, 89],
#        [90, 91, 92, 93, 94, 95, 96, 97, 98, 99]])

now I want to choose rows and columns of a specified by vectors n1 and n2 . As an example:

n1 = range(5)
n2 = range(5)

But when I use:

b = a[n1,n2]
# array([ 0, 11, 22, 33, 44])

Then only the first fifth diagonal elements are chosen, not the whole 5x5 block. The solution I have found is to do it like this:

b = a[n1,:]
b = b[:,n2]
# array([[ 0,  1,  2,  3,  4],
#        [10, 11, 12, 13, 14],
#        [20, 21, 22, 23, 24],
#        [30, 31, 32, 33, 34],
#        [40, 41, 42, 43, 44]])

But I am sure there should be a way to do this simple task in just one command.

Answer 1

You've gotten a handful of nice examples of how to do what you want. However, it's also useful to understand the what's happening and why things work the way they do. There are a few simple rules that will help you in the future.

There's a big difference between "fancy" indexing (ie using a list/sequence) and "normal" indexing (using a slice). The underlying reason has to do with whether or not the array can be "regularly strided", and therefore whether or not a copy needs to be made. Arbitrary sequences therefore have to be treated differently, if we want to be able to create "views" without making copies.

In your case:

import numpy as np

a = np.arange(100).reshape(10,10)
n1, n2 = np.arange(5), np.arange(5)

# Not what you want
b = a[n1, n2]  # array([ 0, 11, 22, 33, 44])

# What you want, but only for simple sequences
# Note that no copy of *a* is made!! This is a view.
b = a[:5, :5]

# What you want, but probably confusing at first. (Also, makes a copy.)
# np.meshgrid and np.ix_ are basically equivalent to this.
b = a[n1[:,None], n2[None,:]]

---

Fancy indexing with 1D sequences is basically equivalent to zipping them together and indexing with the result.

print "Fancy Indexing:"
print a[n1, n2]

print "Manual indexing:"
for i, j in zip(n1, n2):
    print a[i, j]

---

However, if the sequences you're indexing with match the dimensionality of the array you're indexing (2D, in this case), The indexing is treated differently. Instead of "zipping the two together", numpy uses the indices like a mask.

In other words, a[[[1, 2, 3]], [[1],[2],[3]]] is treated completely differently than a[[1, 2, 3], [1, 2, 3]] , because the sequences/arrays that you're passing in are two-dimensional.

In [4]: a[[[1, 2, 3]], [[1],[2],[3]]]
Out[4]:
array([[11, 21, 31],
       [12, 22, 32],
       [13, 23, 33]])

In [5]: a[[1, 2, 3], [1, 2, 3]]
Out[5]: array([11, 22, 33])

---

To be a bit more precise,

a[[[1, 2, 3]], [[1],[2],[3]]]

is treated exactly like:

i = [[1, 1, 1],
     [2, 2, 2],
     [3, 3, 3]])
j = [[1, 2, 3],
     [1, 2, 3],
     [1, 2, 3]]
a[i, j]

In other words, whether the input is a row/column vector is a shorthand for how the indices should repeat in the indexing.

---

np.meshgrid and np.ix_ are just convienent ways to turn your 1D sequences into their 2D versions for indexing:

In [6]: np.ix_([1, 2, 3], [1, 2, 3])
Out[6]:
(array([[1],
       [2],
       [3]]), array([[1, 2, 3]]))

Similarly (the sparse argument would make it identical to ix_ above):

In [7]: np.meshgrid([1, 2, 3], [1, 2, 3], indexing='ij')
Out[7]:
[array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]]),
 array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])]

Answer 2

Another quick way to build the desired index is to use the np.ix_ function:

>>> a[np.ix_(n1, n2)]
array([[ 0,  1,  2,  3,  4],
       [10, 11, 12, 13, 14],
       [20, 21, 22, 23, 24],
       [30, 31, 32, 33, 34],
       [40, 41, 42, 43, 44]])

This provides a convenient way to construct an open mesh from sequences of indices.

Answer 3

You could use np.meshgrid to give the n1 , n2 arrays the proper shape to perform the desired indexing:

In [104]: a[np.meshgrid(n1,n2, sparse=True, indexing='ij')]
Out[104]: 
array([[ 0,  1,  2,  3,  4],
       [10, 11, 12, 13, 14],
       [20, 21, 22, 23, 24],
       [30, 31, 32, 33, 34],
       [40, 41, 42, 43, 44]])

---

Or, without meshgrid:

In [117]: a[np.array(n1)[:,np.newaxis], np.array(n2)[np.newaxis,:]]
Out[117]: 
array([[ 0,  1,  2,  3,  4],
       [10, 11, 12, 13, 14],
       [20, 21, 22, 23, 24],
       [30, 31, 32, 33, 34],
       [40, 41, 42, 43, 44]])

There is a similar example with an explanation of how this integer array indexing works in the docs.

See also the Cookbook recipe Picking out rows and columns .

Answer 4

A nice Trick I've managed to pull (for lazy people only) Is filter + Transpose + filter.

a = np.arange(100).reshape(10,10)
subsetA = [1,3,5,7]
a[subsetA].T[subsetA]

array([[11, 31, 51, 71],
       [13, 33, 53, 73],
       [15, 35, 55, 75],
       [17, 37, 57, 77]])

Answer 5

It seems that a use case for your particular question would deal with image manipulation. To the extent that you are using your example to edit numpy arrays arising from images, you can use the Python Imaging Library (PIL).

# Import Pillow:
from PIL import Image

# Load the original image:
img = Image.open("flowers.jpg")

# Crop the image
img2 = img.crop((0, 0, 5, 5))

The img2 object is a numpy array of the resulting cropped image.

You can read more about image manipulation here with the Pillow package (a user friendly fork on the PIL package):