Selecting multiple slices from a numpy array at once

24,520

Solution 1

You can use the indexes to select the rows you want into the appropriate shape. For example:

 data = np.random.normal(size=(100,2,2,2))

 # Creating an array of row-indexes
 indexes = np.array([np.arange(0,5), np.arange(1,6), np.arange(2,7)])
 # data[indexes] will return an element of shape (3,5,2,2,2). Converting
 # to list happens along axis 0
 data_extractions = list(data[indexes])

 np.all(data_extractions[1] == data[1:6])
 True

The final comparison is against the original data.

Solution 2

stride_tricks can do that

a = np.arange(10)
b = np.lib.stride_tricks.as_strided(a, (3, 5), 2 * a.strides)
b
# array([[0, 1, 2, 3, 4],
#        [1, 2, 3, 4, 5],
#        [2, 3, 4, 5, 6]])

Please note that b references the same memory as a, in fact multiple times (for example b[0, 1] and b[1, 0] are the same memory address). It is therefore safest to make a copy before working with the new structure.

nd can be done in a similar fashion, for example 2d -> 4d

a = np.arange(16).reshape(4, 4)
b = np.lib.stride_tricks.as_strided(a, (3,3,2,2), 2*a.strides)
b.reshape(9,2,2) # this forces a copy
# array([[[ 0,  1],
#         [ 4,  5]],

#        [[ 1,  2],
#         [ 5,  6]],

#        [[ 2,  3],
#         [ 6,  7]],

#        [[ 4,  5],
#         [ 8,  9]],

#        [[ 5,  6],
#         [ 9, 10]],

#        [[ 6,  7],
#         [10, 11]],

#        [[ 8,  9],
#         [12, 13]],

#        [[ 9, 10],
#         [13, 14]],

#        [[10, 11],
#         [14, 15]]])

Solution 3

In this post is an approach with strided-indexing scheme using np.lib.stride_tricks.as_strided that basically creates a view into the input array and as such is pretty efficient for creation and being a view occupies nomore memory space. Also, this works for ndarrays with generic number of dimensions.

Here's the implementation -

def strided_axis0(a, L):
    # Store the shape and strides info
    shp = a.shape
    s  = a.strides

    # Compute length of output array along the first axis
    nd0 = shp[0]-L+1

    # Setup shape and strides for use with np.lib.stride_tricks.as_strided
    # and get (n+1) dim output array
    shp_in = (nd0,L)+shp[1:]
    strd_in = (s[0],) + s
    return np.lib.stride_tricks.as_strided(a, shape=shp_in, strides=strd_in)

Sample run for a 4D array case -

In [44]: a = np.random.randint(11,99,(10,4,2,3)) # Array

In [45]: L = 5      # Window length along the first axis

In [46]: out = strided_axis0(a, L)

In [47]: np.allclose(a[0:L], out[0])  # Verify outputs
Out[47]: True

In [48]: np.allclose(a[1:L+1], out[1])
Out[48]: True

In [49]: np.allclose(a[2:L+2], out[2])
Out[49]: True

Solution 4

You can slice your array with a prepared slicing array

a = np.array(list('abcdefg'))

b = np.array([
        [0, 1, 2, 3, 4],
        [1, 2, 3, 4, 5],
        [2, 3, 4, 5, 6]
    ])

a[b]

However, b doesn't have to generated by hand in this way. It can be more dynamic with

b = np.arange(5) + np.arange(3)[:, None]

Solution 5

In the general case you have to do some sort of iteration - and concatenation - either when constructing the indexes or when collecting the results. It's only when the slicing pattern is itself regular that you can use a generalized slicing via as_strided.

The accepted answer constructs an indexing array, one row per slice. So that is iterating over the slices, and arange itself is a (fast) iteration. And np.array concatenates them on a new axis (np.stack generalizes this).

In [264]: np.array([np.arange(0,5), np.arange(1,6), np.arange(2,7)])
Out[264]: 
array([[0, 1, 2, 3, 4],
       [1, 2, 3, 4, 5],
       [2, 3, 4, 5, 6]])

indexing_tricks convenience methods to do the same thing:

In [265]: np.r_[0:5, 1:6, 2:7]
Out[265]: array([0, 1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6])

This takes the slicing notation, expands it with arange and concatenates. It even lets me expand and concatenate into 2d

In [269]: np.r_['0,2',0:5, 1:6, 2:7]
Out[269]: 
array([[0, 1, 2, 3, 4],
       [1, 2, 3, 4, 5],
       [2, 3, 4, 5, 6]])

In [270]: data=np.array(list('abcdefghijk'))
In [272]: data[np.r_['0,2',0:5, 1:6, 2:7]]
Out[272]: 
array([['a', 'b', 'c', 'd', 'e'],
       ['b', 'c', 'd', 'e', 'f'],
       ['c', 'd', 'e', 'f', 'g']], 
      dtype='<U1')
In [273]: data[np.r_[0:5, 1:6, 2:7]]
Out[273]: 
array(['a', 'b', 'c', 'd', 'e', 'b', 'c', 'd', 'e', 'f', 'c', 'd', 'e',
       'f', 'g'], 
      dtype='<U1')

Concatenating results after indexing also works.

In [274]: np.stack([data[0:5],data[1:6],data[2:7]])

My memory from other SO questions is that relative timings are in the same order of magnitude. It may vary for example with the number of slices versus their length. Overall the number of values that have to be copied from source to target will be the same.

If the slices vary in length, you'd have to use the flat indexing.

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Puchatek
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Puchatek

Software Developer at レキサス

Updated on February 05, 2022

Comments

  • Puchatek
    Puchatek about 2 years

    I'm looking for a way to select multiple slices from a numpy array at once. Say we have a 1D data array and want to extract three portions of it like below:

    data_extractions = []
    
    for start_index in range(0, 3):
        data_extractions.append(data[start_index: start_index + 5])
    

    Afterwards data_extractions will be:

    data_extractions = [
        data[0:5],
        data[1:6],
        data[2:7]
    ]
    

    Is there any way to perform above operation without the for loop? Some sort of indexing scheme in numpy that would let me select multiple slices from an array and return them as that many arrays, say in an n+1 dimensional array?


    I thought maybe I can replicate my data and then select a span from each row, but code below throws an IndexError

    replicated_data = np.vstack([data] * 3)
    data_extractions = replicated_data[[range(3)], [slice(0, 5), slice(1, 6), slice(2, 7)]
    
    • Divakar
      Divakar about 7 years
      What's n there?
    • Paul Panzer
      Paul Panzer about 7 years
      stride_tricks might be a way
    • Puchatek
      Puchatek about 7 years
      @Divakar - dimension. I gave a 1D example for simplicity, but need a generic solution (my real problem is 4D).
  • Puchatek
    Puchatek about 7 years
    That doesn't avoid a for loop ;)
  • Anant Gupta
    Anant Gupta about 7 years
    I agree :) but not a native for loop :)
  • Puchatek
    Puchatek about 7 years
    Nice, I didn't know about np.lib.stride_tricks.as_strided, thank you, Paul.
  • Puchatek
    Puchatek about 7 years
    Damn it, I tried above approach but with indexes as list of ranges as well as list of slices and these would cause IndexErrors. Didn't realize I need to wrap outer list of indexes in a numpy array ^^
  • Paul Panzer
    Paul Panzer about 7 years
    @Puchatek glad to be of help. Just be careful with that stuff. As far as I know it doesn't check ranges, so it will happily allow you to access out-of-range memory etc.
  • Puchatek
    Puchatek about 7 years
    Yep, toyed around with it in Ipython and realized quickly it can blow in my face when used carelessly ^^
  • Divakar
    Divakar about 7 years
    @Puchatek If you are using proper shapes and strides, should be okay.
  • Puchatek
    Puchatek about 7 years
    So I though about this approach but couldn't get it to work because didn't wrap list of lists making indices into a numpy array. Silly me I guess.
  • tmrlvi
    tmrlvi about 7 years
    I think when you put a list into the numpy selector, it tries to filter per axis (i.e., the first item is filter for the first axis etc). Actually, putting it inside another list, as in indexes = [[np.arange(0,5), np.arange(1,6), np.arange(2,7)]] solves it.
  • Puchatek
    Puchatek about 7 years
    Luckily I have to deal with slicing according to a regular pattern. Thank you for the detailed answer :)