Selecting multiple slices from a numpy array at once

问题:

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)] 

回答1:

You can use the the indexes to select you 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] == s[1:6])  True 

回答2:

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 

回答3:

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] 

回答4:

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]]]) 

回答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.

回答6:

We can use list comprehension for this

data=np.array([1,2,3,4,5,6,7,8,9,10]) data_extractions=[data[b:b+5] for b in [1,2,3,4,5]] data_extractions 

Results

[array([2, 3, 4, 5, 6]), array([3, 4, 5, 6, 7]), array([4, 5, 6, 7, 8]), array([5, 6, 7, 8, 9]), array([ 6,  7,  8,  9, 10])] 

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