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This little python code:

   >>> a=np.random.rand(3,4,5)
   >>> a[0][:][3]
   >>> a[0,:,3]

Produces the results:

array([[[ 0.19080354,  0.45701919,  0.17411363,  0.45117827,  0.10413359],
            [ 0.86430848,  0.81831987,  0.27604238,  0.25587538,  0.72733844],
            [ 0.42065355,  0.63994284,  0.64540483,  0.55639512,  0.4455423 ],
            [ 0.04778727,  0.53506934,  0.79615599,  0.24200543,  0.82332594]],
           [[ 0.36535239,  0.5973006 ,  0.71075267,  0.16814739,  0.26409851],
            [ 0.85557313,  0.54190805,  0.65531428,  0.80448208,  0.54959253],
            [ 0.62112884,  0.9159606 ,  0.10186144,  0.14956198,  0.38026561],
            [ 0.70577261,  0.02682898,  0.04136858,  0.15603152,  0.47125989]],
           [[ 0.72864857,  0.09365008,  0.84137507,  0.43887926,  0.26616441],
            [ 0.31022073,  0.54251517,  0.30635049,  0.36270005,  0.85149399],
            [ 0.39371669,  0.38230285,  0.77115029,  0.22647156,  0.57128166],
            [ 0.54906932,  0.87058929,  0.72157733,  0.79480009,  0.033705  ]]])
array([ 0.04778727,  0.53506934,  0.79615599,  0.24200543,  0.82332594])
array([ 0.45117827,  0.25587538,  0.55639512,  0.24200543])

Why do the last two lines, a[0][:][3] and a[0,:,3], not return the same slice of the array? I expected them both to return what came from a[0,:,3].

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  • 1
    a[0].T[:][3] or easier a[0].T[3] this will give you the same output as a[0,:,3] Commented Jun 8, 2018 at 13:43

3 Answers 3

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Each [] is evaluated separately by the Python interpreter, e.g.

In [117]: a=np.random.rand(3,4,5)
In [118]: a[0]
Out[118]: 
array([[0.98688694, 0.77224477, 0.19871568, 0.00552212, 0.81546143],
       [0.70685734, 0.72900717, 0.77127035, 0.07404465, 0.35846573],
       [0.11586906, 0.86310343, 0.62329813, 0.33089802, 0.06355835],
       [0.31098232, 0.32518332, 0.72960618, 0.63755747, 0.88721274]])
In [119]: _[:]
Out[119]: 
array([[0.98688694, 0.77224477, 0.19871568, 0.00552212, 0.81546143],
       [0.70685734, 0.72900717, 0.77127035, 0.07404465, 0.35846573],
       [0.11586906, 0.86310343, 0.62329813, 0.33089802, 0.06355835],
       [0.31098232, 0.32518332, 0.72960618, 0.63755747, 0.88721274]])
In [120]: _[3]
Out[120]: array([0.31098232, 0.32518332, 0.72960618, 0.63755747, 0.88721274])

Making the trailing slices explicit (for clarity to us humans):

In [121]: a[0,:,:][:,:][3,:]
Out[121]: array([0.31098232, 0.32518332, 0.72960618, 0.63755747, 0.88721274])

The middle [:] does not select a dimension from the original a. It operates on the result of a[0,:,:], and does nothing (except make a new array of the same shape and data). The last [3] does not select from the third dimension of a - it selects from the first dimension of the array it got from the [:] step. Note it returns a shape (5,) array, the size of the last dimension of a. a[0,3,:] produces the same thing.

This on the other hand is handled entirely by the numpy indexing, and deals with all 3 dimensions at once:

In [122]: a[0,:,3]
Out[122]: array([0.00552212, 0.07404465, 0.33089802, 0.63755747])

This returns a shape (4,), the middle dimension of a. a[0,:,:][:,3] produces the same thing.

The key point is that numpy operates within the Python interpreter; it does not change Python syntax.

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2

a[0][:][3] says: take 0th (sub)array, third row, and is equivalent to a[0][3]

a[0,:,3] says: take 0th (sub)array, all rows, third element (column)

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That's because [:] doesn't make any change to the array, so a[0][:][3] is same as a[0][3] which equals to a[0, 3, :]

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