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I am experimenting the different dimensions one can have in an array using ndim().

x=np.arange(0,100,1).reshape(1,20,5)

The shape is:

[[[ 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]]]

After, print x.ndim shows the array dimension is 3

I cannot visualize why the dimension is 3.

How does the shapes of respective arrays look like with dimensions 0,1,2,3,4,5......?

share|improve this question
    
If you had done .reshape(20,5) then there would be 2 dimensions, and if you had done .reshape(1,1,20,5) there would be 4 dimensions, etc. –  Spencer Hill Jan 19 at 18:24
    
Your numpy array had 3 dimensions but the content of the array is only 2 dimensional because one of the array dimensions has unit length. –  moarningsun Jan 19 at 22:16
    
@Spencer putting it simply,, would the number of arguments in the reshape function determine the dimension of the array? –  user3211991 Jan 20 at 10:24
    
Yes, that is it！ –  kattern Jan 20 at 13:22
    
@user3211991 Yes. Just keep in mind that the new shape must be compatible with the total number of elements in the original array, as explained in the reshape documentation. –  Spencer Hill Jan 20 at 14:33

1 Answer 1

up vote 2 down vote accepted

A simply way to count dimension is counting [ in the output. One [ for one dimension. Here you have three [s, therefore you have 3 dimension. Since one of the dimension is 1, you may be mislead. Here is another example:

x=np.arange(0,24,1).reshape(2,2,6)

Then, x is 

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]]])

Now, it is clear that x is a 3 dimension array.

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