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I have an array inside an array. Normal array indexing fails to work in this case . One way i can solve this problem is that I can convert 'a' into list and then again convert it into an array --a two step process but i want to know is there any other way to slice it so that the structure of 'a' remains unchanged ? 

a=np.array([[np.arange(3)],[np.arange(6,9)],[np.arange(11,14)]])
array([[[ 0,  1,  2]],

       [[ 6,  7,  8]],

       [[11, 12, 13]]])

a.shape
(3L, 1L, 3L)

a.ndim
3

type(a)
numpy.ndarray

Please help me on this.

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2
  • What makes you think that normal indexing fails, and what exactly do you mean when saying that? Commented Mar 8, 2016 at 12:57
  • if i use a=np.array([np.arange(3),np.arange(6,9),np.arange(11,14)]) instead of my above code then this becomes case of 2d array whose indexing i refer to normal in this case. Commented Mar 8, 2016 at 13:37

2 Answers 2

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Perhaps you have an unnecessary level of nesting there, in case you can initialize your array like this

a=np.array([np.arange(3),np.arange(6,9),np.arange(11,14)])

then you can access its elements with standard notation

a[1,1]

If the level of nesting is not incorrect, then you can still access a's elements like this:

a[1,0,1]

Basically, the problem is that you are wrapping the second-level array inside another array of dimension 1, so when you access its elements you need to keep in mind that.

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6 Comments

Yes i know the other way around of writing but i am just wondering how can one deal with if this happened .
If you look at the result of 'a.shape', you are given the exact dimensions associated with your n-dimensional array. If you don't know in advance the dimension of your array, just query it and use dynamic evaluation (or some other trick). It's quite slower, though.
Can you mention how you indexed 'a' in the above answer more clearly . I mean where do these numbers pointing at and in what order ? like a[rowno,columnno] etc we do in 2d array .Thanks . and if you have any material which you thing is helpful to me for better understanding 3d array please provide me the link.
a's shape is (3, 1, 3), that means that to index any of its elements you need 3 indexes. The first must be in [0,2] because the first dimension is 3, the second must be within [0,0] because the second dimension is 1 and the third must be within [0,2] again because the third dimension is 3. If you don't know in advance what the structure of the array is, you can make a recursive function for iterating over each dimension of the array till you get to the final element. Or you can compose the indexes with a loop cycle as a string and then use eval() to access the location. Is it clear?
Yes somewhat it is clear. I have one more question . what these (3,1,3) represent in 3d array as in 2d array (x,y) represent (row no,colno) . Please explain
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Compare your 2 arrays:

In [39]: a=np.array([[np.arange(3)],[np.arange(6,9)],[np.arange(11,14)]])
In [40]: a1=np.array([np.arange(3),np.arange(6,9),np.arange(11,14)])
In [41]: a
Out[41]: 
array([[[ 0,  1,  2]],

       [[ 6,  7,  8]],

       [[11, 12, 13]]])
In [42]: a.shape       
Out[42]: (3, 1, 3)      # 3d
In [43]: a1
Out[43]: 
array([[ 0,  1,  2],
       [ 6,  7,  8],
       [11, 12, 13]])
In [44]: a1.shape
Out[44]: (3, 3)     # 2d

One way to remove the size 1 dimension is with squeeze:

In [45]: np.squeeze(a)
Out[45]: 
array([[ 0,  1,  2],
       [ 6,  7,  8],
       [11, 12, 13]])

or reshape:

In [46]: a.reshape(3,3)
Out[46]: 
array([[ 0,  1,  2],
       [ 6,  7,  8],
       [11, 12, 13]])

You treat the 3 dimensions of a in exactly the same was the 2 of a1:

In [47]: a1[1,:]
Out[47]: array([6, 7, 8])      # index 1 dim of the 2
In [48]: a[1,:]
Out[48]: array([[6, 7, 8]])    # index 1 dim of the 3
In [49]: a[1,0,:]
Out[49]: array([6, 7, 8])      # index 2 dim of the 3
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4 Comments

thanks for putting efforts to answer the question. But i don't want to change the structure of my 3d array .any other way you know to deal with 3d array without changing its dimensions?
Deal in what way? I showed how to index it. Normal array indexing works the same with 3d as with 2.
like patrick mentioned his way without changing the dimension in the above comments. I am looking for better explaination of the patrick comment . Converting it into 2d array do the stuff i know but i am not sure what will happen to existing 3d array when i convert it to 2d array and other properties or calculations as i dont have real time experience with 3d array . so i just want to preserve the dimension of 3d array and not preferring to change its dimension. I am a newbie in numpy so just learning things. guide me as i stuck. thanks
Just play about with these arrays in an interactive session. You don't have to have 'real time experience' with 3d or 4d arrays; the experience is in the computer, not the physical world.

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