Hi William,

 

You can simply use a for loop for that task:

 

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In [1]: import numpy as np

 

In [2]: a = np.arange(3).reshape((1, 3))

 

In [3]: for x in a.T:

   ...:     print(x)

   ...:

[0]

[1]

[2]

 

Best regards,

Hameer Abbasi

 

From: NumPy-Discussion <numpy-discussion-bounces+einstein.edison=gmail.com@python.org> on behalf of William Ayd <william.ayd@icloud.com>
Reply to: Discussion of Numerical Python <numpy-discussion@python.org>
Date: Tuesday, 19. May 2020 at 01:42
To: "numpy-discussion@python.org" <numpy-discussion@python.org>
Subject: Re: [Numpy-discussion] Using nditer + external_loop to Always Iterate by Column

 

I am trying to use the nditer to traverse each column of a 2D array, returning the column as a 1D array. Consulting the docs, I found this example which works perfectly fine:

 

In [65]: a = np.arange(6).reshape(2,3)                                                                                                                             

 

In [66]: for x in np.nditer(a, flags=['external_loop'], order='F'): 

    ...:     print(x, end=' '

    ...:                                                                                                                                                           

[0 3] [1 4] [2 5] 

 

When changing the shape of the input array to (1, 3) however, this doesnít yield what I am hoping for any more (essentially [0], [1] [2]):

 

In [68]: for x in np.nditer(a, flags=['external_loop'], order='F'): 

    ...:     print(x, end=' '

    ...:                                                                                                                                                           

[0 1 2] 

 

I suspect this may have to do with the fact that the (1, 3) array is both C and F contiguous, and it is trying to return as large of a 1D F-contiguous array as it can. However, I didnít see any way to really force it to go by columns. My best guess was the itershape argument though I couldnít figure out how to get that to work and didnít see much in the documentation.

 

Thanks in advance for the help!

 

- Will