python array manipulation and rearranging

Question

I am trying to manipulate an array and perform operations. My input is an array

a= [['f'  '0'  '1'  '0' '1' '0']
    ['o'  '0'  '0'  '1' '0' '0'] 
    ['o'  '0'  '1'  '0' '1' '1']
    ['!b' '1'  '0'  '1' '0' '0']
    ['a'  '0'  '0'  '1' '0' '0'] 
    ['r'  '0'  '1'  '0' '1' '1']]

If I take the first row, my output should just be the columns in which 1 is present. Similarly, I do for each row and get output. So my output should be an array.

output = [['f' '1' '1' 
           'o' '0' '0'  
           'o' '1' '1' 
          '!b' '0' '0' 
           'a' '0' '0'  
           'r' '1' '1' ]
          ['f' '0' 
           'o' '1'  
           'o' '0' 
          '!b' '1' 
           'a' '1'   
           'r' '0' ]
          ['f' '1' '1' '0'
           'o' '0' '0' '0'
           'o' '1' '1' '1'
          '!b' '0' '0' '0'
           'a' '0' '0' '0'
           'r' '1' '1' '1']
          ['f' '0' '0' 
           'o' '0' '1'  
           'o' '0' '0' 
          '!b' '1' '1' 
           'a' '0' '1'  
           'r' '0' '0' ]
          ['f' '0' 
           'o' '1'  
           'o' '0'  
          '!b' '1'  
           'a' '1'  
           'r' '0' ]
          ['f' '1' '1' '0'
           'o' '0' '0' '0'
           'o' '1' '1' '1'
          '!b' '0' '0' '0'
           'a' '0' '0' '0'
           'r' '1' '1' '1']]

Here's my Code

          output = [] 
           for i in a:
               for j in i:
                   if j == 1:
                      output = a[0:]
                      output.append([n][j]) for n in len(i)
                    else:
                         pass

You'll need to give more information about how the output is derived from the input. — Daniel Roseman
– Daniel Roseman, Commented Nov 23, 2015 at 9:13
Actually it all there. For each row, produce a matrix that has only the columns that have a 1 in this row. — Dan D.
– Dan D., Commented Nov 23, 2015 at 9:17
I understand your question but you need to show what you have tried. — Akshay Hazari
– Akshay Hazari, Commented Nov 23, 2015 at 9:22
@AkshayHazari sorry, I didnt give the code cause I am not well acquainted with array manipulation — Kristy
– Kristy, Commented Nov 23, 2015 at 10:06
@Kristy It is fine about not being acquainted but still some amount of effort needs to be put before asking. Even if you didn't ask, I am not the person who down voted. — Akshay Hazari
– Akshay Hazari, Commented Nov 23, 2015 at 10:11

Dan D. · Accepted Answer · 2015-11-23 09:49:32Z

For each row, produce a matrix that has only the columns that have a '1' in this row.

import numpy as np

a = np.array([['f', '0', '1',  '0', '1', '0'],
     ['o', '0', '0',  '1', '0', '0'], 
     ['o', '0', '1',  '0', '1', '1'],
    ['!b', '1', '0',  '1', '0', '0'],
     ['a', '0', '0',  '1', '0', '0'], 
     ['r', '0', '1',  '0', '1', '1']])

l = []
for r in a:
   l.append(a[:, [i for i, c in enumerate(r) if i == 0 or c == '1']])
print l

This works but perhaps someone more familiar with numpy might be able to do better.

Produces:

[array([['f', '1', '1'],
       ['o', '0', '0'],
       ['o', '1', '1'],
       ['!b', '0', '0'],
       ['a', '0', '0'],
       ['r', '1', '1']], 
      dtype='|S2'), array([['f', '0'],
       ['o', '1'],
       ['o', '0'],
       ['!b', '1'],
       ['a', '1'],
       ['r', '0']], 
      dtype='|S2'), array([['f', '1', '1', '0'],
       ['o', '0', '0', '0'],
       ['o', '1', '1', '1'],
       ['!b', '0', '0', '0'],
       ['a', '0', '0', '0'],
       ['r', '1', '1', '1']], 
      dtype='|S2'), array([['f', '0', '0'],
       ['o', '0', '1'],
       ['o', '0', '0'],
       ['!b', '1', '1'],
       ['a', '0', '1'],
       ['r', '0', '0']], 
      dtype='|S2'), array([['f', '0'],
       ['o', '1'],
       ['o', '0'],
       ['!b', '1'],
       ['a', '1'],
       ['r', '0']], 
      dtype='|S2'), array([['f', '1', '1', '0'],
       ['o', '0', '0', '0'],
       ['o', '1', '1', '1'],
       ['!b', '0', '0', '0'],
       ['a', '0', '0', '0'],
       ['r', '1', '1', '1']], 
      dtype='|S2')]

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