Numpy array, fill empty values for a single column

Question

I have an array in numpy, which was generated using np.array() from a python list so my entries are strings, but some of the values are blank. Here is an example array:

['1', '1', '1', '1']
['1', '1', '', '1']
['1', '1', '1', '1']
['1', '', '1', '1']

There is no 'NaN' or 'None', it is blank. I want to be able to fill all the blank cells in a particular column with the same value.

• aIKid: possible duplicate of numpy array initialization (fill with identical values)

Answer 1

You can use numpy.where() to achieve this.

In [8]: arr = numpy.array(['','1','2','3',''])
In [9]: arr[numpy.where(arr=='')] = '0'
In [10]: arr
Out[10]:
array(['0', '1', '2', '3', '0'],
 dtype='|S1')

Edit As @mgilson pointed out, you could just do:

arr[arr==''] = '0'

• mgilson: I don't think np.where is necessary here (is it?). can't you just do arr[arr==''] = '0'?
• bronstad: I want to fill the blank cells of a particular column, not all blank cells. Say I use the code: arr[arr[0::,1]==''] = '0'. This turns any row with a blank in the 2nd column into all 0 entries, not just the blank.
• qmorgan: You need to index the first data as well. Try: data[:,1][data[:,1]==''] = '0'
• bronstad: Awesome, that worked. Thank you, I tried using numpy.where() before but didn't use the index.