Python output each iteration of loop to a row of an array

Question

I am pretty sure that there is a simple answer to this but I am completely stuck!

I have a list of lists of various numbers of words, and I am trying to see whether I can find these words in a text file. So if the list of words I want to find is:

stem=[[u'travail',u'electr'],[u'la',u'vou']]

Then I am looping through each word of each "row" of the stem list, and trying to find that word in a text file. This code returns the index of the matched position of a word.

for row in stem: 
       for j,i in enumerate(row):
           y=[match.start() for match in re.finditer(re.escape(i), lowe)]
              print y

output:

[669, 2102, 5810]
   [1452, 2120, 5628]
   [1582, 2912, 3109, 5711]
   [605, 761, 882, 948, 1126, 1132, 1357, 1646, 1936, 2011, 2765, 3286, 3316, 3512, 3821, 3839, 3879, 4012, 4052,   4159, 4417, 4457, 4492, 4699, 4813, 4850, 4921, 4966, 4991, 4998, 5008, 5046, 5118, 5201, 5359, 5506, 5680]

How do I get an output so it is like

 [[[669, 2102, 5810], [1452, 2120, 5628]], [[1582, 2912, 3109, 5711], [605, 761, 882, 948, 1126, 1132, 1357, 1646, 1936, 2011, 2765, 3286, 3316, 3512, 3821, 3839, 3879, 4012, 4052, 4159, 4417, 4457, 4492, 4699, 4813, 4850, 4921, 4966, 4991, 4998, 5008, 5046, 5118, 5201, 5359, 5506, 5680]]]

So that the output for each row is in its own list?Thank you!!

Also, what is the point of your enumerate? I don't see you use j anywhere, so it seems like you should just be able to do for i in row: (and then I would probably change the variable name i to column or something a little more clear) — mgilson
– mgilson, Commented Jul 24, 2013 at 7:44
You must use a result list where you will append every sublist. — emigue
– emigue, Commented Jul 24, 2013 at 7:45

mgilson · Accepted Answer · 2013-07-24 07:43:15Z

3

If I understand correctly, something like this should do it:

output = []
for row in stem: 
   current = []
   output.append(current)
   for j,i in enumerate(row):
       y=[match.start() for match in re.finditer(re.escape(i), lowe)]
       current.append(y)

print output

answered Jul 24, 2013 at 7:43

mgilson

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