How to iterate through a list of lists in python?

Question

I have a list of lists like this.

documents = [['Human machine interface for lab abc computer applications','4'],
             ['A survey of user opinion of computer system response time','3'],
             ['The EPS user interface management system','2']]

Now i need to iterate through the above list and output a list of strings, as shown below (without the numbers in the original list)

documents = ['Human machine interface for lab abc computer applications',
             'A survey of user opinion of computer system response time',
             'The EPS user interface management system']

Does this answer your question? [Iterating through list of list in Python](https://stackoverflow.com/questions/6340351/iterating-through-list-of-list-in-python) — Jim G., May 04 '21 at 16:00

machine yearning · Accepted Answer · 2012-03-14T00:01:23.283

35

The simplest solution for doing exactly what you specified is:

documents = [sub_list[0] for sub_list in documents]

This is basically equivalent to the iterative version:

temp = []
for sub_list in documents:
    temp.append(sub_list[0])
documents = temp

This is however not really a general way of iterating through a multidimensional list with an arbitrary number of dimensions, since nested list comprehensions / nested for loops can get ugly; however you should be safe doing it for 2 or 3-d lists.

If you do decide to you need to flatten more than 3 dimensions, I'd recommend implementing a recursive traversal function which flattens all non-flat layers.

edited Mar 14 '12 at 00:01

answered Feb 05 '12 at 17:03

machine yearning

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it just brings down the 1st argument of each sub-array, is not what's being asked in the question! – Anu Nov 24 '18 at 21:05
1

could have simply unpacked the container i.e a list( list of lists!). `*documents`, then you can simply iterate(skip) using range() – Anu Nov 24 '18 at 21:12

score 8 · Answer 2 · answered Feb 05 '12 at 17:14

If you want to simply iterate over the loop and do things with the elements (rather than the specific results requested in the question), you could use a basic for loop

for row in documents:
  #do stuff with the row
  print(row)

  for column in row:
    #do stuff with the columns for a particular row
    print(column)

  if(row[1] > 10):
    print('The value is much too large!!')

This is a language feature known as "flow control".

Note that if you only want the result given in the question, a list comprehension like machine yearning provided is the best way to do it.

documents = [doc[0] for doc in documents]

Note that it discards your original documents list (you are overwriting the original variable) so use the following if you want to have a copy of the first column as well as a copy of your original list:

document_first_row = [doc[0] for doc in documents]

score 5 · Answer 3 · answered Feb 06 '12 at 00:40

As explained in http://docs.python.org/library/operator.html#operator.itemgetter, You can also try with

from operator import itemgetter
documents = map(itemgetter(0), documents)

that should be faster than using an explicit loop.

score 1 · Answer 4 · answered May 12 '17 at 14:56

1

The question is dead but still knowing one more way doesn't hurt:

 documents = [['Human machine interface for lab abc computer applications','4'],
         ['A survey of user opinion of computer system response time','3'],
         ['The EPS user interface management system','2']]

document = []
for first,*remaining in documents:
    document.append(first)

print(document)
['Human machine interface for lab abc computer applications',
 'A survey of user opinion of computer system response time', 
 'The EPS user interface management system'
]

answered May 12 '17 at 14:56

Vaibhav Singh

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6

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it just brings down the 1st argument of each sub-array, is not what's being asked in the question! – Anu Nov 24 '18 at 21:03
1

I don't understand, the user wants a list of only strings as the last output from a list of lists. Can you tell me how did I not understand the question? – Vaibhav Singh Dec 26 '18 at 15:25

KobeJohn · Answer 5 · 2012-02-05T17:31:50.310

1

**edit. thanks DSM. This is wrong as it just flattens the lists. I didn't notice the extra data inside the list after the text that OP wants to ignore.

Ok I'll make it really easy for you!

itertools.chain.from_iterable(documents)

As others have said, it depends on what final behavior your need. So if you need something more complex than that, use recursive traversal or if you are like me, use an iterative traversal. I can help you with that if you need it.

edited Feb 05 '12 at 17:31

answered Feb 05 '12 at 17:21

KobeJohn

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This will flatten the list, not extract the first elements. – DSM Feb 05 '12 at 17:22
indeed. I misread the behavior he wants. offer for help with traversal is still open – KobeJohn Feb 05 '12 at 17:30

score 0 · Answer 6 · answered Jul 23 '13 at 23:34

You can also use zip with argument unpacking to transform a list of "rows" into a list of columns:

rows=[[1,'a','foo'],
      [2,'b','bar'],
      [3,'c','baz']]

columns=zip(*rows)
print columns
#[(1,2,3),
# ('a','b','c'),
# ('foo','bar','baz')]
print columns[0]
#(1,2,3)

the * operator passes all the rows in as seperate arguments to zip

zip(*rows) == zip(row1,row2,row3,...)

zip takes all the rows and assembles columns with one item from each list

score 0 · Answer 7 · answered Feb 15 '14 at 22:23

0

you could use numpy array

for instance

document = [['the quick brown fox', '2' ],['jumped over the lazy fox ','3']]

#

import numpy as np 
document = np.array(document)
document=document[:,0]

answered Feb 15 '14 at 22:23

K Adamu

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How to iterate through a list of lists in python?

7 Answers7

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