One line python code to return all sublists of list containing specific substring in a particluar column

Question

I want to return all the sublists of lists which contains specific substring in a particular column:

For eg:

List=[["2006ab","2005ac"],["2005ab","2004ac"],["2006ab","2005ac"],["2006ab","2003ac"],["2006ab","2005ac"]]

Search Criteria : Return all sublists which contains substring 2005 at the 2nd index .

Output :

[["2006ab","2005ac"],["2006ab","2005ac"],["2006ab","2005ac"]]

I tried using :

  matching = [s for s in List if "2005" in s[1]]

but it returns:

[["2006ab","2005ac"],["2005ab","2004ac"],["2006ab","2005ac"],["2006ab","2005ac"]]

Answer 1

Your list comprehension approach is good and gives the correct result. Are you sure your code is the same from which you pasted the output because it works for me:

>>> List=[["2006ab","2005ac"],["2005ab","2004ac"],["2006ab","2005ac"],["2006ab","2003ac"],["2006ab","2005ac"]]
>>> [sublist for sublist in List if '2005' in sublist[1]]
[['2006ab', '2005ac'], ['2006ab', '2005ac'], ['2006ab', '2005ac']]

which is same as what you desire.

If you are looking for an alternative, you may use filter()

>>> filter(lambda x: '2005' in x[1], List)
[['2006ab', '2005ac'], ['2006ab', '2005ac'], ['2006ab', '2005ac']]