How can I sort the following based on the 1st element?
list1 = [["Value313",1],["Value421",3],["Value234",2]]
Ultimately, I should get the following:
list1 = [["Value234",2],["Value313",1],["Value421",3]]
3 Answers
The default sort order of lists (lexicographically) is already how you want it.
>>> list1 = [["Value313",1],["Value421",3],["Value234",2]]
>>> list1.sort()
>>> print list1
[['Value234', 2], ['Value313', 1], ['Value421', 3]]
Comments
sorted(list1,key=lambda x : x[0])
[['Value234', 2], ['Value313', 1], ['Value421', 3]]
Use sorted and use a key lambda x[0] where x[0] is the first element in each sublist
Or sort in place to avoid creating a new list:
list1 = [["Value313",1],["Value421",3],["Value234",2]]
list.sort(list1,key=lambda x : x[0])
print list1
[['Value234', 2], ['Value313', 1], ['Value421', 3]]
answered Aug 11, 2014 at 16:05
Padraic Cunningham
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1 Comment
Padraic Cunningham
what exactly was the downvote for?
To sort list1 in place (without creating another list) use:
from operator import itemgetter
list1.sort(key=itemgetter(0))
If you want to create another list and leave list1 unsorted use:
from operator import itemgetter
list2 = sorted(list1, key=itemgetter(0))
You could use a lambda, but operator.itemgetter normally has better performance since it's at C level.
1 Comment
enrico.bacis
Sure I did, for for a list of pair of integers and using
sorted, the itemgetter solution on my (old) machine does 207 µs per loop, the lambda 359 µs per loop
list1.sort()will sort list1 (in place) on the first element as a default