I have an array like this:
[['G', 10], ['A', 22], ['S', 1], ['P', 14], ['V', 13], ['T', 7], ['C', 0], ['I', 219]]
I'd like to sort it based on the 2nd element in descending order. An ideal output would be:
[['I', 219], ['A', 22], ['P', 14], ... ]
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4 Answers
list.sort, sorted accept optional key parameter. key function is used to generate comparison key.
>>> sorted(lst, key=lambda x: x[1], reverse=True)
[['I', 219], ['A', 22], ['P', 14], ['V', 13], ['G', 10], ...]
>>> sorted(lst, key=lambda x: -x[1])
[['I', 219], ['A', 22], ['P', 14], ['V', 13], ['G', 10], ...]
>>> import operator
>>> sorted(lst, key=operator.itemgetter(1), reverse=True)
[['I', 219], ['A', 22], ['P', 14], ['V', 13], ['G', 10], ...]
3 Comments
Ishaan
Too late but Can you explain to me what the lambda function does?
falsetru
The return value of the lambda will be used as sort key. About
lambda itself, please read the python documentation: docs.python.org/3/tutorial/…
Antonio SEO
is operator.itemgetter(1) faster than lambda?
Use itemgetter
from operator import itemgetter
a = [[1, 3, 5], [2, 511, 7], [17, 233, 1]]
a = sorted(a, key=itemgetter(1))
Output :
[[1, 3, 5], [17, 233, 1], [2, 511, 7]]
itemgetter can also be used to sort by multiple subarrays.
Comments
Sort the multi-dimensional array in descending order on the basis of 2nd column:
list_name.sort(key=lambda x:x[1],reverse=True)
Comments
x= [[8, 9, 7],
[1, 2, 3],
[5, 4, 3],
[4, 5, 6]]
x.sort(cmp=lambda x,y: cmp(x[0],y[0]))
print x
