Sort multidimensional array based on 2nd element of the subarray
PythonArraysListSortingMultidimensional ArrayPython Problem Overview
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], ... ]
Python Solutions
Solution 1 - Python
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], ...]
Solution 2 - Python
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.
Solution 3 - Python
Do this:
Sort the multi-dimensional array in descending order on the basis of 2nd column:
list_name.sort(key=lambda x:x[1],reverse=True)
Solution 4 - Python
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