# Python Sort Multidimensional Array Based on 2nd Element of Subarray

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### Question :

Python Sort Multidimensional Array Based on 2nd Element of Subarray

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], ... ]
``````

`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], ...]
``````

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.

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)
``````

``````x= [[8, 9, 7],