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

##
Answer #1:

`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], ...]
```

##
Answer #2:

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.

##
Answer #3:

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)
```

##
Answer #4:

```
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
```