How to sort multidimensional array by column?

Question

Is there a way to use the sort() method or any other method to sort a list by column? Lets say I have the list:

[
[John,2],
[Jim,9],
[Jason,1]
]

And I wanted to sort it so that it would look like this:

[
[Jason,1],
[John,2],
[Jim,9],
]

What would be the best approach to do this?

Edit:

Right now I am running into an index out of range error. I have a 2 dimensional array that is lets say 1000 rows b 3 columns. I want to sort it based on the third column. Is this the right code for that?

sorted_list = sorted(list_not_sorted, key=lambda x:x[2])

Asked By: Web Hopeful

||

Source

Answer 1

You can use the sorted method with a key.

sorted(a, key=lambda x : x[1])

Answered By: squiguy

Answer 2

Yes. The sorted built-in accepts a key argument:

sorted(li,key=lambda x: x[1])
Out[31]: [['Jason', 1], ['John', 2], ['Jim', 9]]

note that sorted returns a new list. If you want to sort in-place, use the .sort method of your list (which also, conveniently, accepts a key argument).

or alternatively,

from operator import itemgetter
sorted(li,key=itemgetter(1))
Out[33]: [['Jason', 1], ['John', 2], ['Jim', 9]]

Read more on the python wiki.

Answered By: roippi

Answer 3

You can use list.sort with its optional key parameter and a lambda expression:

>>> lst = [
...     ['John',2],
...     ['Jim',9],
...     ['Jason',1]
... ]
>>> lst.sort(key=lambda x:x[1])
>>> lst
[['Jason', 1], ['John', 2], ['Jim', 9]]
>>>

This will sort the list in-place.

Note that for large lists, it will be faster to use operator.itemgetter instead of a lambda:

>>> from operator import itemgetter
>>> lst = [
...     ['John',2],
...     ['Jim',9],
...     ['Jason',1]
... ]
>>> lst.sort(key=itemgetter(1))
>>> lst
[['Jason', 1], ['John', 2], ['Jim', 9]]
>>>

Answered By: user2555451

Answer 4

The optional key parameter to sort/sorted is a function. The function is called for each item and the return values determine the ordering of the sort

>>> lst = [['John', 2], ['Jim', 9], ['Jason', 1]]
>>> def my_key_func(item):
...     print("The key for {} is {}".format(item, item[1]))
...     return item[1]
... 
>>> sorted(lst, key=my_key_func)
The key for ['John', 2] is 2
The key for ['Jim', 9] is 9
The key for ['Jason', 1] is 1
[['Jason', 1], ['John', 2], ['Jim', 9]]

taking the print out of the function leaves

>>> def my_key_func(item):
...     return item[1]

This function is simple enough to write “inline” as a lambda function

>>> sorted(lst, key=lambda item: item[1])
[['Jason', 1], ['John', 2], ['Jim', 9]]

Answered By: John La Rooy

Answer 5

sorted(list, key=lambda x: x[1])

Note: this works on time variable too.

Answered By: kevin

Answer 6

below solution worked for me in case of required number is float.
Solution:

table=sorted(table,key=lambda x: float(x[5]))
for row in table[:]:
    Ntable.add_row(row)

‘

Answered By: jatin Goyal

How to sort multidimensional array by column?

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