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])

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]) — Web Hopeful
– Web Hopeful, Commented Nov 25, 2013 at 1:11
In response to your edit, since lists are zero indexed, yes x[2] is the third column. The moral of the story is, you can use a key and lambda or an actual function to sort by some stipulation in the sorted and sort functions. — squiguy
– squiguy, Commented Nov 25, 2013 at 2:51

roippi · Accepted Answer · 2013-11-25 01:07:19Z

80

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]]

4 Comments

user2555451 Over a year ago

Might want to mention that this will return a new list.

user2357112 Over a year ago

Indeed. If you want to modify the original list, that would be li.sort(key=whatever).

Abhishek Jain Over a year ago

And how can I get the natural sorting? As in here sorted sorts based on the digit place and '411' will be sorted first then '67'.

bobobobo Over a year ago

@AbhishekJain The type of your data is string, which is why "411" is coming before "67". Convert to number

squiguy · Accepted Answer · 2013-11-25 00:41:24Z

14

You can use the sorted method with a key.

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

answered Nov 25, 2013 at 0:41

squiguy

33.7k8 gold badges63 silver badges67 bronze badges

Comments

score 12 · Accepted Answer · 2013-11-25 01:03:34Z

12

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]]
>>>

edited Nov 25, 2013 at 1:03

answered Nov 25, 2013 at 0:44

user2555451

4 Comments

Web Hopeful Over a year ago

What exactly is the "lambda" key?

user2555451 Over a year ago

@user3024130 - The lambda creates an inline function for the key parameter. I added a link to explain better. Using a lambda would be no different than doing def func(x): return x[1] and then lst.sort(key=func).

Web Hopeful Over a year ago

Okay that makes sense. How would you sort it from highest to lowest instead of lowest to highest?

user2555451 Over a year ago

@user3024130 - Simple. Use this: lst.sort(key=lambda x:x[1], reverse=True). list.sort also takes an optional reverse parameter. If it is set to True, the list is sorted highest to lowest. Otherwise, it is lowest to highest.

kevin · Accepted Answer · 2016-07-30 06:19:44Z

3

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

Note: this works on time variable too.

answered Jul 30, 2016 at 6:19

kevin

1,1771 gold badge13 silver badges17 bronze badges

Comments

John La Rooy · Accepted Answer · 2013-11-25 00:58:49Z

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]]

jatin Goyal · Accepted Answer · 2020-06-27 04:27:05Z

0

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 Jun 27, 2020 at 4:27

jatin Goyal

856 bronze badges

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