Understanding lambda's sort function in python

Question

I am trying to sort the list of lists in python. I have written the following code:

def sort(intervals):
    if intervals == [] : return []  
    intervals.sort(key = lambda x:x.start)
    return intervals

a = [[1,3],[8,10],[15,18],[2,6]]
print(sort(a))

I am getting the following error:

AttributeError: 'list' object has no attribute 'start'

Please can someone explain lambda function for sort and some details about the above error. Thank you!!

Good answers to the question have been posted below. I'd also like to note that the line `if intervals == [] : return return [] ` is unnecessary because `[].sort() == []` so the empty list will be returned even without that line. — Simon, Jul 08 '15 at 23:41

Iron Fist · Accepted Answer · 2015-07-09T00:28:39.127

The reason for your error message is that you are sorting based on an attribute that is not for list (start is not an attribute of a list'), so quick fix is, either to use the sort method of list or use built-in method sorted:

1 - Using sort method of list:

intervals.sort(key = lambda l:l[0])

2 - Using built-in method sorted:

intervals = sorted(intervals, key=lambda l:l[0])

Reading more about sorting list in this wiki post, very interesting.

Sait · Answer 2 · 2015-07-08T23:37:00.603

3

You should use:

intervals.sort(key = lambda x:x[0])

lambda is a fast-way of making functions. For example,

def getFirst(x):
   return x[0]

is equal to:

getFirst = lambda x: x[0]

I guess you should read the official documentation.

PS: Be aware that you are making in place sorting. You can also use sorted(a, key=lambda x:x[0]) which returns another copy of sorted array, if you want otherwise.

edited Jul 08 '15 at 23:37

answered Jul 08 '15 at 23:29

Sait

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score 1 · Answer 3 · edited May 23 '17 at 12:04

1

The use of lambda creates an anonymous function where x is the parameter. Key is basically assigning the result from this function as the basis for the sort. A much more in depth answer can be found here

But it looks like you are looking for something like this:

intervals.sort(key = lambda x:x[0])

Where x[0] is obviously the first element of each list.

edited May 23 '17 at 12:04

Community

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answered Jul 08 '15 at 23:34

user3636636

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score 0 · Answer 4 · answered Jul 08 '15 at 23:38

It isn't actually necessary to use a lambda function to sort a list of lists. Take the following example:

a = [[1,3],[8,10],[15,18],[2,6],[2,5]]
print (sorted(a))

... this gives the output:

[[1, 3], [2, 5], [2, 6], [8, 10], [15, 18]]

... thus it not only sorts on the first element of each sub-list but uses the subsequent elements as keys if the first elements of two sublists are equal.

Understanding lambda's sort function in python

4 Answers4