[FIXED] Is it possible to have multiple statements in a python lambda expression?

Issue

I have a list of lists:

lst = [[567, 345, 234], [253, 465, 756, 2345], [333, 777, 111, 555]]

I want map lst into another list containing only the second smallest number from each sublist. So the result should be [345, 465, 333].

If I were just interested in the smallest number, I could write this as map(lambda x: min(x), lst). To get the second smallest numbers, I thought of sorting and then indexing the results, like map(lambda x: sort(x)[1], lst); however, sort returns None rather than chaining.

If it were possible to use multiple statements in a lambda, I could write map(lambda x: sort(x); x[1], lst), but this is not allowed.

Can I use map to solve the problem without defining a named function? How?

Solution

There are several different answers I can give here, from your specific question to more general concerns. So from most specific to most general:

Q. Can you put multiple statements in a lambda?

A. No. But you don't actually need to use a lambda. You can put the statements in a def instead. i.e.:

def second_lowest(l):
    l.sort()
    return l[1]

map(second_lowest, lst)

Q. Can you get the second lowest item from a lambda by sorting the list?

A. Yes. As alex's answer points out, sorted() is a version of sort that creates a new list, rather than sorting in-place, and can be chained. Note that this is probably what you should be using - it's bad practice for your map to have side effects on the original list.

Q. How should I get the second lowest item from each list in a sequence of lists?

A. sorted(l)[1] is not actually the best way for this. It has O(N log(N)) complexity, while an O(n) solution exists. This can be found in the heapq module.

>>> import  heapq
>>> l = [5,2,6,8,3,5]
>>> heapq.nsmallest(l, 2)
[2, 3]

So just use:

map(lambda x: heapq.nsmallest(x,2)[1],  list_of_lists)

It's also usually considered clearer to use a list comprehension, which avoids the lambda altogether:

[heapq.nsmallest(x,2)[1] for x in list_of_lists]

Answered By - Brian