Sorting a list (with lambda) with a variable number of arguments

Question

Given:

a = [[1,2,3], [1,2], [1,1,1,1]]

How to sort a in place based on the sum of the values?

I tried using a lambda, but then it errors with a lambda not allowing a variable amount of arguments:

a.sort(lambda x: sum(x))
TypeError: <lambda>() takes exactly 1 arguments (2 given)

The previous was a simplified example; I am actually trying to use another function that takes multiple arguments... this does change the problem.

a.sort(lambda x: len(my_function("123", x)))

Hopefully the answer can be done in base python.

Answer 1

You can do it as:

a = [[1,2,3], [1,2], [1,1,1,1]]

a = sorted(a, key=lambda x: sum(x))

>>> print a
[[1,2], [1,1,1,1], [1,2,3]]

Answer 2

a.sort(key=sum)  # Need the 'key' keyword, and don't bother with lambda here.

is what you want.

Output:

[[1, 2], [1, 1, 1, 1], [1, 2, 3]]

EDIT:

To fix the multiple argument example, just once again specify the key argument:

a.sort(key=lambda arg1, arg2, etc: function(arg1, arg2, etc))

Answer 3

Try this ,

>>> a = [[1,2,3], [1,2], [1,1,1,1]]
>>> print sorted(a, key=sum)
[[1, 2], [1, 1, 1, 1], [1, 2, 3]]
>>>