python sort without lambda expressions

Question

I often do sorts in Python using lambda expressions, and although it works fine, I find it not very readable, and was hoping there might be a better way. Here is a typical use case for me.

I have a list of numbers, e.g., x = [12, 101, 4, 56, ...]

I have a separate list of indices: y = range(len(x))

I want to sort y based on the values in x, and I do this:

y.sort(key=lambda a: x[a])

Is there a good way to do this without using lambda?

Answer 1

You can use the __getitem__ method of the list x. This behaves the same as your lambda and will be much faster since it is implemented as a C function instead of a python function:

>>> x = [12, 101, 4, 56]
>>> y = range(len(x))
>>> sorted(y, key=x.__getitem__)
[2, 0, 3, 1]

Answer 2

Not elegantly, but:

[a for (v, a) in sorted((x[a], a) for a in y)]

BTW, you can do this without creating a separate list of indices:

[i for (v, i) in sorted((v, i) for (i, v) in enumerate(x))]

Answer 3

I suppose if I wanted to create another function, I could do it something like this (not tested):

def sortUsingList(indices, values):
    return indices[:].sort(key=lambda a: values[a])

Though I think I prefer to use lambda instead to avoid having to create an extra function.

Answer 4

I'm not sure if this is the kind of alternative you meant, but you could define the key function with a def:

def sort_key(value):
    return x[value]

y.sort(key = sort_key)

Personally, I think this is worse than the lambda as it moves the sort criteria away from the line of code doing the sort and it needlessly adds the sort_key function into your namespace.