How to use key in the sort function in this situation (Python)?

Question

I have a list like this:

l=[('a',2),('d',0),('c',1),('b',4)]

Now I want to sort it, but by the number, not by the word. If I just use l.sort() it just sorts by alphabetically. My desired output should be sth like (d,0)(c,1)(a,2)(b,4)

I tried l.sort(key = l[0][1]) but it won't work too (l[0][1] actually refer to number 2, which is the second value)

The key argument expects a callable, i.e. a function that can be applied to every element in l in order to determine its rank compared to other elements. — timgeb
– timgeb, Commented Feb 17, 2022 at 14:47

Julia · Accepted Answer · 2022-02-17 14:47:02Z

2

You need to pass a lambda as the key that does the right thing:

>>> l = [("a", 2), ("d", 0), ("c", 1), ("b", 4)]
>>> l.sort(key=lambda val: val[1])
>>> l
[("d", 0), ("c", 1), ("a", 2), ("b", 4)]

answered Feb 17, 2022 at 14:47

Julia

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6 Comments

timgeb Over a year ago

Of course you could also pass operator.itemgetter(1) or def get_1(val): return val[1], does not have to be a lambda.

Quang Hoàng Minh Over a year ago

Thank you. I saw people use this lamda thing in some StackOverflow posts before but I didn't understand it. I'm doing some research on this thing right now. Still, may I ask that after "key", it must be a function, right?

chepner Over a year ago

A callable, not necessarily a function. operator.itemgetter is a class whose instances are callable. For example, itemgetter(1)(x) is equivalent to (lambda y: y[1])(x).

timgeb Over a year ago

@chepner but at OP's level it may be less confusing to call everything that's callable a function. At least in this context.

chepner Over a year ago

I don't think it's ever too early to emphasize the distinction between functions and callables (or the fact that a lambda expression defines a function, rather than being something special).

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