How do I sort a zipped list in Python?

39
votes

What's the Pythonic way to sort a zipped list?

code :

names = list('datx')
vals  = reversed(list(xrange(len(names))))
zipped = zip(names, vals)

print zipped

The code above prints [('d', 3), ('a', 2), ('t', 1), ('x', 0)]

I want to sort zipped by the values. So ideally it would end up looking like this [('x', 0), ('t', 1), ('a', 2), ('d', 3)].

6
pythonlistsortingzip
The first two answers show off a pet peeve of mine about Python: the presence of both sorted and sort. - JasonFruit
There's good reason for both: .sort() sorts a list in-place. And sorted works on any iterator, but needs to use additional storage to accomplish the task. - Ned Batchelder
@JasonFruit: If you don't know the difference, I highly recommend you find out. - Matt Joiner
@Matt Joiner, Ned Batchelder: I know the difference; I think I've not made myself clear. What bugs me is that sort() is a method, and sorted is a built-in function. It doesn't seem consistent. - JasonFruit
@JasonFruit (I know this is from a while back), but sorted is consistent with other special python syntax, including in, len, and reversed, which depend upon the __contains__, __len__, and __getitem__ + __len__ respectively (I think sorted needs __getitem__ and __len__ but I'm not sure). In many ways, it's also similar to the syntax for [] which is based on __setitem__ and __getitem__, or () which initializes __call__. They're builtin functions that translate special internal functions into clear external syntax. - Jeff Tratner

6 Answers

38
votes
sorted(zipped, key = lambda t: t[1])
61
votes

Quite simple:

sorted(zipped, key=lambda x: x[1])
9
votes
import operator
sorted(zipped, key=operator.itemgetter(1))

If you want it a little bit more faster, do ig = operator.itemgetter(1) and use ig as key function.

4
votes

It's simpler and more efficient to zip them in order in the first place (if you can). Given your example it's pretty easy:

>>> names = 'datx'
>>> zip(reversed(names), xrange(len(names)))
<<< [('x', 0), ('t', 1), ('a', 2), ('d', 3)]
4
votes

In your case you don't need to sort at all because you just want an enumerated reversed list of your names:

>>> list(enumerate(names[::-1]))      # reverse by slicing
[(0, 'x'), (1, 't'), (2, 'a'), (3, 'd')]

>>> list(enumerate(reversed(names)))  # but reversed is also possible
[(0, 'x'), (1, 't'), (2, 'a'), (3, 'd')]

But if you need to sort it then you should use sorted (as provided by @utdemir or @Ulrich Dangel) because it will work on Python2 (zip and itertools.zip) and Python3 (zip) and won't fail with an AttributeError like .sort(...) (which only works on Python2 zip because there zip returns a list):

>>> # Fails with Python 3's zip:
>>> zipped = zip(names, vals)
>>> zipped.sort(lambda x: x[1])
AttributeError: 'zip' object has no attribute 'sort'

>>> # Fails with Python 2's itertools izip:
>>> from itertools import izip
>>> zipped = izip(names, vals)
>>> zipped.sort(lambda x: x[1])
AttributeError: 'itertools.izip' object has no attribute 'sort'

But sorted does work in each case:

>>> zipped = izip(names, vals)
>>> sorted(zipped, key=lambda x: x[1])
[('x', 0), ('t', 1), ('a', 2), ('d', 3)]

>>> zipped = zip(names, vals)  # python 3
>>> sorted(zipped, key=lambda x: x[1])
[('x', 0), ('t', 1), ('a', 2), ('d', 3)]
0
votes

Sort feature importance in a classifier (dtc=decision_tree):

for name, importance in sorted(zip(X_train.columns, 
                dtc.feature_importances_),key=lambda x: x[1]):
    print(name, importance)