How do I sort a zipped list in Python?

Question

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)].

Asked By: rectangletangle

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Source

Answer 1

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

Answered By: Owen

Answer 2

Quite simple:

sorted(zipped, key=lambda x: x[1])

Answered By: Ulrich Dangel

Answer 3

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.

Answered By: utdemir

Answer 4

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)]

Answered By: zeekay

Answer 5

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)]

Answered By: MSeifert

Answer 6

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)

Answered By: Max Kleiner

How do I sort a zipped list in Python?

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