如何在 Python 中对压缩列表进行排序？

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)] .

Answer 1

非常简单：

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

Answer 2

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

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.

Answer 4

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)]

Answer 5

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)]

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)