翻转键和字典python中的值 [英] flip keys and values in dictionary python

问题描述

我有一个名为z的字典，看起来像这样

---

{0：[0.28209479177387814，0.19947114020071635,0.10377687435514868,0.03377133158686996 ]，
-1：[0.28209479177387814，0.19947114020071635,0.10377687435514868,07338133158686996]}。

我想翻转值和键以使

{0.28209479177387814：0，0.19947114020071635：0，0.10377687435514868：0，0.07338133158686996：0，
0.28209479177387814：-1，0.19947114020071635：-1,0.10377687435514868：-1，0.07338133158686996：-1 }

似乎有效的代码是：

  for a in z：
 newdict = dict.fromkeys（z [a]，a）
  

但它只适用于z中的其中一个键，并返回：$ / $ $ $ $ $ $ $ $ $ $ $ $ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1}

我做错了什么？

解决方案

你c在字典中没有重复键，但您可以使用元组以有意义的方式将它们组合在一起。

  from itertools import产品，链
 
 tuples = chain.from_iterable（product（vs，[k]）for k，vs in orig_dict.items（））
＃注意与...相同：
＃＃tuples = [] 
＃＃for k，vs in orig_dict.items（）：
＃＃for [（v，k）for v in vs]：
＃ ＃tuples.append（tup）
  

这将产生：

pre> [（0.28209479177387814,0），（0.19947114020071635,0），
（0.10377687435514868,0），（0.07338133158686996,0），
（0.28209479177387814， -1），（0.19947114020071635，-1），
（0.10377687435514868，-1），（0.07338133158686996，-1）]


现在，如果你真的想要一些有趣的东西，你可以排序，并将它们分组在一起。

  from itertools import groupby 
 
 groups = groupby（sorted（tuples），key = lambda kv：kv [0]）
  

创建如下：

  [（0.07338133158686996，[（0.07338133158686996,0，
 0.07338133158686996，-1]），
 ...] 
  

您可以通过以下方式将它们转化为dict：

  final_dict = {k：[v [1] for v in vs] for k，vs in groups} 
  

哪个应该最终给出：

  {0.07338133158686996：[0，-1]，
 ...} 
  

I have a dictionary called z that looks like this

---

{0: [0.28209479177387814, 0.19947114020071635, 0.10377687435514868, 0.07338133158686996], -1: [0.28209479177387814, 0.19947114020071635, 0.10377687435514868, 0.07338133158686996]}.

I want to flip the values and keys to have

{0.28209479177387814:0, 0.19947114020071635:0, 0.10377687435514868:0, 0.07338133158686996:0, 0.28209479177387814:-1, 0.19947114020071635:-1, 0.10377687435514868:-1, 0.07338133158686996:-1}

The piece of code that seems to work is :

for a in z:
     newdict=dict.fromkeys(z[a],a)

but it only works for one of the keys in z and returns this:

{0.28209479177387814: -1, 0.07338133158686996: -1, 0.10377687435514868: -1, 0.19947114020071635: -1}

what am I doing wrong?

解决方案

You can't have duplicate keys in a dictionary, but you can pair them together using tuples in a meaningful way.

from itertools import product, chain

tuples = chain.from_iterable(product(vs, [k]) for k, vs in orig_dict.items())
# note this is identical to:
# # tuples = []
# # for k, vs in orig_dict.items():
# #     for tup in [(v, k) for v in vs]:
# #         tuples.append(tup)

That will produce:

[(0.28209479177387814, 0), (0.19947114020071635, 0),
 (0.10377687435514868, 0), (0.07338133158686996, 0),
 (0.28209479177387814, -1), (0.19947114020071635, -1),
 (0.10377687435514868, -1), (0.07338133158686996, -1)]

Now if you really wanted something interesting, you could sort that and group it together.

from itertools import groupby

groups = groupby(sorted(tuples), key=lambda kv: kv[0])

That creates something like:

[(0.07338133158686996, [(0.07338133158686996, 0,
                         0.07338133158686996, -1] ),
 ... ]

You could toss those into a dict by doing:

final_dict = {k: [v[1] for v in vs] for k, vs in groups}

Which should finally give:

{0.07338133158686996: [0, -1],
 ... }

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