字典Python中重复键值的列表 [英] List of values for duplicate keys in dictionary Python
问题描述
如果这个问题已经在这里进行了探讨,我们提前道歉 - 我在这里查看了不同的答案,但找不到我需要的。
我的目标是创建这样的字典 - {'a':[10,9,10,10],'b':[10,9,1,0],'c':[0,5, 0,1]等等}
我所拥有的是重复键的多个字典(每个其他字典中的相同的键),一些东西像这样
{'a':10,'b':0,'c':2}
{' a':4,'c':4}
{'a':4,'b':5,'c' :3}
我无法知道这些字典的数量,或者如果钥匙持续到'f '或'g',但我知道这些键是重复的。我试过 defaultdict
,但我得到的是 -
defaultdict (< type'list'> {'a':[10]})
defaultdict(< type'list'> {'a':[10],'b' ]})
defaultdict(< type'list'>,{'a':[10],'b':[3],'c':[0]})
然后下一个字典相同的东西 -
defaultdict(< type'list'> {'a':[4]})
defaultdict(< type'list'> {'a' 4],'b':[5]})
defaultdict(< type'list'>,{'a':[4],'b':[5],'c' ]})
上面输出的代码是 -
d = collections.defaultdict(list)
for k,v in z.iteritems():
d [k] .append v)
c = d.items()
打印d
如果我一个 print c
(打印 d.items()
)我得到 -
[('a',[10])]
pre>
[('a',[10]),('b',[3] )]
[('a',[10]),('c',[0]),('b',[3])]
每个字典再次重复。如何获得1个持有所有键的值,值 -
{'a':[10,0,.. ],'b':[4,3,4,...]等}
我还应该补充一点,我是一个for循环的结果,而不是单独存储一个唯一的变量。
解决方案如果我正确理解你的想法,你正在尝试合并各种字典。一种使用内置功能的方法(我确定很快有人会给你一个
numpy
和集合
答案)可以看像这样:ds = [
{'a':10,'b':0,'c' :2},
{'a':7,'b':4,'c':4},
{'a':4,'b':5,'c' }]
merged = {}
d中的d:
为k,v在d.items()中:
如果k未合并:合并[ k] = []
合并[k] .append(v)
打印(合并)
(为了清楚起见,请详细填写)
编辑:
阅读完您的评论后,我想要的结果是值/键列表,您可以在结果合并的字典中使用:print( [(v,k)for k,v in merged.items()])
[([10,7,4],'a'),([2,4,3] '),([0,4,5],'b')]
Apologies in advance if this question has already been explored here - I looked at different answers here but couldn't find what I need.
My goal is to create a dictionary like this --
{'a':[10, 9, 10, 10], 'b':[10, 9, 1, 0], 'c':[0, 5, 0, 1], and so on}
What I have is multiple dictionaries with duplicate keys (same keys in every other dictionary), something like this
{'a':10, 'b': 0, 'c': 2}
{'a':7, 'b': 4, 'c': 4}
{'a':4, 'b': 5, 'c': 3}
I have no way of knowing the number of such dictionaries, or if there are keys continuing up to 'f', or a 'g' in them but I know that the keys are duplicated. I've tried
defaultdict
but what I get is--defaultdict(<type 'list'>, {'a': [10]}) defaultdict(<type 'list'>, {'a': [10], 'b': [3]}) defaultdict(<type 'list'>, {'a': [10], 'b': [3], 'c': [0]})
and then the same thing for the next dictionary --
defaultdict(<type 'list'>, {'a': [4]}) defaultdict(<type 'list'>, {'a': [4], 'b': [5]}) defaultdict(<type 'list'>, {'a': [4], 'b': [5], 'c': [1]})
The code I have for the above output is --
d = collections.defaultdict(list) for k, v in z.iteritems(): d[k].append(v) c = d.items() print d
If I do a
print c
instead (to print thed.items()
) I get --[('a', [10])] [('a', [10]), ('b', [3])] [('a', [10]), ('c', [0]), ('b', [3])]
which is again repeated for each dictionary. How do I get 1 dict holding all the keys, values --
{'a':[10,0,..], 'b':[4, 3, 4,..], etc.} ?
I should also add that the dicts I have are the result of a for loop and not stored individually in a unique variable.
解决方案If I understand correctly your attetion, you are trying to merge various dictionaries. One way using built-ins (I am sure that soon someone will give you a
numpy
andcollections
answer) could look like this:ds = [ {'a':10, 'b': 0, 'c': 2}, {'a':7, 'b': 4, 'c': 4}, {'a':4, 'b': 5, 'c': 3} ] merged = {} for d in ds: for k, v in d.items (): if k not in merged: merged [k] = [] merged [k].append (v) print (merged)
(Quite verbose for clarity)
EDIT: After having read your comment stating "The result I want is a list of values/key.", you can use this on the resulting merged dictionary:
print ( [ (v, k) for k, v in merged.items () ] )
This yields:
[([10, 7, 4], 'a'), ([2, 4, 3], 'c'), ([0, 4, 5], 'b')]
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