Question

我有一个唯一键字典，其中一些键共享相同的值。

例如：

D = {'ida':{'key':'1'},'idb':{'key':'2'},'idc':{'key':'3'},'idd':{'key':'3'},'ide':{'key':'4'},'idf':{'key':'4'},'idg':{'key':'4'}}

我想要一个与其他键共享相同值的键列表。

在这种情况下，它将是

l = ['idc','idd','ide','idf','idg']

但是，我想从共享相同值的所有键集中排除一个键。

例如，我想拥有密钥

l = ['idd','idf','idg']

排除'idc'和'ide'

或者可能是

l = ['idc','ide','idf']

排除'idd'和'idg'。

Answer 1

如果值出现多次，则list comp将添加该键。

dup_keys = [k for k in D if sum(D[k] in x for x in D.iteritems()) > 1 ]

['idf', 'idg', 'idd', 'ide', 'idc']

dup_keys[1:]
['idg', 'idd', 'ide', 'idc']
dup_keys[1:-1]
['idg', 'idd', 'ide']

if sum(D[k] in x for x in D.iteritems()) > 1检查值> 1次。

要忽略某些键添加一些and条件，我不确定您想要忽略哪些键。

Answer 2

# aDictionary = { <aKey>: <aValue>, ... }             python dictionary constructor

  aDictionary = { 'ida': '1', \                  
                  'idb': '2', \
                  'idc': '3', \
                  'idd': '3', \
                  'ide': '4', \
                  'idf': '4', \
                  'idg': '4'  \
                  }

# the OP used aDictionaryOfDictionaries construction instead.
# aDictOfDICTs = {'ida':{'key':'1'},'idb':{'key':'2'}, ... }

# The algorithmisation of the processing task is different.

# for aDictionary, a inverseDictionary may be assembled 
# to hold for each <aValue>,
# a referring <aKey>, incl. those, where are more than one primary <aKey>-s,
# ( stored asListOfPrimaryKEYs or aSetOfPrimaryKEYs ), which len() is easy to test

公平＆amp;适当的anInverseDICTIONARY构造的强大解决方案 可以在＆＃34;快速功能解决方案中找到非双射地图（值不唯一）：＆＃34;在＆gt;＆gt;＆gt; https://stackoverflow.com/a/22235665/3666197

请注意，并非所有帖子都能在那里产生强大的解决方案。

<强>最后

>>> aListOfDUP_KEYs = [ anInvDictKEY for anInvDictKEY in anInverseDICTIONARY.keys() if ( len( anInverseDICTIONARY[anInvDictKEY] ) > 1 )  ]

    [ 3, 4 ]

''' as anInverseDICTIONARY = { 1: set( [ 'ida' ] ),
                               2: set( [ 'idb' ] ),
                               3: set( [ 'idd', 'idc' ] ),
                               4: set( [ 'idg', 'idf', 'ide' ] )
                               }
'''

通过DUP_KEY从 anInverseDICTIONARY 中移除任何一个DUP-e，或通过列表理解从原始的 aDictionary 移除

>>> [ aDictionary.pop( anInverseDICTIONARY[aDUP_KEY].pop() ) for aDUP_KEY in aListOfDUP_KEYs ]

    [ 'idd', 'idf' ]

如何获取与同一字典中的另一个键共享值的键列表？

2 个答案: