python - 如何按值过滤字典？

Question

这里是新手问题，所以请耐心等待。

假设我有一本如下所示的字典:

a = {"2323232838": ("first/dir", "hello.txt"),
     "2323221383": ("second/dir", "foo.txt"),
     "3434221": ("first/dir", "hello.txt"),
     "32232334": ("first/dir", "hello.txt"),
     "324234324": ("third/dir", "dog.txt")}

我希望将所有彼此相等的值移动到另一个字典中。

matched = {"2323232838": ("first/dir", "hello.txt"),
           "3434221":    ("first/dir", "hello.txt"),
           "32232334":   ("first/dir", "hello.txt")}

剩下的不匹配项应该是这样的:

remainder = {"2323221383": ("second/dir", "foo.txt"),
             "324234324":  ("third/dir", "dog.txt")}

在此先感谢，如果您提供示例，请尽可能多地发表评论。

Answer 1

下面的代码将产生两个变量，matches 和 remainders。 matches 是一个字典数组，其中来自原始字典的匹配项目将有一个对应的元素。 remainder 将包含一个包含所有不匹配项的字典。

请注意，在您的示例中，只有一组匹配值:('first/dir', 'hello.txt')。如果有多个集合，则每个集合在 matches 中都有对应的条目。

import itertools

# Original dict
a = {"2323232838": ("first/dir", "hello.txt"),
     "2323221383": ("second/dir", "foo.txt"),
     "3434221": ("first/dir", "hello.txt"),
     "32232334": ("first/dir", "hello.txt"),
     "324234324": ("third/dir", "dog.txt")}

# Convert dict to sorted list of items
a = sorted(a.items(), key=lambda x:x[1])

# Group by value of tuple
groups = itertools.groupby(a, key=lambda x:x[1])

# Pull out matching groups of items, and combine items   
# with no matches back into a single dictionary
remainder = []
matched   = []

for key, group in groups:
   group = list(group)
   if len(group) == 1:
      remainder.append( group[0] )
   else:
      matched.append( dict(group) )
else:
   remainder = dict(remainder)

输出:

>>> matched
[
  {
    '3434221':    ('first/dir', 'hello.txt'), 
    '2323232838': ('first/dir', 'hello.txt'), 
    '32232334':   ('first/dir', 'hello.txt')
  }
]

>>> remainder
{
  '2323221383': ('second/dir', 'foo.txt'), 
  '324234324':  ('third/dir', 'dog.txt')
}

作为新手，您可能会在上面的代码中了解到一些不熟悉的概念。以下是一些链接: