Question

新手问题，请耐心等待。

假设我的字典看起来像这样：

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")}

我想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")}

其余不匹配的项目应如下所示：

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

提前致谢，如果您提供示例，请尽可能多地评论。

Answer 1

以下代码将生成两个变量matches和remainders。 matches是一个字典数组，其中原始字典中的匹配项将具有相应的元素。 remainder将包含一个包含所有不匹配项的字典，如您的示例所示。

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，则每个集合都会在matches中有相应的条目。

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')
}

作为一个新手，你可能会在上面的代码中介绍一些不熟悉的概念。以下是一些链接：

Answer 2

您所要求的内容称为“倒置索引” - 不同的项目仅使用一系列密钥记录一次。

>>> from collections import defaultdict
>>> 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")}
>>> invert = defaultdict( list )
>>> for key, value in a.items():
...     invert[value].append( key )
... 
>>> invert
defaultdict(<type 'list'>, {('first/dir', 'hello.txt'): ['3434221', '2323232838', '32232334'], ('second/dir', 'foo.txt'): ['2323221383'], ('third/dir', 'dog.txt'): ['324234324']})

反向字典具有与1个或更多键的列表相关联的原始值。

现在，从中获取修改后的词典。

过滤

>>> [ invert[multi] for multi in invert if len(invert[multi]) > 1 ]
[['3434221', '2323232838', '32232334']]
>>> [ invert[uni] for uni in invert if len(invert[uni]) == 1 ]
[['2323221383'], ['324234324']]

扩展

>>> [ (i,multi) for multi in invert if len(invert[multi]) > 1 for i in invert[multi] ]
[('3434221', ('first/dir', 'hello.txt')), ('2323232838', ('first/dir', 'hello.txt')), ('32232334', ('first/dir', 'hello.txt'))]
>>> dict( (i,multi) for multi in invert if len(invert[multi]) > 1 for i in invert[multi] )
{'3434221': ('first/dir', 'hello.txt'), '2323232838': ('first/dir', 'hello.txt'), '32232334': ('first/dir', 'hello.txt')}

类似（但更简单）的处理适用于出现一次的物品。

Answer 3

迭代字典与迭代python中的列表没有区别：

for key in dic:
    print("dic[%s] = %s" % (key, dic[key]))

这将打印字典的所有键和值。

Answer 4

我认为你的唯一身份证是关键可能不是很漂亮，但会返回一个带有你独特价值的字典：

>>> dict_ = {'1': ['first/dir', 'hello.txt'],
'3': ['first/dir', 'foo.txt'], 
'2': ['second/dir', 'foo.txt'], 
'4': ['second/dir', 'foo.txt']}  
>>> dict((v[0]+v[1],k) for k,v in dict_.iteritems())  
{'second/dir/foo.txt': '4', 'first/dir/hello.txt': '1', 'first/dir/foo.txt': '3'}

我见过你更新了你的帖子：

>>> a
{'324234324': ('third/dir', 'dog.txt'), 
'2323221383': ('second/dir', 'foo.txt'), 
'3434221': ('first/dir', 'hello.txt'), 
'2323232838': ('first/dir', 'hello.txt'), 
'32232334': ('first/dir', 'hello.txt')}
>>> dict((v[0]+"/"+v[1],k) for k,v in a.iteritems())
{'second/dir/foo.txt': '2323221383', 
'first/dir/hello.txt': '32232334', 
'third/dir/dog.txt': '324234324'}

Answer 5

如果您知道要过滤的值：

known_tuple = 'first/dir','hello.txt'
b = {k:v for k, v in a.items() if v == known_tuple}

然后a将成为：

a = dict(a.items() - b.items())

这是py3k表示法，但我确信在旧版本中可以实现类似的功能。如果您不知道known_tuple是什么，那么您需要先找到它。例如：

c = list(a.values())
for i in set(c):
    c.remove(i)
known_tuple = c[0]

如何按值过滤字典？

5 个答案: