保留除一个密钥 python 之外的所有条目

Question

我有一个python字典。为了给出上下文，我正在尝试编写自己的简单交叉验证单元。

所以基本上我想要的是获取除给定键之外的所有值。根据输入，它返回字典中的所有值，除了给定的值。

因此，如果输入是 2 和 5，那么输出值没有键 2 和 5 的值？

Answer 1

for key, value in your_dict.items():
    if key not in your_blacklisted_set:
        print value

美妙之处在于这个伪代码示例是有效的 Python 代码。

它也可以表示为列表理解：

resultset = [value for key, value in your_dict.items() if key not in your_blacklisted_set]

Answer 2

如果您的目标是返回一个新字典，其中包含除一个或几个之外的所有键/值，请使用以下命令：

exclude_keys = ['exclude', 'exclude2']
new_d = {k: d[k] for k in set(list(d.keys())) - set(exclude_keys)}

其中'exclude'可以替换为应排除的(列表(键。

Answer 3

只是为了玩套装

keys = set(d.keys())
excludes = set([...])
for key in keys.difference(excludes):
    print d[key]

Answer 4

keys = ['a', 'b']
a_dict = {'a':1, 'b':2, 'c':3, 'd':4}
[a_dict.pop(key) for key in keys]

弹出要丢弃的密钥后，a_dict将保留您想要的密钥。

Answer 5

给定字典说

d = {
     2: 2, 5: 16, 6: 5,
     7: 6, 11: 17, 12: 9,
     15: 18, 16: 1, 18: 16,
     19: 17, 20: 10
     }

那么简单的理解示例将达到您可能想要的

[v for k,v in d.iteritems() if k not in (2,5)]

此示例列出所有不带键 {2,5} 的值

例如，上述理解的 O/P 是

[5, 6, 1, 17, 9, 18, 1, 16, 17, 10]

Answer 6

此外，作为使用集合的列表推导：

d = dict(zip(range(9),"abcdefghi"))
blacklisted = [2,5]
outputs = [d[k] for k in set(d.keys())-set(blacklisted)]

Answer 7

以下几行怎么样：

In [7]: d = dict((i,i+100) for i in xrange(10))
In [8]: d
Out[8]: 
{0: 100,
 1: 101,
 2: 102,
 3: 103,
 4: 104,
 5: 105,
 6: 106,
 7: 107,
 8: 108,
 9: 109}
In [9]: exc = set((2, 5))
In [10]: for k, v in d.items():
   ....:     if k not in exc:
   ....:         print v
   ....:         
   ....:         
100
101
103
104
106
107
108
109

Answer 8

你也可以使用 pydash - https://pydash.readthedocs.io/en/latest/

from pydash import omit
d = {'a': 1, 'b': 2, 'c': 3}
c = omit(d, 'a')
print(c) # {'b': 2, 'c': 3}

Answer 9

使用异常处理

facebook_posts = [
    {'Likes': 21, 'Comments': 2}, 
    {'Likes': 13, 'Comments': 2, 'Shares': 1}, 
    {'Likes': 33, 'Comments': 8, 'Shares': 3}, 
    {'Comments': 4, 'Shares': 2}, 
    {'Comments': 1, 'Shares': 1}, 
    {'Likes': 19, 'Comments': 3}
]
total_likes = 0
for post in facebook_posts:
    try:
        total_likes = total_likes + post['Likes']
    except KeyError:
        pass
print(total_likes)

您还可以使用：

except KeyError: 'My Key'

但可能只适合一次性使用

Answer 10

如果要修改生成的字典，请小心使用浅拷贝。

import copy
def copy_dict_except_keys_shallow(d:dict, exclude_keys):
  return {k: d[k] for k in set(list(d.keys())) - set(exclude_keys)}
def copy_dict_except_keys_deep(d:dict, exclude_keys):
  return {k: copy.deepcopy(d[k]) for k in set(list(d.keys())) - set(exclude_keys)}
original = {'a': 1, 'b': [1, 2, 3]}
deep = copy_dict_except_keys_deep(original, ['a'])
deep['b'].append(4)
print("Modifying the deep copy preserves the original")
print(f"original: {original}")
print(f"deep copy: {deep}")
print()
shallow = copy_dict_except_keys_shallow(original, 'a')
shallow['b'].append(4)
print("Modifying the shallow copy changes the original")
print(f"original: {original}")
print(f"shallow copy: {shallow}")

Modifying the deep copy preserves the original
original: {'a': 1, 'b': [1, 2, 3]}
deep copy: {'b': [1, 2, 3, 4]}
Modifying the shallow copy changes the original
original: {'a': 1, 'b': [1, 2, 3, 4]}
shallow copy: {'b': [1, 2, 3, 4]}

Answer 11

这是我

的：

def dict_exclude_keys(d: dict, excludes: list[str]) -> dict:
    keys = set(list(d.keys())) - set(excludes)
    return {k: d[k] for k in keys if k in d}