关于python：如何遍历字典以删除与字符串匹配的值？


How to loop through a dictionary to remove values that match a string?

如果值包含特定的字符串，我希望从字典中删除这些值，并随后删除任何具有空列表作为值的键。

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mydict = {
    'Getting links from: https://www.foo.com/':
    [
        '+---OK--- http://www.this.com/',
        '+---OK--- http://www.is.com/',
        '+-BROKEN- http://www.broken.com/',
        '+---OK--- http://www.set.com/',
        '+---OK--- http://www.one.com/'
    ],
    'Getting links from: https://www.bar.com/':
    [
        '+---OK--- http://www.this.com/',
        '+---OK--- http://www.is.com/',
        '+-BROKEN- http://www.broken.com/'
    ],
    'Getting links from: https://www.boo.com/':
    [
        '+---OK--- http://www.this.com/',
        '+---OK--- http://www.is.com/'
    ]
}

val ="is"

for k, v in mydict.iteritems():
   if v.contains(val):
     del mydict[v]

我想要的结果是：

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{
    'Getting links from: https://www.foo.com/':
    [
        '+-BROKEN- http://www.broken.com/',
        '+---OK--- http://www.set.com/',
        '+---OK--- http://www.one.com/'
    ],
    'Getting links from: https://www.bar.com/':
    [
        '+-BROKEN- http://www.broken.com/'
    ]
}

号

如何删除所有包含字符串的字典值，然后删除所有没有值的键？

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简单循环：

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val ="is"

new_dict = dict()
for k, v in mydict.items():
    values = [i for i in v if val not in i]
    if values: new_dict[k] = values

print(new_dict)

输出：

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{'Getting links from: https://www.foo.com/': ['+-BROKEN- http://www.broken.com/', '+---OK--- http://www.set.com/', '+---OK--- http://www.one.com/'], 'Getting links from: https://www.bar.com/': ['+-BROKEN- http://www.broken.com/']}

。

您可以在字典理解中使用列表理解。在迭代字典时，不应该更改字典中的项数。

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res = {k: [x for x in v if 'is' not in x] for k, v in mydict.items()}

# {'Getting links from: https://www.foo.com/': ['+-BROKEN- http://www.broken.com/',
#                                               '+---OK--- http://www.set.com/',
#                                               '+---OK--- http://www.one.com/'],
#  'Getting links from: https://www.bar.com/': ['+-BROKEN- http://www.broken.com/'],
#  'Getting links from: https://www.boo.com/': []}

。

如果要删除列表值为空的项，可以在后续步骤中执行以下操作：

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res = {k: v for k, v in res.items() if v}

。

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这是一条单行线：

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{k: [e for e in v if val not in e] for k, v in mydict.items() if any([val not in e for e in v])}

。

预期产量：

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Out[1]: {
    'Getting links from: https://www.bar.com/':
    [
        '+-BROKEN- http://www.broken.com/'
    ],
    'Getting links from: https://www.foo.com/':
    [
        '+-BROKEN- http://www.broken.com/',
        '+---OK--- http://www.set.com/',
        '+---OK--- http://www.one.com/'
    ]
}
相关讨论

使用dict comprehension可以尝试以下操作：

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import re

val = 'is'

# step 1 - remove line having is
mydict = {k:[re.sub(r'.*is*.', '', x) for x in v] for k,v in mydict.items()}

# filtering out keys if there is no value - if needed
mydict = {k:v for k,v in mydict.items() if len(v) > 0}

print(mydict)

{'Getting links from: https://www.foo.com/': ['com/',
  'com/',
  '+-BROKEN- http://www.broken.com/',
  '+---OK--- http://www.set.com/',
  '+---OK--- http://www.one.com/'],
 'Getting links from: https://www.bar.com/': ['com/',
  'com/',
  '+-BROKEN- http://www.broken.com/'],
 'Getting links from: https://www.boo.com/': ['com/', 'com/']}

有几种方法可以做到。一个使用regex，另一个不使用regex。

如果您不熟悉regex，可以尝试以下方法：

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for key, value in mydict.items():
    if val in value:
        mydict.pop(key)

输出为：

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    mydict = {'Getting links from: https://www.bar.com/': ['+---OK--- http://www.this.com/',
  '+---OK--- http://www.is.com/',
  '+-BROKEN- http://www.broken.com/'],
 'Getting links from: https://www.boo.com/': ['+---OK--- http://www.this.com/',
  '+---OK--- http://www.is.com/'],
 'Getting links from: https://www.foo.com/': ['+---OK--- http://www.this.com/',
  '+---OK--- http://www.is.com/',
  '+-BROKEN- http://www.broken.com/',
  '+---OK--- http://www.set.com/',
  '+---OK--- http://www.one.com/']}

号

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