从包含python中字典的列表中删除元素

我有一个python脚本，它将字典存储在列表变量“data”中：

[  
   {  
      'hp':'cdldfdbc07s001:10000',
      'n':'cdldfdbc07s001',
      'rsid':'cdl-dit1',
      'v':'3.0.12'
   },
   {  
      'hp':'cdldfdbc07s002:10000',
      'n':'cdldfdbc07s002',
      'rsid':'cdl-dit1',
      'v':'3.0.12'
   },
   {  
      'hp':'cdldfdbc07s001:11000',
      'n':'cdldfdbc07s001',
      'rsid':'cdl-fit1',
      'v':'3.0.12'
   },
   {  
      'hp':'cdldfdbc07s002:11000',
      'n':'cdldfdbc07s002',
      'rsid':'cdl-fit1',
      'v':'3.0.12'
   },
   {  
      'hp':'cdldvjassvp073:10000',
      'n':'cdldvjassva073',
      'rsid':'mobile',
      'v':'2.6.3'
   },
   {  
      'hp':'cdldvjassva072:10000',
      'n':'cdldvjassva072',
      'rsid':'mobile',
      'v':'2.6.3'
   },
   {  
      'hp':'cdldvjassva074:11002',
      'n':'cdldvjassva074',
      'rsid':'cache2',
      'v':'2.6.3'
   },
   {  
      'hp':'cdldfdbc05s002:11000',
      'n':'cdldfdbc05s002',
      'rsid':'dit2',
      'v':'3.0.9'
   }
]

我想

1.遍历列表中的所有元素，并删除“hp”中包含“jass”的元素

我的代码似乎没有删除元素，我的len（data）返回的数字与以前相同

s=0   
while s< len(data):
     for i in ((data[s]['hp'])):
            if "jass" in i:
                    data.remove(s)
            s += 1
print len(data)

---

要删除python字典条目--

---

要删除python字典条目--

使用

remove

从列表中删除元素：

print(len(data))  # prints 8


for d in data[:]:  # iterate over a COPY of the list[:]
    if 'jass' in d['hp']:
        data.remove(d)

 print(len(data))  # prints 5

您也可以使用列表理解在一行中完成此操作：

[item for item in data if "jass" not in item["hp"]]

[数据。删除数据[：]中d的（d），如果d['hp']中的'jass'。]

使用

remove

从列表中删除元素：

print(len(data))  # prints 8


for d in data[:]:  # iterate over a COPY of the list[:]
    if 'jass' in d['hp']:
        data.remove(d)

 print(len(data))  # prints 5

您也可以使用列表理解在一行中完成此操作：

[item for item in data if "jass" not in item["hp"]]

---

[data.remove（d）for d in data[：]如果d['hp']中的'jass'，

解决这些筛选问题的最佳方法是使用列表理解来重建DICT列表，其中的值中没有jass：

data = [{'hp': 'cdldfdbc07s001:10000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, {'hp': 'cdldfdbc07s002:10000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, {'hp': 'cdldfdbc07s001:11000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, {'hp': 'cdldfdbc07s002:11000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-fit1', 'v': '3.0.12'},
{'hp': 'cdldvjassvp073:10000', 'n': 'cdldvjassva073', 'rsid': 'mobile', 'v': '2.6.3'}, {'hp': 'cdldvjassva072:10000', 'n': 'cdldvjassva072', 'rsid': 'mobile', 'v': '2.6.3'}, {'hp': 'cdldvjassva074:11002', 'n': 'cdldvjassva074', 'rsid': 'cache2', 'v': '2.6.3'},
 {'hp': 'cdldfdbc05s002:11000', 'n': 'cdldfdbc05s002', 'rsid': 'dit2', 'v': '3.0.9'}]

data_without_jass = [d for d in data if all("jass" not in v for v in d.values())]

print(data_without_jass)

结果:

[{'hp': 'cdldfdbc07s001:10000', 'rsid': 'cdl-dit1', 'v': '3.0.12', 
'n': 'cdldfdbc07s001'}, {'hp': 'cdldfdbc07s002:10000', 'rsid': 'cdl-dit1', 'v': '3.0.12', 'n': 'cdldfdbc07s002'}, {'hp': 'cdldfdbc07s001:11000',
 'rsid': 'cdl-fit1', 'v': '3.0.12', 'n': 'cdldfdbc07s001'}, {'hp': 'cdldfdbc07s002:11000', 'rsid': 'cdl-fit1', 'v': '3.0.12', 'n': 'cdldfdbc07s002'}, 
{'hp': 'cdldfdbc05s002:11000', 'rsid': 'dit2', 'v': '3.0.9', 'n': 'cdldfdbc05s002'}]

这里的关键是条件：

all("jass" not in v for v in d.values())

测试每个子目录的所有值是否不包含“jass” （或使用反向逻辑：测试是否每个子目录的值都不包含“jass”）

我现在注意到，该条件仅检查hp密钥，因此：

[d for d in data if "jass" not in d["hp"]]

---

更准确

在这些过滤问题中，最好的方法是使用列表理解来重建DICT列表，而不使用值中的JAS：

data = [{'hp': 'cdldfdbc07s001:10000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, {'hp': 'cdldfdbc07s002:10000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, {'hp': 'cdldfdbc07s001:11000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, {'hp': 'cdldfdbc07s002:11000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-fit1', 'v': '3.0.12'},
{'hp': 'cdldvjassvp073:10000', 'n': 'cdldvjassva073', 'rsid': 'mobile', 'v': '2.6.3'}, {'hp': 'cdldvjassva072:10000', 'n': 'cdldvjassva072', 'rsid': 'mobile', 'v': '2.6.3'}, {'hp': 'cdldvjassva074:11002', 'n': 'cdldvjassva074', 'rsid': 'cache2', 'v': '2.6.3'},
 {'hp': 'cdldfdbc05s002:11000', 'n': 'cdldfdbc05s002', 'rsid': 'dit2', 'v': '3.0.9'}]

data_without_jass = [d for d in data if all("jass" not in v for v in d.values())]

print(data_without_jass)

结果:

[{'hp': 'cdldfdbc07s001:10000', 'rsid': 'cdl-dit1', 'v': '3.0.12', 
'n': 'cdldfdbc07s001'}, {'hp': 'cdldfdbc07s002:10000', 'rsid': 'cdl-dit1', 'v': '3.0.12', 'n': 'cdldfdbc07s002'}, {'hp': 'cdldfdbc07s001:11000',
 'rsid': 'cdl-fit1', 'v': '3.0.12', 'n': 'cdldfdbc07s001'}, {'hp': 'cdldfdbc07s002:11000', 'rsid': 'cdl-fit1', 'v': '3.0.12', 'n': 'cdldfdbc07s002'}, 
{'hp': 'cdldfdbc05s002:11000', 'rsid': 'dit2', 'v': '3.0.9', 'n': 'cdldfdbc05s002'}]

这里的关键是条件：

all("jass" not in v for v in d.values())

测试每个子目录的所有值是否不包含“jass” （或使用反向逻辑：测试是否每个子目录的值都不包含“jass”）

我现在注意到，该条件仅检查hp密钥，因此：

[d for d in data if "jass" not in d["hp"]]

---

更准确

您需要从列表中删除项目，而不是从列表中的dict中删除项目：

l = [{'hp': 'cdldfdbc07s001:10000',  'n': 'cdldfdbc07s001',  'rsid': 'cdl-dit1', 'v': '3.0.12'},  
     {'hp': 'cdldfdbc07s002:10000',  'n': 'cdldfdbc07s002',  'rsid': 'cdl-dit1', 'v': '3.0.12'},  
     {'hp': 'cdldfdbc07s001:11000',  'n': 'cdldfdbc07s001',  'rsid': 'cdl-fit1', 'v': '3.0.12'},  
     {'hp': 'cdldfdbc07s002:11000',  'n': 'cdldfdbc07s002',  'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
     {'hp': 'cdldvjassvp073:10000',  'n': 'cdldvjassva073',  'rsid': 'mobile',   'v': '2.6.3'},  
     {'hp': 'cdldvjassva072:10000',  'n': 'cdldvjassva072',  'rsid': 'mobile',   'v': '2.6.3'},  
     {'hp': 'cdldvjassva074:11002',  'n': 'cdldvjassva074',  'rsid': 'cache2',   'v': '2.6.3'}, 
     {'hp': 'cdldfdbc05s002:11000',  'n': 'cdldfdbc05s002',  'rsid': 'dit2', 'v': '3.0.9'}
]

result = [x for x in l  if 'jass' not in x["hp"]] # the if condition lets only dicts in
                                                  # that do not have 'jass' in its key 'hp'

print (l)

print(result)

输出：

[{'hp': 'cdldfdbc07s001:10000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:10000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-di
t1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s001:11000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:11000', 'n': 'cdldfdbc07s0
02', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldvjassvp073:10000', 'n': 'cdldvjassva073', 'rsid': 'mobile', 'v': '2.6.3'}, 
 {'hp': 'cdldvjassva072:10000', '
n': 'cdldvjassva072', 'rsid': 'mobile', 'v': '2.6.3'}, 
 {'hp': 'cdldvjassva074:11002', 'n': 'cdldvjassva074', 'rsid': 'cache2', 'v': '2.6.3'}, 
 {'hp': 'cdldfdbc05
s002:11000', 'n': 'cdldfdbc05s002', 'rsid': 'dit2', 'v': '3.0.9'}]

[{'hp': 'cdldfdbc07s001:10000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:10000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-di
t1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s001:11000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:11000', 'n': 'cdldfdbc07s0
02', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc05s002:11000', 'n': 'cdldfdbc05s002', 'rsid': 'dit2', 'v': '3.0.9'}]

---

您需要从列表中删除项目，而不是从列表中的dict中删除项目：

l = [{'hp': 'cdldfdbc07s001:10000',  'n': 'cdldfdbc07s001',  'rsid': 'cdl-dit1', 'v': '3.0.12'},  
     {'hp': 'cdldfdbc07s002:10000',  'n': 'cdldfdbc07s002',  'rsid': 'cdl-dit1', 'v': '3.0.12'},  
     {'hp': 'cdldfdbc07s001:11000',  'n': 'cdldfdbc07s001',  'rsid': 'cdl-fit1', 'v': '3.0.12'},  
     {'hp': 'cdldfdbc07s002:11000',  'n': 'cdldfdbc07s002',  'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
     {'hp': 'cdldvjassvp073:10000',  'n': 'cdldvjassva073',  'rsid': 'mobile',   'v': '2.6.3'},  
     {'hp': 'cdldvjassva072:10000',  'n': 'cdldvjassva072',  'rsid': 'mobile',   'v': '2.6.3'},  
     {'hp': 'cdldvjassva074:11002',  'n': 'cdldvjassva074',  'rsid': 'cache2',   'v': '2.6.3'}, 
     {'hp': 'cdldfdbc05s002:11000',  'n': 'cdldfdbc05s002',  'rsid': 'dit2', 'v': '3.0.9'}
]

result = [x for x in l  if 'jass' not in x["hp"]] # the if condition lets only dicts in
                                                  # that do not have 'jass' in its key 'hp'

print (l)

print(result)

输出：

[{'hp': 'cdldfdbc07s001:10000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:10000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-di
t1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s001:11000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:11000', 'n': 'cdldfdbc07s0
02', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldvjassvp073:10000', 'n': 'cdldvjassva073', 'rsid': 'mobile', 'v': '2.6.3'}, 
 {'hp': 'cdldvjassva072:10000', '
n': 'cdldvjassva072', 'rsid': 'mobile', 'v': '2.6.3'}, 
 {'hp': 'cdldvjassva074:11002', 'n': 'cdldvjassva074', 'rsid': 'cache2', 'v': '2.6.3'}, 
 {'hp': 'cdldfdbc05
s002:11000', 'n': 'cdldfdbc05s002', 'rsid': 'dit2', 'v': '3.0.9'}]

[{'hp': 'cdldfdbc07s001:10000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-dit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:10000', 'n': 'cdldfdbc07s002', 'rsid': 'cdl-di
t1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s001:11000', 'n': 'cdldfdbc07s001', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc07s002:11000', 'n': 'cdldfdbc07s0
02', 'rsid': 'cdl-fit1', 'v': '3.0.12'}, 
 {'hp': 'cdldfdbc05s002:11000', 'n': 'cdldfdbc05s002', 'rsid': 'dit2', 'v': '3.0.9'}]

如果您选择肯定（考虑保留哪些项目）而不是否定（删除某些项目），您可以使用列表：

[item for item in data if "jass" not in item["hp"]]

如果您选择肯定（考虑保留哪些项目）而不是否定（删除某些项目），您可以使用列表：

[item for item in data if "jass" not in item["hp"]]

---

发布预期结果post预期结果Top希望从列表中删除项而不是从dictOP中删除项而不是从dictOP中删除项您不应该使用列表理解来产生副作用，也就是说。我不确定我是否理解，哪些副作用？您正在创建一个没有任何副作用的列表，你不应该对副作用使用列表理解，也就是说。我不确定我是否理解，哪些副作用？你在创建一个毫无意义的列表，而不是循环。