关于python：如何从字典列表中删除k，v条目

How to remove k,v entries from a list of dictionaries

我想从现有的JSON文件中删除不需要的值：

1	{"items": [ {"id":"abcd","h": 2,"x": 0,"level": 4 }, {"id":"dfgg","h": 7,"x": 5,"level": 30 } ] }

我已经尝试在适当的位置删除这些值，但得到了"迭代期间字典更改的大小"。

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with open('inventory2.json', 'r') as inf:
data = json.load(inf)
inf.close()

keysiwant = ['x', 'h']
for dic in data['items']:
for k, v in dic.items():
if k not in keysiwant:
dic.pop(k, None)

号

相关讨论

问题3：在Python dict.items()只是一景附注(a copy of the字典的项目-你不能随迭代的变化信息。＜／P >

但是你可以把《dict.items()迭代器为A(这list()副本和计算机网络decouples从词典，这条路的)，那么你可以反复交叉的copy of the dict.items()吧：＜／P >

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import json

t ="""{"items": [ {"id":"abcd","h": 2,"x": 0,"level": 4 },
{"id":"dfgg","h": 7,"x": 5,"level": 30 } ] }"""

data = json.loads(t) # loads is better for SO-examples .. it makes it a mcve
keysiwant = ['x', 'h']
for dic in data['items']:
for k, v in list(dic.items()):
if k not in keysiwant:
dic.pop(k, None)

print(data)

输出：＜／P >

1	{'items': [{'h': 2, 'x': 0}, {'h': 7, 'x': 5}]}

更多的在线python2 / python3 dict.items()答案：这是什么是差分之间的dict.items()和()dict.iteritems吗？＜／P >

相关讨论

请尝试这个。信息技术的使用，因为它fewer iterations设置超时群岛一线，在送他们到流行音乐/解除。也，它只使用Keys(list(DLC))，而不是(键/值的元组。＜／P >

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import json

t ="""{"items": [ {"id":"abcd","h": 2,"x": 0,"level": 4 },
{"id":"dfgg","h": 7,"x": 5,"level": 30 } ] }"""

data = json.loads(t)
keysiwant = ["x","h"]

for dic in data["items"]:
for k in (k for k in list(dic) if k not in keysiwant):
dic.pop(k, None)

print(data)

输出：＜／P >

1	{'items': [{'h': 2, 'x': 0}, {'h': 7, 'x': 5}]}