Question

我有一个包含这样的字典的列表：

json_obj = [[{'id': None},{'id': '5b98d01c0835f23f538cdcab'},{'id': '5b98d0440835f23f538cdcad'},{'id': '5b98d0ce0835f23f538cdcb9'}],[{'id': None},{'id': '5b98d01c0835f23f538cd'},{'id': '5b98d0440835f23f538cd'},{'id': '5b98d0ce0835f23f538cdc'}]]

我希望它存储在这样的列表列表中：

y=[['None','5b98d01c0835f23f538cdcab','5b98d0440835f23f538cdcad','5b98d0ce0835f23f538cdcb9'],['None','5b98d01c0835f23f538cd','5b98d0440835f23f538cd','5b98d0ce0835f23f538cdc']]

要尝试从字典中读取ID

for d in json_obj:
    print(d['id'])

但是我看到上面的代码出现此错误：

TypeError: list indices must be integers or slices, not str

Answer 1

您有一个嵌套的列表列表。有时可以明显地观察到这一点，请注意嵌套的[]语法：

json_obj = [[{'id': None}, {'id': 'abc'}, {'id': 'def'}, {'id': 'ghi'}],
            [{'id': None}, {'id': 'jkl'}, {'id': 'mno'}, {'id': 'pqr'}]]

您的语法将适用于单个列表：

json_obj = [{'id': None}, {'id': 'abc'}, {'id': 'def'}, {'id': 'ghi'},
            {'id': None}, {'id': 'jkl'}, {'id': 'mno'}, {'id': 'pqr'}]

for d in json_obj:
    print(d['id'])

对于嵌套列表，您可以使用标准库中的itertools.chain.from_iterable：

json_obj = [[{'id': None}, {'id': 'abc'}, {'id': 'def'}, {'id': 'ghi'}],
            [{'id': None}, {'id': 'jkl'}, {'id': 'mno'}, {'id': 'pqr'}]]

from itertools import chain

for d in chain.from_iterable(json_obj):
    print(d['id'])

或者，如果没有导入，则可以使用嵌套的for循环：

for L in json_obj:
    for d in L:
        print(d['id'])

Answer 2

使用嵌套列表理解。

json_obj = [[{'id': None},{'id': '5b98d01c0835f23f538cdcab'},{'id': '5b98d0440835f23f538cdcad'},{'id': '5b98d0ce0835f23f538cdcb9'}],[{'id': None},{'id': '5b98d01c0835f23f538cd'},{'id': '5b98d0440835f23f538cd'},{'id': '5b98d0ce0835f23f538cdc'}]]
print( [[j["id"] for j in i] for i in json_obj] )

或

for i in json_obj:
    for j in i:
        print(j["id"])

输出：

[[None, '5b98d01c0835f23f538cdcab', '5b98d0440835f23f538cdcad', '5b98d0ce0835f23f538cdcb9'], [None, '5b98d01c0835f23f538cd', '5b98d0440835f23f538cd', '5b98d0ce0835f23f538cdc']]

在Python列表中访问字典

2 个答案: