我有一個列表，其中包含這樣的字典：json_obj = [[{'id': None},{'id': '5b98d01c0835f23f538cdcab'},{'id': '5b98d0440835f23f538cdcad'},{'id': '5b98d0ce0835f23f538cdcb9'}],[{'id': None},{'id': '5b98d01c0835f23f538cd'},{'id': '5b98d0440835f23f538cd'},{'id': '5b98d0ce0835f23f538cdc'}]]我希望它存儲在這樣的列表列表中：y=[['None','5b98d01c0835f23f538cdcab','5b98d0440835f23f538cdcad','5b98d0ce0835f23f538cdcb9'],['None','5b98d01c0835f23f538cd','5b98d0440835f23f538cd','5b98d0ce0835f23f538cdc']]為了從我嘗試過的字典中讀取 idfor d in json_obj: print(d['id'])但是我在上面的代碼中看到了這個錯誤：TypeError: list indices must be integers or slices, not str
2 回答
心有法竹
TA貢獻1866條經驗 獲得超5個贊
您有一個嵌套的列表列表。有時有助于明顯地觀察這一點，注意嵌套[]語法：
json_obj = [[{'id': None}, {'id': 'abc'}, {'id': 'def'}, {'id': 'ghi'}],
[{'id': None}, {'id': 'jkl'}, {'id': 'mno'}, {'id': 'pqr'}]]
您的語法適用于單個列表：
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'])
對于嵌套列表，您可以使用itertools.chain.from_iterable標準庫中的：
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，如果沒有導入，您可以使用嵌套for循環：
for L in json_obj:
for d in L:
print(d['id'])
慕后森
TA貢獻1802條經驗 獲得超5個贊
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] )
或者
for i in json_obj:
for j in i:
print(j["id"])
輸出：
[[None, '5b98d01c0835f23f538cdcab', '5b98d0440835f23f538cdcad', '5b98d0ce0835f23f538cdcb9'], [None, '5b98d01c0835f23f538cd', '5b98d0440835f23f538cd', '5b98d0ce0835f23f538cdc']]
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