這是我試圖轉換為字典的元組：
rule_tuple = tuple((('rule1', 'col1', 'val1'), ('rule1', 'col2', 'val2'), ('rule1', 'col3', 'val3'), ('rule2', 'col1', 'val1'), ('rule2', 'col2', 'val2')))
這是預期的輸出：
{'rule1': {'col1': 'val1', 'col2': 'val2', 'col3': 'val3'},
'rule2': {'col1': 'val1', 'col2': 'val2'}}
這是我嘗試過的：
dict((rule, (dict((c, v) for c, v in (col, val)))) for rule, col, val in rule_tuple)
uj5u.com熱心網友回復：
您可以回圈并將外部鍵的默認值設定為空字典，然后分配：
rule_tuple = (('rule1', 'col1', 'val1'), ('rule1', 'col2', 'val2'), ('rule1', 'col3', 'val3'), ('rule2', 'col1', 'val1'), ('rule2', 'col2', 'val2'))
d = {}
for k1, k2, v in rule_tuple:
d.setdefault(k1, {})[k2] = v
留給你d：
{'rule1': {'col1': 'val1', 'col2': 'val2', 'col3': 'val3'},
'rule2': {'col1': 'val1', 'col2': 'val2'}}
uj5u.com熱心網友回復：
tuple()定義rule_tuple. _ 修復此問題后：
from collections import defaultdict
result = defaultdict(dict)
rule_tuple = (('rule1', 'col1', 'val1'), ('rule1', 'col2', 'val2'), ('rule1', 'col3', 'val3'), ('rule2', 'col1', 'val1'), ('rule2', 'col2', 'val2'))
for rule, col, val in rule_tuple:
result[rule].update({col:val})
print(result)
輸出：
defaultdict(<class 'dict'>, {'rule1': {'col1': 'val1', 'col2': 'val2', 'col3': 'val3'}, 'rule2': {'col1': 'val1', 'col2': 'val2'}})
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