我想知道是否有一種Python式的方法來(lái)壓縮這個(gè)嵌套的for循環(huán)：dict = { "keyA": { "subkey1": { "A1a": "frog", "A1b": "dog", "A1c": "airplane" } }, "keyA": { "subkey2": { "A2a": "cat" } }, "keyB": { "subkey1": { "B1a": "Zorba", "B1q": ["popcorn", -34] } }, "keyB": { "subkey2": { "B2z": "A Man A Plan A Canal", "B2e": "armadillo", "B2w": [1, 3, "jump"] } }, "keyC": { "subkey1": { "C1a": 3.14, "C1z": { "aaa": "dishwater", "bbb": "Dishwalla" }, "C1x": "bat" } }, "keyC": { "subkey2": { "C2a": None, "C2b": 123 } }}for key in dict.keys(): for subsubkey in dict[key]["subkey2"].keys(): print(key+":"+subsubkey)輸出：keyA:A2akeyB:B2zkeyB:B2ekeyB:B2wkeyC:C2akeyC:C2b
1 回答
阿晨1998
TA貢獻(xiàn)2037條經(jīng)驗(yàn) 獲得超6個(gè)贊
解決這個(gè)問(wèn)題的一種 Pythonic 方法是使用列表理解。這允許您在一行中定義一個(gè)列表，遵循for您已經(jīng)布置的循環(huán)結(jié)構(gòu)。工作版本可能類似于：
final_keys?=?[(first_key,?second_key)?for?first_key?in?dict.keys()?for?second_key?in?dict[first_key]['subkey2'].keys()]
輸出（從您的數(shù)據(jù)集）：
[('keyA',?'A2a'),?('keyB',?'B2z'),?('keyB',?'B2e'),?('keyB',?'B2w'),?('keyC',?'C2a'),?('keyC',?'C2b')]添加回答
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