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饮歌长啸
使用嵌套元组理解和isinstance:output = [tuple(j for j in i if not isinstance(j, str)) for i in ListTuples]输出:[(100,), (80,), (20,), (40,), (40,)]请注意,元组中有尾随逗号以将它们与例如(100)相同的100.
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慕侠2389804
由于提取每个元组的第一项就足够了,您可以解包并使用列表理解。对于元组列表:res = [(value,) for value, _ in ListTuples] # [(100,), (80,), (20,), (40,), (40,)]如果您只需要一个整数列表:res = [value for value, _ in ListTuples] # [100, 80, 20, 40, 40]对于后者的功能替代品,您可以使用operator.itemgetter:from operator import itemgetterres = list(map(itemgetter(0), ListTuples)) # [100, 80, 20, 40, 40]
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汪汪一只猫
尝试这个ListTuples = [(100, 'AAA'), (80, 'BBB'), (20, 'CCC'), (40, 'DDD'), (40, 'EEE')]ListInt = []ListStr = []for item in ListTuples: strAux = '' intAux = '' for i in range(0, len(item)): if(isinstance(item[i], str)): strAux+=item[i] else: intAux+=str(item[i]) ListStr.append(strAux) ListInt.append(intAux)print(ListStr)'''TO CONVERT THE STR LIST TO AN INT LIST'''output= list(map(int, ListInt)) print(output)输出是 [100, 80, 20, 40, 40]