慕少森
您可以尝试以下操作:>>> df = pd.DataFrame([[1, 2, 4], [4, 5, 6], [7, '[8', 9]])>>> df = df.astype('str')>>> df 0 1 20 1 2 41 4 5 62 7 [8 9>>> df[df.columns[[df[i].str.contains('[', regex=False).any() for i in df.columns]]] 10 21 52 [8>>>
慕码人2483693
请试试这个,希望这对你有用df = pd.DataFrame([['a', 'b', 'c','[d'], ['e','[f','g','h'],['i','j','k','l'],['m','n','o','[p']],columns=['col1','col2','col3','col4'])cols = []for col in df.columns: if df[col].str.contains('[',regex=False).any() == True: cols.append(col)df[cols]输出 col2 col40 b [d1 [f h2 j l3 n [p