我必须删除一些列,有时列名是要键入的hart,所以我希望获得一个具有相应序列号的列表或元组或数组,然后我可以使用df1.drop(df1.columns[[0, 1, 3]], axis=1)删除它们。
对熊猫来说,像这样做的最快方法是什么?
In [2]: df1.columns
Out[2]: Index(['Unnamed: 0', 'Unnamed: 1', 'Unnamed: 2', 'Unnamed: 3', '上班', '下班',
'上班.1', '下班.1', 'Unnamed: 8', 'Unnamed: 9', 'Unnamed: 10',
'Unnamed: 11'],
dtype='object'
In [3]: a = df1.columns.tolist()
b = list(range(len(df1.columns)))
tuple(zip(a, b))
Out[3]: (('Unnamed: 0', 0),
('Unnamed: 1', 1),
('Unnamed: 2', 2),
('Unnamed: 3', 3),
('上班', 4),
('下班', 5),
('上班.1', 6),
('下班.1', 7),
('Unnamed: 8', 8),
('Unnamed: 9', 9),
('Unnamed: 10', 10),
('Unnamed: 11', 11))
In [4]: df1.drop(df1.columns[6:], axis=1)发布于 2016-09-30 17:34:00
最后,我找到了一种又快又简单的方法:
for i, element in enumerate(df.columns):
print(i, element)https://stackoverflow.com/questions/39360107
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