只是为了扩展nitin的答案

Just to expand on nitin's answer set_index :

In [100]:

df.set_index('head0')

Out[100]:
       head1  head2  head3
head0                     
bar       32      3    100
bix       22    NaN    NaN
foo       11      1    NaN
qux      NaN     10    NaN
xoo      NaN      2     20

请注意，这将返回df，因此您必须像df = df.set_index('head0')那样分配回df或设置参数inplace=True:df.set_index('head0', inplace=True)

Note that this returns the df, so you either have to assign back to the df like: df = df.set_index('head0') or set param inplace=True: df.set_index('head0', inplace=True)

您也可以直接分配给索引:

You can also directly assign to the index:

In [99]:

df.index = df['head0']
df
Out[99]:
      head0  head1  head2  head3
head0                           
bar     bar     32      3    100
bix     bix     22    NaN    NaN
foo     foo     11      1    NaN
qux     qux    NaN     10    NaN
xoo     xoo    NaN      2     20

请注意，执行上述操作将需要删除多余的"head0"列，这可以通过调用

Note that doing the above will require you to drop the extraneous 'head0' column which can be done by calling drop like so: df.drop('head0', axis=1)