giu*_*deb 9 python dataframe pandas
我有一个数据帧,我只选择包含索引值的行到df1.index.
例如:
In [96]: df
Out[96]:
A B C D
1 1 4 9 1
2 4 5 0 2
3 5 5 1 0
22 1 3 9 6
Run Code Online (Sandbox Code Playgroud)
和这些索引
In[96]:df1.index
Out[96]:
Int64Index([ 1, 3, 4, 5, 6, 7, 22, 28, 29, 32,], dtype='int64', length=253)
Run Code Online (Sandbox Code Playgroud)
我想要这个输出:
In [96]: df
Out[96]:
A B C D
1 1 4 9 1
3 5 5 1 0
22 1 3 9 6
Run Code Online (Sandbox Code Playgroud)
谢谢
jez*_*ael 23
用途isin:
df = df[df.index.isin(df1.index)]
Run Code Online (Sandbox Code Playgroud)
或者获取所有交叉索引并选择loc:
df = df.loc[df.index & df1.index]
df = df.loc[np.intersect1d(df.index, df1.index)]
df = df.loc[df.index.intersection(df1.index)]
Run Code Online (Sandbox Code Playgroud)
print (df)
A B C D
1 1 4 9 1
3 5 5 1 0
22 1 3 9 6
Run Code Online (Sandbox Code Playgroud)
编辑:
我尝试了解决方案:df = df.loc [df1.index].你认为这个解决方案是正确的吗?
解决方案不正确:
df = df.loc[df1.index]
print (df)
A B C D
1 1.0 4.0 9.0 1.0
3 5.0 5.0 1.0 0.0
4 NaN NaN NaN NaN
5 NaN NaN NaN NaN
6 NaN NaN NaN NaN
7 NaN NaN NaN NaN
22 1.0 3.0 9.0 6.0
28 NaN NaN NaN NaN
29 NaN NaN NaN NaN
32 NaN NaN NaN NaN
C:/Dropbox/work-joy/so/_t/t.py:23: FutureWarning:
Passing list-likes to .loc or [] with any missing label will raise
KeyError in the future, you can use .reindex() as an alternative.
See the documentation here:
http://pandas.pydata.org/pandas-docs/stable/indexing.html#deprecate-loc-reindex-listlike
print (df)
Run Code Online (Sandbox Code Playgroud)
Spc*_*ond 11
现在可以将索引传递给 .loc 的行索引器/切片器,您只需要确保也指定列,即:
df = df.loc[df1.index, :] # works
Run Code Online (Sandbox Code Playgroud)
并不是
df = df.loc[df1.index] # won't work
Run Code Online (Sandbox Code Playgroud)
IMO 这与 .loc 的预期用法更简洁/一致
| 归档时间: |
|
| 查看次数: |
12956 次 |
| 最近记录: |