pandas 根据行中的值对熊猫数据框的列进行排序
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Order columns of a pandas dataframe according to the values in a row
提问by Curious
How do I order columns according to the values of the last row? In the example below, my final dataframe should have columns in the following order: 'ddd' 'aaa' 'ppp' 'fff'.
如何根据最后一行的值对列进行排序？在下面的示例中，我的最终数据框应具有按以下顺序排列的列：'ddd' 'aaa' 'ppp' 'fff'。
>>> df = DataFrame(np.random.randn(10, 4), columns=['ddd', 'fff', 'aaa', 'ppp'])
>>> df
ddd fff aaa ppp
0 -0.177438 0.102561 -1.318710 1.321252
1 0.980348 0.786721 0.374506 -1.411019
2 0.405112 0.514216 1.761983 -0.529482
3 1.659710 -1.017048 -0.737615 -0.388145
4 -0.472223 1.407655 -0.129119 -0.912974
5 1.221324 -0.656599 0.563152 -0.900710
6 -1.816420 -2.898094 -0.232047 -0.648904
7 2.793261 0.568760 -0.850100 0.654704
8 -2.180891 2.054178 -1.050897 -1.461458
9 -1.123756 1.245987 -0.239863 0.359759
回答by DSM
[updated to simplify]
[更新以简化]
tl;dr:
tl;博士：
In [29]: new_columns = df.columns[df.ix[df.last_valid_index()].argsort()]
In [30]: df[new_columns]
Out[30]:
aaa ppp fff ddd
0 0.328281 0.375458 1.188905 0.503059
1 0.305457 0.186163 0.077681 -0.543215
2 0.684265 0.681724 0.210636 -0.532685
3 -1.134292 1.832272 0.067946 0.250131
4 -0.834393 0.010211 0.649963 -0.551448
5 -1.032405 -0.749949 0.442398 1.274599
Some explanation follows. First, build the DataFrame:
一些解释如下。首先，构建DataFrame：
In [24]: df = pd.DataFrame(np.random.randn(6, 4), columns=['ddd', 'fff', 'aaa', 'ppp'])
In [25]: df
Out[25]:
ddd fff aaa ppp
0 0.503059 1.188905 0.328281 0.375458
1 -0.543215 0.077681 0.305457 0.186163
2 -0.532685 0.210636 0.684265 0.681724
3 0.250131 0.067946 -1.134292 1.832272
4 -0.551448 0.649963 -0.834393 0.010211
5 1.274599 0.442398 -1.032405 -0.749949
Get the last row:
获取最后一行：
In [26]: last_row = df.ix[df.last_valid_index()]
Get the indices that would sort it:
获取对其进行排序的索引：
In [27]: last_row.argsort()
Out[27]:
ddd 2
fff 3
aaa 1
ppp 0
Name: 5, Dtype: int32
Use this to index df:
使用它来索引df：
In [28]: df[last_row.argsort()]
Out[28]:
aaa ppp fff ddd
0 0.328281 0.375458 1.188905 0.503059
1 0.305457 0.186163 0.077681 -0.543215
2 0.684265 0.681724 0.210636 -0.532685
3 -1.134292 1.832272 0.067946 0.250131
4 -0.834393 0.010211 0.649963 -0.551448
5 -1.032405 -0.749949 0.442398 1.274599
Profit!
利润！
回答by DSM
The sort_valuesmethod does this directly when given axis=1argument.
该sort_values方法在给定axis=1参数时直接执行此操作。
sorted_df = df.sort_values(df.last_valid_index(), axis=1)
So, it is no longer necessary to transpose the dataframe to sort by a row. Also, the sortmethod is now deprecated.
因此，不再需要将数据帧转置为按行排序。此外，该sort方法现已弃用。
回答by sanguineturtle
I would use transpose and the sort method (which works on columns):
我会使用转置和排序方法（适用于列）：
df = pd.DataFrame(np.random.randn(10, 4), columns=['ddd', 'fff', 'aaa', 'ppp'])
last_row_name = df.index[-1]
sorted_df = df.T.sort(columns=last_row_name).T
You might suffer a performance hit but it is quick and easy.
您可能会遭受性能损失，但它既快速又简单。
回答by husimu
df=df[df.iloc[-1,:].sort_values().index]
This works
这有效
