根据元组索引排序行

huangapple go评论69阅读模式
英文:

Sort Rows Based on Tuple Index

问题

Lambda函数如下,它可以根据第3个元组索引元素的相应'Num'值进行排序:

dataTable = dataTable.reindex(sorted(dataTable.index, key=lambda x: (not isinstance(x, tuple), x, -dataTable.loc[x, 'Num'])))
英文:

Overview:

Pandas dataframe with a tuple index and corresponding 'Num' column:

Index                               Num

('Total', 'A')                      23

('Total', 'A', 'Pandas')            3

('Total', 'A', 'Row')               7

('Total', 'A', 'Tuple')             13

('Total', 'B')                      35

('Total', 'B', 'Rows')              12

('Total', 'B', 'Two')               23

('Total', 'C')                      54

('Total', 'C', 'Row')               54

Total                               112

The index and 'Num' column are already sorted with a lambda function by Alphabetical Order and based on the length of tuple elements:

dataTable = dataTable.reindex(sorted(dataTable.index, key=lambda x: (not isinstance(x, tuple), x)))

Problem:

Now, I want to sort only the 3rd tuple index element based on it's corresponding 'Num' value. Here would be an updated example of the dataframe:

Index                               Num

('Total', 'A')                      23

('Total', 'A', 'Tuple')             13

('Total', 'A', 'Row')               7

('Total', 'A', 'Pandas')            3

('Total', 'B')                      35

('Total', 'B', 'Two')               23

('Total', 'B', 'Rows')              12

('Total', 'C')                      54

('Total', 'C', 'Row')               54

Total                               112

Question:

What Lambda function can achieve this?

答案1

得分: 1

您可以尝试

```py
def fn(x):
    vals = x.sort_values(by='Num', ascending=False)
    df.loc[x.index] = vals.values

m = df['Index'].apply(len).eq(3)
df[m].groupby(df.loc[m, 'Index'].str[1], group_keys=False).apply(fn)

print(df)

打印:

                Index  Num
0          (Total, A)   23
1   (Total, A, Tuple)   13
2     (Total, A, Row)    7
3  (Total, A, Pandas)    3
4          (Total, B)   35
5     (Total, B, Two)   23
6    (Total, B, Rows)   12
7          (Total, C)   54
8     (Total, C, Row)   54
9               Total  112

初始 df

                Index  Num
0          (Total, A)   23
1  (Total, A, Pandas)    3
2     (Total, A, Row)    7
3   (Total, A, Tuple)   13
4          (Total, B)   35
5    (Total, B, Rows)   12
6     (Total, B, Two)   23
7          (Total, C)   54
8     (Total, C, Row)   54
9               Total  112

<details>
<summary>英文:</summary>

You can try:

```py
def fn(x):
    vals = x.sort_values(by=&#39;Num&#39;, ascending=False)
    df.loc[x.index] = vals.values

m = df[&#39;Index&#39;].apply(len).eq(3)
df[m].groupby(df.loc[m, &#39;Index&#39;].str[1], group_keys=False).apply(fn)

print(df)

Prints:

                Index  Num
0          (Total, A)   23
1   (Total, A, Tuple)   13
2     (Total, A, Row)    7
3  (Total, A, Pandas)    3
4          (Total, B)   35
5     (Total, B, Two)   23
6    (Total, B, Rows)   12
7          (Total, C)   54
8     (Total, C, Row)   54
9               Total  112

Initial df:

                Index  Num
0          (Total, A)   23
1  (Total, A, Pandas)    3
2     (Total, A, Row)    7
3   (Total, A, Tuple)   13
4          (Total, B)   35
5    (Total, B, Rows)   12
6     (Total, B, Two)   23
7          (Total, C)   54
8     (Total, C, Row)   54
9               Total  112

huangapple
  • 本文由 发表于 2023年6月2日 01:05:51
  • 转载请务必保留本文链接:https://go.coder-hub.com/76384189.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定