对列中的所有行对执行操作 [英] Perform an operation on all pairs of rows in a column

问题描述

假设以下DataFrame:

Assume the following DataFrame:

id    A   
1     0
2     10
3     200
4     3000

我想在所有行到所有其他行之间进行计算.
例如，如果计算为lambda r1, r2: abs(r1-r2)，则输出将为(按某种顺序)

I would like to make a calculation betweeen all rows to all other rows.
For example, if the calculation were lambda r1, r2: abs(r1-r2), then the output would be (in some order)

id       col_name
1        10
2        200
3        3000
4        190
5        2990
6        2800

问题:

1. 如何仅获得上述输出?
2. 如何以最像熊猫一样"的方式将结果与创作者联系起来?

我想将所有内容尽可能地保留在单个表中，以仍然支持合理查找的方式.

I would like to keep everything in a single table as much as possible, in a way that still supports reasonable lookup.

我的数据量并不大，而且永远也不会.

The size of my data is not large, and never will be.

回答我的问题2的一种方法是

One way that would answer my question 2 would be

id       col_name    origin1    origin2
1        10          1          2
2        200         1          3
3        3000        1          4
4        190         2          3
5        2990        2          4
6        2800        3          4

我想知道这是否是标准的，是否有内置的方法，或者是否还有其他/更好的方法

And I would like to know if this is standard, and has a built in way of doing this, or if there is another/better way

推荐答案

IIUC itertools

import itertools

s=list(itertools.combinations(df.index, 2)) 
pd.Series([df.A.loc[x[1]]-df.A.loc[x[0]] for x in s ])
Out[495]: 
0      10
1     200
2    3000
3     190
4    2990
5    2800
dtype: int64

更新

s=list(itertools.combinations(df.index, 2)) 

pd.DataFrame([x+(df.A.loc[x[1]]-df.A.loc[x[0]],) for x in s ])
Out[518]: 
   0  1     2
0  0  1    10
1  0  2   200
2  0  3  3000
3  1  2   190
4  1  3  2990
5  2  3  2800

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