从python中的数据中查找并消除对倒数

这是一个令人费解的问题.

这段代码应该定位日期中现有的对等对，以便我可以处理它们(在另一个函数中).它必须返回往复运动的位置，以及它们发生的df线.相互意义(a，b)==(b，a).

Explaining my scenario:

我使用"random"来模拟这样一个事实:在我所在的代码点上，一些倒数已经通过其他过滤器/处理消除了.

然后，我现在使用的函数"reciprocals\u locator"应该会将我指向数据帧中仍然存在交互并需要进一步处理的行.

我正在try 重构的原始解决方案是一个复杂的循环，它在循环中以数据帧的方式进行操作.所有这些我都不喜欢在我的最终代码中.

So here goes question (1):是否有更好的方法/算法或甚至外部/导入的函数以更精简的方式完成这一技巧？

如下图所示，我知道行(1,5)中存在重复项，但它也显示了倒数解(5,1)！我最初认为这是一个简单的"挑选一半"结果列表的例子，但这可能不会起作用，因为itertools.product的工作方式，坦率地说，我还没有深入研究.

在我最初的循环代码实现中也会发生同样的情况.也许我现在知道有一个更简单的itertools-based个解决方案.

**希望下面的所有代码都足够简单

### Problem scenario recreation

import itertools as it
import pandas as pd
import random as rd

## Original dataset
sourcelst = ['a','b','c','d','e']
pairs = [list(perm) for perm in it.permutations(sourcelst,2)]
df = pd.DataFrame(pairs, columns=['y','x'])

## Dataset after some sort of processing, some reciprocals are eliminated,
## but some stay and we nee to process them separately
drop_rows = [rd.randint(0, len(pairs)-1) for _ in range(2)]
df.drop(index=drop_rows, inplace=True)
df.reset_index(inplace=True, drop=True)


## Finding reciprocals

### Original LOOPS implementation, this one shows the actual pairs
### for ease of analysis
def reciprocal_pairs_orig(df):
    reciprocals = []
    row_y = 0
    while row_y < len(df.index):
        for row_x in range(len(df.index)):
            if (df['x'].iloc[row_x] == df['y'].iloc[row_y]) and (df['y'].iloc[row_x] == df['x'].iloc[row_y]):
                
                reciprocals.append([[df['y'].iloc[row_x], df['x'].iloc[row_x]],[df['y'].iloc[row_y], df['x'].iloc[row_y]]])
        row_y += 1
    return reciprocals

### List comprehension refactor, showing the pairs
def reciprocal_pairs_refactor(df):
    return [[[df['y'].iloc[row_x], df['x'].iloc[row_x]], [df['y'].iloc[row_y], df['x'].iloc[row_y]]]
        for row_y, row_x in it.product(range(len(df.index)), range(len(df.index)))
        if (df['x'].iloc[row_x] == df['y'].iloc[row_y]) and (df['y'].iloc[row_x] == df['x'].iloc[row_y])]


### This is the actual function that I want to use
def reciprocal_locator_orig(df):
    reciprocals = []
    row_y = 0
    while row_y < len(df.index):
        for row_x in range(len(df.index)):
            if (df['x'].iloc[row_x] == df['y'].iloc[row_y]) and (df['y'].iloc[row_x] == df['x'].iloc[row_y]):
                
                reciprocals.append([row_y, row_x])
        row_y += 1
    return reciprocals

### List comprehension refactor
def reciprocal_locator_refactor(df):
    return [[row_y, row_x]
        for row_y, row_x in it.product(range(len(df.index)), range(len(df.index)))
        if (df['x'].iloc[row_x] == df['y'].iloc[row_y]) and (df['y'].iloc[row_x] == df['x'].iloc[row_y])]