python - 如何找到 DataFrame 行的所有组合？

Question

如果这个问题与论坛中其他人提出的问题相似，我很抱歉，但我找不到足够相似的问题。我有一个包含 9 列和 3 行的 df，我想找到这些行之间所有可能的组合。我尝试使用 itertools 包中的组合，但我似乎无法使其工作。我想要的输出将是所有可能组合的列表。谢谢你，如果它与其他问题相似，我们深表歉意。

import pandas as pd
from itertools import combinations

df1 = pd.DataFrame({"Main1": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main2": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main3": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main4": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main5": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main6": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main7": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main8": ["Outcome1", "Outcome2", "Outcome3"],
                    "Main9": ["Outcome1", "Outcome2", "Outcome3"]})

    Main1   Main2   Main3   Main4   Main5   Main6   Main7   Main8   Main9
0   Outcome1    Outcome1    Outcome1    Outcome1    Outcome1    Outcome1    Outcome1    Outcome1    Outcome1
1   Outcome2    Outcome2    Outcome2    Outcome2    Outcome2    Outcome2    Outcome2    Outcome2    Outcome2
2   Outcome3    Outcome3    Outcome3    Outcome3    Outcome3    Outcome3    Outcome3    Outcome3    Outcome3

all_combinations = list(combinations(df1, 3))

编辑:较小的样本和所需的输出:

df1 = pd.DataFrame({"Main1": ["Outcome1", "Outcome2", "Outcome3"], "Main2": ["Outcome1", "Outcome2", "Outcome3"]}) 

期望的输出是这样的:

[["Outcome1","Outcome1"], ["Outcome1","Outcome2"], ["Outcome1","Outcome3"], ["Outcome2","Outcome1"], ["Outcome2","Outcome2"], ["Outcome2","Outcome3"], ["Outcome3","Outcome1"], ["Outcome3","Outcome2"], ["Outcome3","Outcome3"]] 

Answer 1

您正在寻找列表自身的笛卡尔积。

from itertools import product

options = ['Outcome1', 'Outcome2', 'Outcome3']

result = product(options, options)
print(*result, sep='\n')

输出

('Outcome1', 'Outcome1')
('Outcome1', 'Outcome2')
('Outcome1', 'Outcome3')
('Outcome2', 'Outcome1')
('Outcome2', 'Outcome2')
('Outcome2', 'Outcome3')
('Outcome3', 'Outcome1')
('Outcome3', 'Outcome2')
('Outcome3', 'Outcome3')