如何使用python itertools.compositions

我试图从所有供应商组合中找出可能的产品列表。 找出组合中的任何一个都可以生产出产品

import itertools
import pandas as pd
import numpy as np

Column = {'ID':['1','2','3','4','5'],'Supplier 1':['B','B','A','B','B'],'Supplier 2':['A','NaN','B','NaN','A']}
df=pd.DataFrame(Column)
df

# Define all Supplier Columns
cols = [c for c in df.columns if "Supplier" in c]
# get unique suppliers
suppl = np.unique(np.concatenate([df[c].dropna() for c in cols]))
result = []
for sn in range(len(suppl)):
# generate combinations of suppliers
for combi in itertools.combinations(suppl, sn+1):
result.append({combi:......

从

Desire（任一供应商均可生产）：

新代码：

from itertools import combinations, chain
import pandas as pd
import numpy as np

df = {'ID':['1','2','3','4','5'],'Supplier 1':['B','B','A','B','B'],'Supplier 2':['A',np.nan,'B',np.nan,'A']}
df=pd.DataFrame(Column)
from itertools import combinations, chain

g1 = df.groupby(['Supplier 1'])['ID'].apply(list)
g2 = df.groupby(['Supplier 2'])['ID'].apply(list)

res = (g1 + g2).to_dict()
res = [[','.join(comb), ','.join(sorted(set(chain.from_iterable([res[k] for k in comb]))))]
       for x in range(1, len(res) + 1) for comb in combinations(res.keys(), x)]

df2 = pd.DataFrame(res, columns=['Supplier', 'ID'])
print(df2)

---

这不是一个有效的解决方案，但会奏效

from itertools import combinations, chain

g1 = df.groupby(['Supplier 1'])['ID'].apply(list)
g2 = df.groupby(['Supplier 2'])['ID'].apply(list)

res = (g1 + g2).to_dict()
res = [[','.join(comb), ','.join(sorted(set(chain.from_iterable([res[k] for k in comb]))))]
       for x in range(1, len(res) + 1) for comb in combinations(res.keys(), x)]

df2 = pd.DataFrame(res, columns=['Supplier', 'ID'])
print(df2)

输出：

  Supplier         ID
0        A      1,3,5
1        B  1,2,3,4,5
2      A,B  1,2,3,4,5

---

Supplier 2

在您的字典中有

C

，但您的示例没有显示

C

@deadshot，我错了。桌子上只有A和B。感谢感谢@deadshot，但它弹出了一个错误。这种事发生在你身上吗

-->12用于范围（1，len（res）+1）内的x，用于组合中的梳（res.keys（），x）]类型错误：“float”对象不可编辑

不，我没有收到任何错误。将“NaN”替换为仍显示错误的np.nanIt。我已经把完整的代码贴在上面了。你能帮我看一下吗。非常感谢。删除np周围的引号。但同样的错误。我只是完全复制了你的代码，没有修改。不知道发生了什么

from itertools import combinations, chain

g1 = df.groupby(['Supplier 1'])['ID'].apply(list)
g2 = df.groupby(['Supplier 2'])['ID'].apply(list)

res = (g1 + g2).to_dict()
res = [[','.join(comb), ','.join(sorted(set(chain.from_iterable([res[k] for k in comb]))))]
       for x in range(1, len(res) + 1) for comb in combinations(res.keys(), x)]

df2 = pd.DataFrame(res, columns=['Supplier', 'ID'])
print(df2)

  Supplier         ID
0        A      1,3,5
1        B  1,2,3,4,5
2      A,B  1,2,3,4,5