我的“数据框”包含两列，第一列是SKU编号，第二列是每个SKU编号的部件号。有些SKU共享相同的部件号，如何找到这些共享部件号的SKU？
import pandas as pd
table_teste = pd.read_csv("table.csv")
print(table_teste)[see in the picture attached here the screenshot of the input vales][1]
Output:
SKU Part Number
0 4679343 126420
1 4679343 489136
2 4679343 490202
3 4679343 490282
4 4679343 491971
5 4679343 492963
6 4679343 626681
7 4679343 627996
8 4679343 628361
9 4679343 628379
10 4679343 628379
11 4679343 628408
12 4679343 628531
13 4679343 1105601
14 4679343 1140073
15 4679343 2169104
16 4679343 2169104
17 4679343 2169142
18 4679343 2185762
19 4679343 2194712
20 4679343 2195058
21 4679343 2256086
22 4679343 2315522
23 4679343 2315522
24 4679343 2319835
25 4679343 8314101
26 4679343 8314102
27 4679343 8314229
28 4679343 8314231
29 4679343 8314232
... ... ...
73953 WRO80CKDWA W11234774
73954 WRO80CKDWA W11239503
73955 WRO80CKDWA W11240332
73956 WRO80CKDWA W11240358
73957 WRO80CKDWA W11240361
73958 WRO80CKDWA W11240362
73959 WRO80CKDWA W11240363
73960 WRO80CKDWA W11282632
73961 WRO80CKDWA W11282632
73962 WRO80CKDWA W11293453
73963 WRO80CKDWA W11294381
73964 WRO80CKDWA W11294503
73965 WRO80CKDWA W11298984
73966 WRO80CKDWA W11308860
73967 WRO80CKDWA W11308879
73968 WRO80CKDWA W11314128
73969 WRO80CKDWA W11317776
73970 WRO80CKDWA W11323281
73971 WRO80CKDWA W11323282
73972 WRO80CKDWA W11323283
73973 WRO80CKDWA W11323284
73974 WRO80CKDWA W11366199
73975 WRO80CKDWA W11366205
73976 WRO80CKDWA W11366209
73977 WRO80CKDWA W11366214
73978 WRO80CKDWA W11366215
73979 WRO80CKDWA W11370412
73980 WRO80CKDWA W11370419
73981 WRO80CKDWA W11370494
73982 WRO80CKDWA ZCOMP_FREIGHT
现在，我需要生成一个矩阵，该矩阵在行中具有SKU编号，在列中具有相同的SKU编号，并且矩阵中包含SKU编号1和SKU编号2的组合所共享的零件编号的数量相同。 SKU编号2与SKU编号3相同，依此类推。共有182个SKU编号。
谢谢
答案 0 :(得分：1)
查找具有1个以上SKU的所有部件号：
partNumber_w_dupSKU = data %>%
group_by(partNumber) %>%
summarize(n_SKU = n_distinct(SKU)) %>%
ungroup() %>%
filter(n_SKU > 1)
找到与这些部件号关联的所有SKU：
data %>%
arrange(SKU) %>%
filter(partNumber %in% partNumber_w_dupSKU$partNumber)
答案 1 :(得分：0)
您可以在零件号上使用groupby（），从而将数据框相应地分组 如果对SKU编号进行分组，则它将显示具有SKU编号的数据框，该数据框共享一个公共零件号 反之亦然
答案 2 :(得分：0)
您可以尝试使用groupby，将组转换为列表并重置索引。
# dictionary of sku number as key and value as part number
# I'm assuming this is how the df might look like
d = {1: 2, 2: 3, 3: 2, 4: 2, 5: 3, 6: 2, 7: 3}
# making a dataframe out of the dict to resemble df in que
df = pd.DataFrame(d.items(), columns=['SKU Number', 'Part Number'])
df
Output:
SKU Number Part Number
0 1 2
1 2 3
2 3 2
3 4 2
4 5 3
5 6 2
6 7 3
# first groupby part numbers
g = df.groupby('Part Number')
# convert groups to list of two-SKU combinations and then reset index to create a new data
x = g['SKU Number'].apply(itertools.combinations(x, 2))).reset_index(name='SKU numbers')
x
Output:
Part Number SKU numbers
0 2 [(1, 3), (1, 4), (1, 6), (3, 4), (3, 6), (4, 6)]
1 3 [(2, 5), (2, 7), (5, 7)]
^现在，每个零件号都有所有两个成员的SKU组合。让我们在SKU编号列中展开列表。
x = x.explode('SKU numbers')
x
out:
Part Number SKU numbers
0 2 (1, 3)
0 2 (1, 4)
0 2 (1, 6)
0 2 (3, 4)
0 2 (3, 6)
0 2 (4, 6)
1 3 (2, 5)
1 3 (2, 7)
1 3 (5, 7)
现在，我们需要对SKU号对进行分组并计算与之关联的零件号
x = x.groupby('SKU numbers').count().reset_index()
x
out:
SKU numbers Part Number
0 (1, 3) 1
1 (1, 4) 1
2 (1, 6) 1
3 (2, 5) 1
4 (2, 7) 1
5 (3, 4) 1
6 (3, 6) 1
7 (4, 6) 1
8 (5, 7) 1
^现在，每个SKU对都有零件编号的计数。让我们构造矩阵。
import numpy as np
indexes = x['SKU numbers'].values
part_number_counts = x['Part Number'].values
# in my small case, we have 7 unique SKUs
unique_SKUs = 7
# creating a zero matrix so that we can populate part num counts
# for each SKU pair
a = np.zeros((unique_SKUs, unique_SKUs))
a
out:
array([[0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0.]])
# split [(x1, y1), (x2, y2) ...] to rows -> [x1, x2 ...]
# columns -> [y1, y2 ....]
rows, columns = map(np.array , zip(*indexes))
# rows-1, columns-1 are done to make index 0-based
a[rows-1, columns-1] = part_number_counts
a
out:
array([[0., 0., 1., 1., 0., 1., 0.],
[0., 0., 0., 0., 1., 0., 1.],
[0., 0., 0., 1., 0., 1., 0.],
[0., 0., 0., 0., 0., 1., 0.],
[0., 0., 0., 0., 0., 0., 1.],
[0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0.]])
对于最后一部分，我使用我的索引（SKU对），将它们转换为基于0的索引，并将其对应的part_number_counts更新为零矩阵，以获得结果矩阵。
结果矩阵的形状为（unique_SKU_numbers，unique_SKU_numbers），值i，j对应于part_number_counts