循环遍历 2 个数据帧并采用特定列的模式

Question

给定的是 df1（其中包含每个商店销售最多和最少的产品）：

id   most_sold_A  most_sold_B  most_sold_C  least_sold_A  least_sold_B  least_sold_C
1     1             0           0             0            1             1
2     0             1           0             1            0             0
3     0             1           1             1            0             0

并且df2（包含2个店铺之间的距离）也给出了：

id1   id2   distance 
1     2      0.5
1     3      3.0
2     3      0.2

生成的数据帧应该

检查哪些 shop_ids 在每个 shopid 的 1k 距离内
采用most_sold_product模式，1k以内压倒所有参赛者
取least_sold_product模式1k以内的所有参赛者

结果 df:

id   most_sold_A  most_sold_B  most_sold_C  least_sold_A  least_sold_B  least_sold_C    /
1     1             0           0             0            1             1
2     0             1           0             1            0             0
3     0             1           1             1            0             0

most_sold_competition_within_1k   least_sold_competition_within_1k
B                                    A
[A,B,C]                              [A,B,C]
B                                    A

编辑

df1 = pd.DataFrame([[1,1,0,0,0,1,1],
[2,0,1,0,1,0,0],
[3,0,1,1,1,0,0]],columns = ["id","most_sold_A","most_sold_B","most_sold_C","least_sold_A","least_sold_B","least_sold_C"])
df2 = pd.DataFrame([[1,2,0.5],
[1,3,3.0],
[2,3,0.2]], columns = ["id1","id2","distance"])

Answer 1

似乎“棘手”的部分是为每家商店找到相关的竞争对手。我敢肯定还有更多优雅的解决方案，但 straight-forward 一个是：

def find_competitors(x, df2):
    shops = np.unique(df2[(df2.id1==x.id) | (df2.id2 == x.id)][['id1','id2']])
    competitors = np.delete(shops, np.argwhere(shops == x.id))
    return competitors

df2 = df2[df2.distance<=1]
df1['competitors'] = df1.apply(lambda x: find_competitors(x, df2),axis=1)

现在，对于每家商店，您现在是相关的竞争对手，您可以通过简单地遍历每家商店的竞争对手来找到其他 2 个问题的答案（竞争对手的销量最高和销量最低的产品）。我希望已经足够清楚了。

更新

为了找到竞争对手least/most的产品，您可以使用：

most_cols = [col for col in df1.columns if 'most' in col]

def find_competitors_by_metric(x, metric_cols):
    competitors_metric = df1[df1.id.isin(x.competitors)][metric_cols]
    return competitors_metric.T[competitors_metric.any()].T.columns

most_for_competitors = df1.apply(lambda x: find_competitors_by_metric(x,most_cols),axis=1)

现在您可以向该函数发送您想为商店的竞争对手计算哪些指标（假设这些指标存在于数据框中）。

Answer 2

我想出了一些东西，但我认为它可以进一步优化。这个想法是首先过滤范围内的竞争对手，然后加入，然后用 .apply():

计算结果

import numpy as np
import pandas as pd

df1 = pd.DataFrame([[1,1,0,0,0,1,1],
[2,0,1,0,1,0,0],
[3,0,1,1,1,0,0]],columns = ["id","most_sold_A","most_sold_B","most_sold_C","least_sold_A","least_sold_B","least_sold_C"])
df2 = pd.DataFrame([[1,2,0.5],
[1,3,3.0],
[2,3,0.2]], columns = ["id1","id2","distance"])

df2 = pd.concat([df2,df2[["id2","id1","distance"]].rename(columns = {"id2":"id1","id1":"id2"})]).reset_index()[["id1","id2","distance"]]
df2["id2"] = df2["id2"].astype(str)
df2 = df2[df2["distance"]<1][["id1","id2"]].groupby("id1").agg({'id2': ','.join}).reset_index()

df3 = pd.merge(df1,df2,how = 'left',left_on="id", right_on="id1")

most_cols = [col for col in df3.columns if 'most' in col]
least_cols = [col for col in df3.columns if 'least' in col]

df3["most_sold_competition_within_1k"] = df3.apply(lambda x: [df3[df3["id"]==int(elem)][most_cols].columns[[df3[df3["id"]==int(elem)][most_cols].values == 1][0][0]] for elem in x["id2"].split(",")],axis = 1)
df3["least_sold_competition_within_1k"] = df3.apply(lambda x: [df3[df3["id"]==int(elem)][least_cols].columns[[df3[df3["id"]==int(elem)][least_cols].values == 1][0][0]] for elem in x["id2"].split(",")],axis = 1)

df3 = df3[["id"]+most_cols+least_cols+["most_sold_competition_within_1k","least_sold_competition_within_1k"]]

df3

输出：

    id  most_sold_A most_sold_B most_sold_C least_sold_A    least_sold_B    least_sold_C    most_sold_competition_within_1k   least_sold_competition_within_1k
0   1   1           0           0           0               1               1              [[most_sold_B]]              [[least_sold_A]]
1   2   0           1           0           1               0               0    [[most_sold_B, most_sold_C], [most_sold_A]  [[least_sold_A], [least_sold_B, least_sold_C]]
2   3   0           1           1           1               0               0      [[most_sold_B]]                            [[least_sold_A]]

循环遍历 2 个数据帧并采用特定列的模式

Looping through 2 dataframes and take mode of specific columns

python

loops

nested

mode

pandas