Question

我有如下餐厅销售明细。

+----------+------------+---------+----------+
| Location | Units Sold | Revenue | Footfall |
+----------+------------+---------+----------+
| Loc - 01 |        100 | 1,150   |       85 |
+----------+------------+---------+----------+

我想从下表的餐厅数据中找到与以上餐厅最相关的餐厅

+----------+------------+---------+----------+
| Location | Units Sold | Revenue | Footfall |
+----------+------------+---------+----------+
| Loc - 02 |        100 | 1,250   |       60 |
| Loc - 03 |         90 | 990     |       90 |
| Loc - 04 |        120 | 1,200   |       98 |
| Loc - 05 |        115 | 1,035   |       87 |
| Loc - 06 |         89 | 1,157   |       74 |
| Loc - 07 |        110 | 1,265   |       80 |
+----------+------------+---------+----------+

请指导我如何使用python或pandas完成此操作。注意：-相关性指的是Units Sold，Revenue和Footfall上最匹配/相似的餐厅。

Answer 1

如果应将您的相关性描述为最小欧氏距离，则解决方案是：

#convert columns to numeric
df1['Revenue'] = df1['Revenue'].str.replace(',','').astype(int)
df2['Revenue'] = df2['Revenue'].str.replace(',','').astype(int)

#distance of all columns subtracted by first row of first DataFrame
dist = np.sqrt((df2['Units Sold']-df1.loc[0, 'Units Sold'])**2 + 
               (df2['Revenue']- df1.loc[0, 'Revenue'])**2 + 
               (df2['Footfall']- df1.loc[0, 'Footfall'])**2)

print (dist)
0    103.077641
1    160.390149
2     55.398556
3    115.991379
4     17.058722
5    115.542200
dtype: float64

#get index of minimal value and select row of second df
print (df2.loc[[dist.idxmin()]])
   Location  Units Sold  Revenue  Footfall
4  Loc - 06          89     1157        74

Answer 2

可能是执行此操作的更好方法，但是我认为这很有效，因为它很冗长，所以我尝试使代码保持干净和可读性：

首先，让我们使用this帖子中的自定义numpy函数。

import numpy as np
import pandas as pd


def find_nearest(array, value):
    array = np.asarray(array)
    idx = (np.abs(array - value)).argmin()
    return array[idx]

然后使用数据框的数组，传入第一个数据框的值以找到最接近的匹配项。

us = find_nearest(df2['Units Sold'],df['Units Sold'][0])
ff = find_nearest(df2['Footfall'],df['Footfall'][0])
rev = find_nearest(df2['Revenue'],df['Revenue'][0])

print(us,ff,rev,sep=',')
100,87,1157

然后返回具有所有三个条件的数据帧

    new_ df = (df2.loc[
    (df2['Units Sold'] == us) |
    (df2['Footfall'] == ff) |
    (df2['Revenue'] == rev)])

这给了我们：

    Location    Units Sold  Revenue Footfall
0   Loc - 02    100         1250    60
3   Loc - 05    115         1035    87
4   Loc - 06    89          1157    74

Answer 3

修复数据

对于数字列。我可能对此概括了太多。另外，我将索引设置为'Location'列

def fix(d):
    d.update(
        d.astype(str).replace(',', '', regex=True)
         .apply(pd.to_numeric, errors='ignore')
    )
    d.set_index('Location', inplace=True)

fix(df1)
fix(df2)

曼哈顿距离

df2.loc[[df2.sub(df1.loc['Loc - 01']).abs().sum(1).idxmin()]]

          Units Sold Revenue  Footfall
Location                              
Loc - 06          89    1157        74

欧几里得距离

df2.loc[[df2.sub(df1.loc['Loc - 01']).pow(2).sum(1).pow(.5).idxmin()]]

          Units Sold Revenue  Footfall
Location                              
Loc - 06          89    1157        74

寻找最相关的项目

3 个答案:

修复数据

曼哈顿距离

欧几里得距离