Question

我需要帮助创建Python函数来实现以下目的：

1）将3个Pandas数据帧作为输入（包含索引列，以及第二列中的关联整数或浮点值）。这些定义如下：

import pandas as pd

df1=pd.DataFrame([['placementA',2],['placementB',4]],columns=
['placement','value'])
df1.set_index('placement',inplace=True)

df2=pd.DataFrame([['strategyA',1],['strategyB',5],['strategyC',6]],columns=
['strategy','value'])
df2.set_index('strategy',inplace=True)

df3=pd.DataFrame([['categoryA',1.5],['categoryB',2.5]],columns=
['category','value'])
df3.set_index('category',inplace=True)

2）使用这三个数据帧，创建一个新的数据帧（'df4'），用于组织前3列中3个索引的所有可能组合;

3）在第4列中，附加三个源数据帧中所有相关“值”的数学乘积。因此，函数的DataFrame输出应如下所示：https://ibb.co/cypEY6

非常感谢您的帮助。

科林

Answer 1

使用所有索引和列的product并按构造函数创建DataFrame，对于多个列，使用prod：

from  itertools import product

names = ['placement','strategy','category']
mux = pd.MultiIndex.from_product([df1.index, df2.index, df3.index], names=names)
df = (pd.DataFrame(list(product(df1['value'], df2['value'], df3['value'])), index=mux)
       .prod(1).reset_index(name='mult'))
print (df)
     placement   strategy   category  mult
0   placementA  strategyA  categoryA   3.0
1   placementA  strategyA  categoryB   5.0
2   placementA  strategyB  categoryA  15.0
3   placementA  strategyB  categoryB  25.0
4   placementA  strategyC  categoryA  18.0
5   placementA  strategyC  categoryB  30.0
6   placementB  strategyA  categoryA   6.0
7   placementB  strategyA  categoryB  10.0
8   placementB  strategyB  categoryA  30.0
9   placementB  strategyB  categoryB  50.0
10  placementB  strategyC  categoryA  36.0
11  placementB  strategyC  categoryB  60.0

备选方案是列表理解的所有值multiple：

import operator
import functools
from  itertools import product

names = ['placement','strategy','category']
a = list(product(df1.index, df2.index, df3.index))

b = product(df1['value'], df2['value'], df3['value'])
data = [functools.reduce(operator.mul, x, 1) for x in b]

df = pd.DataFrame(a, columns=names).assign(mult=data)
print (df)
     placement   strategy   category  mult
0   placementA  strategyA  categoryA   3.0
1   placementA  strategyA  categoryB   5.0
2   placementA  strategyB  categoryA  15.0
3   placementA  strategyB  categoryB  25.0
4   placementA  strategyC  categoryA  18.0
5   placementA  strategyC  categoryB  30.0
6   placementB  strategyA  categoryA   6.0
7   placementB  strategyA  categoryB  10.0
8   placementB  strategyB  categoryA  30.0
9   placementB  strategyB  categoryB  50.0
10  placementB  strategyC  categoryA  36.0
11  placementB  strategyC  categoryB  60.0

包含DataFrames列表的动态解决方案，每个列表中只需要相同的列名value：

dfs = [df1, df2, df3]

names = ['placement','strategy','category']
a = list(product(*[x.index for x in dfs]))
b = list(product(*[x['value'] for x in dfs]))
data = pd.DataFrame(b).product(1)

df = pd.DataFrame(a, columns=names).assign(mult=data)
print (df)
     placement   strategy   category  mult
0   placementA  strategyA  categoryA   3.0
1   placementA  strategyA  categoryB   5.0
2   placementA  strategyB  categoryA  15.0
3   placementA  strategyB  categoryB  25.0
4   placementA  strategyC  categoryA  18.0
5   placementA  strategyC  categoryB  30.0
6   placementB  strategyA  categoryA   6.0
7   placementB  strategyA  categoryB  10.0
8   placementB  strategyB  categoryA  30.0
9   placementB  strategyB  categoryB  50.0
10  placementB  strategyC  categoryA  36.0
11  placementB  strategyC  categoryB  60.0

将Python Pandas数据帧相乘以获得列中的值的乘积

1 个答案: