Python 将数据帧相乘,得到列中值的乘积
我需要帮助创建Python函数以实现以下目标: 1) 将3个数据帧作为输入(包含索引列,在第二列中包含关联的整数或浮点值)。这些定义如下:Python 将数据帧相乘,得到列中值的乘积,python,pandas,recursion,dataframe,iteration,Python,Pandas,Recursion,Dataframe,Iteration,我需要帮助创建Python函数以实现以下目标: 1) 将3个数据帧作为输入(包含索引列,在第二列中包含关联的整数或浮点值)。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]
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”),该数据帧组织前3列中3个索引的所有可能组合
3) 在第4列中,附加来自三个源数据帧的所有相关“值”的数学积。
因此,函数的DataFrame输出应如下所示:
非常感谢你的帮助
Colin使用所有索引和列的
product
,并通过构造函数创建DataFrame
,对于多个所有列使用:
备选方案是按列表理解的多个所有值:
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
具有数据帧列表的动态解决方案
,只需在每个数据帧中使用相同的列名值
:
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
@Bharath猜测
itertools
最好确保itertools
和pd。多索引
生成相同的产品,或者匹配的值错误。先生,产品的多索引非常好。Op在输出中添加了C类,但没有提出问题。感谢耶兹雷尔-我注意到这与我的预期输出并不完全匹配。看起来好像我的“C类”在输出组合中丢失了。我想这是一个简单的改变?此外,我想我还可以提到,我希望将此解决方案扩展到可变数量的输入数据帧(它不会像本例中那样总是3帧)。@ColinBlyth这是一个非常好的解决方案,现在您可以投票了,所以请投票并感谢回答者。
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