Python的熊猫GROUPBY，排名，然后根据自定义排名

Question

问题设置

大熊猫据帧

df = pd.DataFrame({'Group': ['A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A'], 'Subgroup': ['Group 1', 'Group 1', 'Group 1', 'Group 1', 'Group 1', 'Group 1', 'Group 2', 'Group 2', 'Group 2'], 'Keyword': ['kw 1', 'kw 1', 'kw 1', 'kw 2', '+kw +2', 'kw 2', 'kw 3', 'kw 3', 'kw 3'], 'Normalized': ['kw 1', 'kw 1', 'kw 1', 'kw 2', 'kw 2', 'kw 2', 'kw 3', 'kw 3', 'kw 3'], 'Criterion Type': ['Exact', 'Phrase', 'Broad', 'Phrase', 'Broadified', 'Exact', 'Broad', 'Exact', 'Phrase'], 'Max CPC': [1.62, 1.73, 0.87, 1.70, 0.85, 1.60, 0.99, 1.58, 1.68], 'CPC Rank': [2, 1, 3, 1, 3, 2, 3, 2, 1], 'Type Rank': [1, 2, 3, 2, 3, 1, 3, 1, 2]}) 

该值分配给获得在正确的位置的列：

df = df[['Group', 'Subgroup', 'Keyword', 'Normalized', 'Criterion Type', 'Max CPC', 'CPC Rank', 'Type Rank']] 

目标

groupby['Group', 'Subgroup', 'Normalized']，然后rankMax CPC s。接下来，我要地图关联到CPC Rank到Type Rank这是基于Criterion Type确定，我自己的自定义排名的Max CPC： {'Exact':1, 'Phrase':2, 'Broadified':3, 'Broad':4}

的结果将是New CPC列其适当Max CPC。

Answer 1

import pandas as pd 
import numpy as np 

df = pd.DataFrame({'Group': ['A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A'], 'Subgroup': ['Group 1', 'Group 1', 'Group 1', 'Group 1', 'Group 1', 'Group 1', 'Group 2', 'Group 2', 'Group 2'], 'Keyword': ['kw 1', 'kw 1', 'kw 1', 'kw 2', '+kw +2', 'kw 2', 'kw 3', 'kw 3', 'kw 3'], 'Normalized': ['kw 1', 'kw 1', 'kw 1', 'kw 2', 'kw 2', 'kw 2', 'kw 3', 'kw 3', 'kw 3'], 'Criterion Type': ['Exact', 'Phrase', 'Broad', 'Phrase', 'Broadified', 'Exact', 'Broad', 'Exact', 'Phrase'], 'Max CPC': [1.62, 1.73, 0.87, 1.70, 0.85, 1.60, 0.99, 1.58, 1.68], 'CPC Rank': [2, 1, 3, 1, 3, 2, 3, 2, 1], 'Type Rank': [1, 2, 3, 2, 3, 1, 3, 1, 2]}) 
df = df[['Group', 'Subgroup', 'Keyword', 'Normalized', 'Criterion Type', 'Max CPC', 'CPC Rank', 'Type Rank']] 

#Sort by custom priority based on their Criterion Type 
df = df.sort(['Group', 'Subgroup', 'Normalized', 'Type Rank']) 
#Reset index and drop old one 
df = df.reset_index(drop=True) 
print(df) 
#Create df1 which is a Series of the Max CPC column in its correctly ranked order 
df1 = df.sort(['Group', 'Subgroup', 'Normalized', 'CPC Rank'])['Max CPC'] 
#Reset index and drop old one 
df1 = df1.reset_index(drop=True) 
print(df1) 

#Add the df1 Series to df and name the column New CPC 
df['New CPC'] = df1 

print(df) 

这是迄今为止最有效的解决了这个问题。困难的部分是意识到我可以通过Type Ranksortdf行，所以Criterion Type行按排名排序。这意味着我希望最高的Max CPC适用于第一个，第二个最高的Max CPC适用于第二个，依此类推。

然后我所要做的就是创建一个Max CPCSeries排序方式CPC Rank。

最后，将此Series添加到现有的df。

Answer 2

我已经对每个组内的值进行了排序并使用索引分配了排序后的值。 这是你想要的吗？

df['new CPC'] = -1 
parts = [] 
grouped = df.groupby(['Group', 'Subgroup', 'Normalized']) 
for name, group in grouped: 
    type_rank_index = group.sort(columns='Type Rank').index 
    cpc_rank_index = group.sort(columns='CPC Rank').index 
    group.loc[type_rank_index, 'new CPC'] = group.loc[cpc_rank_index, 'Max CPC'] 
    parts.append(group) 

result = pd.concat(parts) 

Answer 3

试试这个

def group_rank(df): 
    # first of all you've to rank according to `Max CPC` 
    df['CPC Rank'] = df['Max CPC'].rank(ascending = False) 
    # create the mapping 
    mapping = pd.Series(data=df['Max CPC'].values , index= df['CPC Rank'].values) 
    # create new column according to your ranking 
    df['New CPC'] = df['Type Rank'].map(mapping) 
    return df 

df.groupby(['Group', 'Subgroup', 'Normalized']).apply(group_rank)