我从这样的输入数据开始
df1 = pandas.DataFrame( {
"Name" : ["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"] ,
"City" : ["Seattle", "Seattle", "Portland", "Seattle", "Seattle", "Portland"] } )
印刷出来时是这样的:
City Name
0 Seattle Alice
1 Seattle Bob
2 Portland Mallory
3 Seattle Mallory
4 Seattle Bob
5 Portland Mallory
分组非常简单:
g1 = df1.groupby( [ "Name", "City"] ).count()
打印产生一个GroupBy对象:
City Name
Name City
Alice Seattle 1 1
Bob Seattle 2 2
Mallory Portland 2 2
Seattle 1 1
但我最终想要的是另一个DataFrame对象，它包含GroupBy对象中的所有行。换句话说，我想得到以下结果:
City Name
Name City
Alice Seattle 1 1
Bob Seattle 2 2
Mallory Portland 2 2
Mallory Seattle 1 1
我不太清楚如何在pandas文档中实现这一点。欢迎任何提示。
我发现这对我很有用。
import numpy as np
import pandas as pd
df1 = pd.DataFrame({
"Name" : ["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"] ,
"City" : ["Seattle", "Seattle", "Portland", "Seattle", "Seattle", "Portland"]})
df1['City_count'] = 1
df1['Name_count'] = 1
df1.groupby(['Name', 'City'], as_index=False).count()
这些解决方案只部分适用于我，因为我正在进行多个聚合。下面是我分组的一个输出示例，我想转换为一个数据框架:
Because I wanted more than the count provided by reset_index(), I wrote a manual method for converting the image above into a dataframe. I understand this is not the most pythonic/pandas way of doing this as it is quite verbose and explicit, but it was all I needed. Basically, use the reset_index() method explained above to start a "scaffolding" dataframe, then loop through the group pairings in the grouped dataframe, retrieve the indices, perform your calculations against the ungrouped dataframe, and set the value in your new aggregated dataframe.
df_grouped = df[['Salary Basis', 'Job Title', 'Hourly Rate', 'Male Count', 'Female Count']]
df_grouped = df_grouped.groupby(['Salary Basis', 'Job Title'], as_index=False)
# Grouped gives us the indices we want for each grouping
# We cannot convert a groupedby object back to a dataframe, so we need to do it manually
# Create a new dataframe to work against
df_aggregated = df_grouped.size().to_frame('Total Count').reset_index()
df_aggregated['Male Count'] = 0
df_aggregated['Female Count'] = 0
df_aggregated['Job Rate'] = 0
def manualAggregations(indices_array):
temp_df = df.iloc[indices_array]
return {
'Male Count': temp_df['Male Count'].sum(),
'Female Count': temp_df['Female Count'].sum(),
'Job Rate': temp_df['Hourly Rate'].max()
}
for name, group in df_grouped:
ix = df_grouped.indices[name]
calcDict = manualAggregations(ix)
for key in calcDict:
#Salary Basis, Job Title
columns = list(name)
df_aggregated.loc[(df_aggregated['Salary Basis'] == columns[0]) &
(df_aggregated['Job Title'] == columns[1]), key] = calcDict[key]
如果字典不是你的东西，计算可以内联应用在for循环中:
df_aggregated['Male Count'].loc[(df_aggregated['Salary Basis'] == columns[0]) &
(df_aggregated['Job Title'] == columns[1])] = df['Male Count'].iloc[ix].sum()
