在没有聚合的 pandas 数据透视表中重复条目并重命

Duplicate entries and rename column row in pandas pivot table without aggregation(在没有聚合的 pandas 数据透视表中重复条目并重命名列行)

问题描述

我正在尝试将这个示例数据框从长格式改成宽格式，而不聚合任何数据.

I'm trying to reshape this sample dataframe from long to wide format, without aggregating any of the data.

import numpy as np
import pandas as pd

df = pd.DataFrame({'SubjectID': ['A', 'A', 'A', 'B', 'B', 'C', 'A'], 'Date': 
['2010-03-14', '2010-03-15', '2010-03-16', '2010-03-14', '2010-05-15', 
'2010-03-14', '2010-03-14'], 'Var1': [1 , 12, 4, 7, 90, 1, 9], 'Var2': [ 0, 
0, 1, 1, 1, 0, 1], 'Var3': [np.nan, 1, 0, np.nan, 0, 1, np.nan]})

df['Date'] = pd.to_datetime(df['Date']); df

    Date    SubjectID   Var1    Var2    Var3
0   2010-03-14  A   1   0   NaN
1   2010-03-15  A   12  0   1.0
2   2010-03-16  A   4   1   0.0
3   2010-03-14  B   7   1   NaN
4   2010-05-15  B   90  1   0.0
5   2010-03-14  C   1   0   1.0
6   2010-03-14  A   9   1   NaN

为了避免重复值，我按 "Date" 列进行分组并获取每个值的累积计数.然后我做一个数据透视表

To get around the duplicate values, I'm grouping by the "Date" column and getting the cumulative count for each value. Then I make a pivot table

df['idx'] = df.groupby('Date').cumcount()

dfp = df.pivot_table(index = 'SubjectID', columns = 'idx'); dfp 

    Var1    Var2    Var3
idx 0   1   2   3   0   1   2   3   0   2
SubjectID                                       
A   5.666667    NaN NaN 9.0 0.333333    NaN NaN 1.0 0.5 NaN
B   90.000000   7.0 NaN NaN 1.000000    1.0 NaN NaN 0.0 NaN
C   NaN NaN 1.0 NaN NaN NaN 0.0 NaN NaN 1.0

但是，我希望 idx 列索引是 "Date" 列中的值，并且我不想聚合任何数据.预期的输出是

However, I want the idx column index to be the values from the "Date" column and I don't want to aggregate any data. The expected output is

     Var1_2010-03-14 Var1_2010-03-14 Var1_2010-03-15 Var1_2010-03-16 Var1_2010-05-15 Var2_2010-03-14    Var2_2010-03-15 Var2_2010-03-16 Var2_2010-05-15 Var3_2010-03-14 Var3_2010-03-15 Var3_2010-03-16 Var3_2010-05-15
SubjectID                                       
A   1   9   12  4   NaN 0   1   0    1.0    NaN NaN NaN 1.0 0.0 NaN
B   7.0 NaN NaN NaN 90  1   NaN NaN  1.0    NaN NaN NaN NaN NaN 0.0
C   1   NaN NaN NaN NaN 0   NaN NaN  NaN    NaN 1.0 NaN NaN NaN NaN

我该怎么做?最后，我将通过 dfp.columns = [col[0]+ '_' + str(col[1]) for col in dfp.columns] 合并两个列索引.

How can I do this? Eventually, I'll merge the two column indexes by dfp.columns = [col[0]+ '_' + str(col[1]) for col in dfp.columns].

推荐答案

你在正确的道路上:

# group
df['idx'] = df.groupby('Date').cumcount()

# set index and unstack
new = df.set_index(['idx','Date', 'SubjectID']).unstack(level=[0,1])

# drop idx column
new.columns = new.columns.droplevel(1)
new.columns = [f'{val}_{date}' for val, date in new.columns]

我认为这是您的预期输出

I think this is your expected output

使用map貌似会快一点:

df['idx'] = df.groupby('Date').cumcount()
df['Date'] = df['Date'].astype(str)
new = df.set_index(['idx','Date', 'SubjectID']).unstack(level=[0,1])
new.columns = new.columns.droplevel(1)
#new.columns = [f'{val}_{date}' for val, date in new.columns]
new.columns = new.columns.map('_'.join)

这是一个 50,000 行的测试示例:

Here is a 50,000 row test example:

#data
data = pd.DataFrame(pd.date_range('2000-01-01', periods=50000, freq='D'))
data['a'] = list('abcd')*12500
data['b'] = 2
data['c'] = list('ABCD')*12500
data.rename(columns={0:'date'}, inplace=True)

# list comprehension:
%%timeit -r 3 -n 200
new = data.set_index(['a','date','c']).unstack(level=[0,1])
new.columns = new.columns.droplevel(0)
new.columns = [f'{x}_{y}' for x,y in new.columns]

# 98.2 ms ± 13.3 ms per loop (mean ± std. dev. of 3 runs, 200 loops each)

# map with join:
%%timeit -r 3 -n 200
data['date'] = data['date'].astype(str)
new = data.set_index(['a','date','c']).unstack(level=[0,1])
new.columns = new.columns.droplevel(0)
new.columns = new.columns.map('_'.join)

# 84.6 ms ± 3.87 ms per loop (mean ± std. dev. of 3 runs, 200 loops each)

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