重复条目并重命名 pandas 数据透视表中的列行而不进行聚合

我正在尝试将这个示例数据帧从长格式重塑为宽格式，而不聚合任何数据。

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

为了解决重复值，我按"Date"列并获取每个值的累积计数。然后我制作一个数据透视表

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，我想要idx列索引是来自的值"Date"列，我不想聚合任何数据。预期输出是

     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

我怎样才能做到这一点？最终，我将合并两个列索引dfp.columns = [col[0]+ '_' + str(col[1]) for col in dfp.columns].

您走在正确的道路上：

# 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]

我认为这是你的预期输出

Using map看起来会快一点：

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 行测试示例：

#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)