蟒蛇：熊猫枢表多列在一旦其中有重复的价值观

Question 1

有一只熊猫dataframme列名、学校和标记

name  school  marks

tom     HBS     55
tom     HBS     55
tom     HBS     14
mark    HBS     28
mark    HBS     19
lewis   HBS     88

如何转换成这样

name  school  marks_1 marks_2 marks_3

tom     HBS     55     55       14
mark    HBS     28     19
lewis   HBS     88

试图这样的：

df = df.pivot_table(index='name', values='marks', columns='school') \
    .reset_index() \
    .rename_axis(None, axis=1)

print(df)

df = df.pivot('name','marks','school')

检查这些链接

https://stackoverflow.com/questions/22798934/pandas-long-to-wide-reshape-by-two-variables
https://stackoverflow.com/questions/62391419/pandas-group-by-and-convert-rows-into-multiple-columns
https://stackoverflow.com/questions/60698109/pandas-multiple-rows-to-single-row-with-multiple-columns-on-2-indexes

得到这一错误，由于重复的价值观。如何处理如果重复存在，我们必须保持他们

ValueError: Index contains duplicate entries, cannot reshape

Question 2

尝试使用 set_index 和 unstack 与 groupby 和 cumcount:

df_out = df.set_index(['name',
                       'school',
                       df.groupby(['name','school'])\
           .cumcount() +1]).unstack()
df_out.columns = [f'{i}_{j}' for i, j in df_out.columns]
df_out = df_out.reset_index()
df_out

输出：

    name school  marks_1  marks_2  marks_3
0  lewis    HBS     88.0      NaN      NaN
1   mark    HBS     28.0     19.0      NaN
2    tom    HBS     55.0     55.0     14.0

Question 3

的 cumcount 功能允许一个以创建独特的指标之前，枢转。这种建立在同样的想法，作为@ScottBoston；但是， pivot 功能用这里:

index = ['name', 'school']

                  # create an extra column for uniqueness          
temp = (df.assign(counter = df.groupby(index)
                              .cumcount()
                              .add(1)
                              .astype(str))
          .pivot(index = index, columns = 'counter')
        )

# flatten the columns
temp.columns = temp.columns.map('_'.join)

temp.reset_index()

    name school  marks_1  marks_2  marks_3
0  lewis    HBS     88.0      NaN      NaN
1   mark    HBS     28.0     19.0      NaN
2    tom    HBS     55.0     55.0     14.0

或者，可以使用的 pivot_wider 功能，从 pyjanitor，这是语法糖周围 pd.pivot有些助手：

# pip install pyjanitor
import pandas as pd
import janitor
(df.assign(counter = df.groupby(index)
                       .cumcount()
                       .add(1))                              
   .pivot_wider(index = index, 
                names_from = 'counter', 
                names_sep = '_')
)

    name school  marks_1  marks_2  marks_3
0  lewis    HBS     88.0      NaN      NaN
1   mark    HBS     28.0     19.0      NaN
2    tom    HBS     55.0     55.0     14.0

Scott Boston · Answer 1 · 2021-11-23T02:27:52

尝试使用 set_index 和 unstack 与 groupby 和 cumcount:

df_out = df.set_index(['name',
                       'school',
                       df.groupby(['name','school'])\
           .cumcount() +1]).unstack()
df_out.columns = [f'{i}_{j}' for i, j in df_out.columns]
df_out = df_out.reset_index()
df_out

输出：

    name school  marks_1  marks_2  marks_3
0  lewis    HBS     88.0      NaN      NaN
1   mark    HBS     28.0     19.0      NaN
2    tom    HBS     55.0     55.0     14.0

sammywemmy · Answer 2 · 2021-11-23T03:14:53

的 cumcount 功能允许一个以创建独特的指标之前，枢转。这种建立在同样的想法，作为@ScottBoston；但是， pivot 功能用这里:

index = ['name', 'school']

                  # create an extra column for uniqueness          
temp = (df.assign(counter = df.groupby(index)
                              .cumcount()
                              .add(1)
                              .astype(str))
          .pivot(index = index, columns = 'counter')
        )

# flatten the columns
temp.columns = temp.columns.map('_'.join)

temp.reset_index()

    name school  marks_1  marks_2  marks_3
0  lewis    HBS     88.0      NaN      NaN
1   mark    HBS     28.0     19.0      NaN
2    tom    HBS     55.0     55.0     14.0

或者，可以使用的 pivot_wider 功能，从 pyjanitor，这是语法糖周围 pd.pivot有些助手：

# pip install pyjanitor
import pandas as pd
import janitor
(df.assign(counter = df.groupby(index)
                       .cumcount()
                       .add(1))                              
   .pivot_wider(index = index, 
                names_from = 'counter', 
                names_sep = '_')
)

    name school  marks_1  marks_2  marks_3
0  lewis    HBS     88.0      NaN      NaN
1   mark    HBS     28.0     19.0      NaN
2    tom    HBS     55.0     55.0     14.0

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