Python : Pandy kontingenčná tabuľka pre viacerých stĺpcov naraz, ktorá má duplicitné hodnoty

0

Otázka

majú pandy dataframme so stĺpcami meno , školu a známky

name  school  marks

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

Ako transponovať a previesť do takhle

name  school  marks_1 marks_2 marks_3

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

snažil tento:

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')

kontrolovať tieto odkazy

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

ako túto chybu kvôli duplicitné hodnoty. ako zvládnuť ak duplicitné existuje a máme ich udržať

ValueError: Index contains duplicate entries, cannot reshape
dataframe group-by pandas pivot
2021-11-23 02:17:12
2

Najlepšiu odpoveď

2

Skúste použiť set_index a unstack s groupby a 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

Výstup:

    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
2021-11-23 02:27:52
1

Na cumcount funkcia umožňuje vytvoriť jedinečné indexy pred výkyvná. Toto stavia na rovnakú myšlienku ako @ScottBoston; avšak, pivot funkcia sa využíva tu:

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

Alternatívne môžete použiť pivot_wider funkciu zo pyjanitor, ktorý je syntaktická cukru v okolí pd.pivot- niektorí pomocníci:

# 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
2021-11-23 03:14:53

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