如何在 Pandas 中使用 groupby 創建新列？

Question

我有一個像這樣的熊貓數據框，

>>> data = {
    'hotel_code': [1, 1, 1, 1, 1],
    'feed': [1, 1, 1, 1, 2],
    'price_euro': [100, 200, 250, 120, 130],
    'client_nationality': ['fr', 'us', 'ru,de', 'gb', 'cn,us,br,il,fr,gb,de,ie,pk,pl']
}
>>> df = pd.DataFrame(data)
>>> df
   hotel_code  feed  price_euro             client_nationality
0           1     1         100                             fr
1           1     1         200                             us
2           1     1         250                          ru,de
3           1     1         120                             gb
4           1     2         130  cn,us,br,il,fr,gb,de,ie,pk,pl

這是預期的輸出，

>>> data = {
    'hotel_code': [1, 1],
    'feed': [1, 2],
    'cluster1': ['fr', 'cn,us,br,il,fr,gb,de,ie,pk,pl'],
    'cluster2': ['us', np.nan],
    'cluster3': ['ru,de', np.nan],
    'cluster4': ['gb', np.nan],
}
>>> df = pd.DataFrame(data)
>>> df
   hotel_code  feed                       cluster1 cluster2 cluster3 cluster4
0           1     1                             fr       us    ru,de       gb
1           1     2  cn,us,br,il,fr,gb,de,ie,pk,pl      NaN      NaN      NaN

我想通過唯一的hotel_code和feed創建集群列，但我不知道。 簇號是可變的。 任何的想法？ 提前致謝。

Answer 1

使用GroupBy.cumcount作為每個組的計數器，通過帶有feed和計數器Series的hotel_code創建 MultiIndex 並通過Series.unstack重塑，最后rename列，並將DataFrame.reset_index的MultiIndex rename為列：

g = df.groupby(["hotel_code", "feed"]).cumcount()

df1 = (df.set_index(["hotel_code", "feed", g])['client_nationality']
         .unstack()
         .rename(columns = lambda x: f'cluster_{x+1}')
         .reset_index())
print (df1)
   hotel_code  feed                      cluster_1 cluster_2 cluster_3  \
0           1     1                             fr        us     ru,de   
1           1     2  cn,us,br,il,fr,gb,de,ie,pk,pl       NaN       NaN   

  cluster_4  
0        gb  
1       NaN  

Answer 2

您可以使用集群創建一個新的數據框：

clusters = pd.DataFrame(
    df.groupby(["hotel_code", "feed"])
    .agg(list)
    .reset_index()
    .client_nationality.tolist()
)
clusters.columns = [f"cluster_{i}" for i in range(1, clusters.shape[1] + 1)]

然后，

pd.concat(
    [
        df.drop(["price_euro", "client_nationality"], axis=1)
        .drop_duplicates(["hotel_code", "feed"])
        .reset_index(drop=True),
        clusters,
    ],
    axis=1,
)

會回來

   hotel_code  feed                      cluster_1 cluster_2 cluster_3 cluster_4
0           1     1                             fr        us     ru,de        gb
1           1     2  cn,us,br,il,fr,gb,de,ie,pk,pl      None      None      None

Answer 3

hotel_code在hotel_code和feed ，然后在client_nationality聚合，最后拆分和擴展。

更新具有所需后綴的列。

df.groupby(['hotel_code', 'feed'])['client_nationality']
  .agg(' '.join)
  .str.split(' ', expand=True)
  .rename(columns = lambda x: f'cluster_{x+1}')

輸出

                                     cluster_1 cluster_2 cluster_3 cluster_4
hotel_code feed                                                             
1          1                                fr        us     ru,de        gb
           2     cn,us,br,il,fr,gb,de,ie,pk,pl      None      None      None