我有一个这样的熊猫数据框,
>>> 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
这是预期的输出,
>>> 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
我想通过唯一的hotel_code
和feed
创建集群列,但我不知道。群集号是可变的。任何想法?预先感谢。
答案 0 :(得分:5)
将GroupBy.cumcount
用于每个组的计数器,使用hotel_code
和feed
创建Series
的MultiIndex,然后使用Series.unstack
进行整形,最后rename
列,DataFrame.reset_index
代表MultiIndex
到列:
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
答案 1 :(得分:3)
您可以使用集群创建一个新的数据框:
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,
)
会返回
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
答案 2 :(得分:3)
在hotel_code
和feed
上分组,然后在client_nationality
上聚合,最后拆分并扩展。
使用所需的后缀更新列。
df.groupby(['hotel_code', 'feed'])['client_nationality']
.agg(' '.join)
.str.split(' ', expand=True)
.rename(columns = lambda x: f'cluster_{x+1}')
输出
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