在新列的单元格级别计算数据框列表中的连续元素

Question

我有以下df：

df6 = pd.DataFrame({'name':['Sara',  'John', 'Jack'],
                   'places': ['UK,UK,UK,UK,US,CA', 'US,US,US,CA,CA,CA', 'Mexico,AUS,AUS,Mexico,Mexico']
                   })

df6

看起来像：

    name    places
0   Sara    UK,UK,UK,UK,US,CA
1   John    US,US,US,CA,CA,CA
2   Jack    Mexico,AUS,AUS,Mexico,Mexico

地点列仅关注 5 个国家/地区。我要做的是找出每个国家连续访问的次数。所以基本上输出会是这样的：

    name    UK   US   CA   Mexico   AUS    
0   Sara    4    0    0       0      0
1   John    0    3    3       0      0  
2   Jack    0    0    0       2      2

到目前为止我所做的步骤是：

df6['consecutive'] = df6.places.map(lambda x: [Counter(group[1]) for group in groupby(x.split(','))])

这给了我一个list of dicts：

    name    places                        consecutive
0   Sara    UK,UK,UK,UK,US,CA             [{'UK': 4}, {'US': 1}, {'CA': 1}]
1   John    US,US,US,CA,CA,CA             [{'US': 3}, {'CA': 3}]
2   Jack    Mexico,AUS,AUS,Mexico,Mexico  [{'Mexico': 1}, {'AUS': 2}, {'Mexico': 2}]

现在我坚持如何遍历连续列中的每个单元格以找到每个单元格的 values > 1 并将 df6 重塑为最终输出：

    name    UK   US   CA   Mexico   AUS    
0   Sara    4    0    0       0      0
1   John    0    3    3       0      0  
2   Jack    0    0    0       2      2

Answer 1

或者你可以使用pivot_table：

import pandas as pd

df6 = pd.DataFrame({'name':['Sara',  'John', 'Jack'],
                   'places': ['UK,UK,UK,UK,US,CA', 'US,US,US,CA,CA,CA', 'Mexico,AUS,AUS,Mexico,Mexico']
               })

df6['places'] = df6.places.str.split(',')
df6 = df6.explode('places')
df6['lag_places'] = df6.places.shift(1)
df6 = df6.query('places == lag_places').pivot_table(index = 'name', columns = 'places',  aggfunc = 'count')
df6.loc[:, df6.columns != 'places'] = df6.loc[:, df6.columns != 'places'].apply(lambda x: x+1) # add 1 according to your definition
df6.columns = [x[1] for x in df6.columns]
df6.fillna(0, inplace = True)

#      AUS   CA  Mexico   UK   US
#name                            
#Jack  2.0  0.0     2.0  0.0  0.0
#John  0.0  3.0     0.0  0.0  3.0
#Sara  0.0  0.0     0.0  4.0  0.0

Answer 2

我们可以str.split 和explode places。然后使用groupby size 和unstack 来获得带有loc 的连续计数过滤器，以仅包括大于1 次连续访问。然后groupby sum 减少到每个名称的一行，join 回到原来的 DataFrame：

places = df6["places"].str.split(',').explode()  # Each place in own row

df7 = df6[['name']].join(
    places.groupby(
        [df6['name'],  # Name
         places,  # Places
         # consecutive duplicates in separate groups
         places.ne(places.shift()).groupby(df6['name']).cumsum()]
    ).size()  # Count how many in each group
        .loc[lambda x: x > 1]  # Filter to include only > 1 visits
        .unstack(1, fill_value=0)  # Make places columns
        .groupby(level=0).sum(),  # Get single row per name
    on='name'  # join back on name column
)

df7:

   name  AUS  CA  Mexico  UK  US
0  Sara    0   0       0   4   0
1  John    0   3       0   0   3
2  Jack    2   0       2   0   0

Answer 3

你可以使用pd.crosstab:

df6["places"] = df6["places"].apply(lambda x: x.split(","))
df6 = df6.explode("places")

out = pd.crosstab(df6["name"], df6["places"])
out.index.name = None
out.columns.name = None
print(out)

打印：

      AUS  CA  Mexico  UK  US
Jack    2   0       3   0   0
John    0   3       0   0   3
Sara    0   1       0   4   1

---

编辑：总结consecutive 列（对于连续值> 1）：

from itertools import groupby
from collections import Counter

df6["consecutive"] = df6.places.map(
    lambda x: [
        {k: v for k, v in Counter(group[1]).items() if v > 1}
        for group in groupby(x.split(","))
    ]
)

df6 = df6.explode("consecutive").reset_index(drop=True)
out = (
    pd.concat([df6, pd.DataFrame(df6.pop("consecutive").tolist())], axis=1)
    .groupby("name")
    .sum()
)
print(out)

打印：

       UK   US   CA  AUS  Mexico
name                            
Jack  0.0  0.0  0.0  2.0     2.0
John  0.0  3.0  3.0  0.0     0.0
Sara  4.0  0.0  0.0  0.0     0.0