ValueError：值的长度与长度索引不匹配

Question

import pandas as pd
dict1 = {'id_game': [112, 113, 114], 'game_name' : ['x','z','y'],'id_category':[1,2,3], 'id_players':[[588,589,590],[589],[588,589]]}
dict2 = {'id_player': [588, 589, 590],'player_name' : ['fff','aaa','ccc'] ,'indication':['mmm x ggg sdg y', 'uuu x fdb y kfnkjq z', 'fffre x']}
game_df = pd.DataFrame(dict1)
player_df = pd.DataFrame(dict2)

这是我拥有的数据样本，我正在寻找一种解决方案，用于根据game_df['id_players']和player_df['id_player']或game_df['game_name']之间的关系在第二个 dataframe game_df game_df['game_name']和drug_df['indication']

在以下脚本中，我使用game_name和indication值：

new_list = []
for i in range(len(game_df)):
    for j in range(len(player_df)):
        if game_df['game_name'][i] in player_df['indication'][j]:
            new_list.append(game_df['id_category'][i])
            print(new_list)
            
player_df['categories_id'] = new_list 

错误：

--> 747         raise ValueError(
    748             "Length of values "
    749             f"({len(data)}) "

ValueError: Length of values (6) does not match length of index (3)

Answer 1

您的代码可以通过在print(new_list)之后添加break来修复，具有相同的缩进。

...
if game_df['game_name'][i] in player_df['indication'][j]:
    new_list.append(game_df['id_category'][i])
    print(new_list)
    break

---

话虽如此，非常不鼓励迭代数据帧，因为它很慢并且很快就会变得笨拙。 解决此类问题的规范方法是merge id_player(s)上的数据帧，即将id_players中的 id 分解为单独的行，

>>> game_df = game_df.explode("id_players").rename(columns={"id_players": "id_player"})
>>> game_df
   id_game game_name  id_category id_player
0      112         x            1       588
0      112         x            1       589
0      112         x            1       590
1      113         z            2       589
2      114         y            3       588
2      114         y            3       589

所以你可以.merge它与game_df ，

>>> df = game_df.merge(player_df, on="id_player")
>>> df
   id_game game_name  id_category id_player player_name            indication
0      112         x            1       588         fff       mmm x ggg sdg y
1      114         y            3       588         fff       mmm x ggg sdg y
2      112         x            1       589         aaa  uuu x fdb y kfnkjq z
3      113         z            2       589         aaa  uuu x fdb y kfnkjq z
4      114         y            3       589         aaa  uuu x fdb y kfnkjq z
5      112         x            1       590         ccc               fffre x

这将使分析变得相当简单，例如检查game_name是否在indication中

df.apply(lambda row: row.game_name in row.indication, axis=1)

虽然它为所有这些都返回 True，所以我不确定这是否真的是你想要的。