Python 在一行中打印多级索引pd.DataFrame的每个级别中的所有行

我有一个数据帧，在执行了

groupby（）

和之后，它被转换为多索引数据帧

对于团队中的每个球员，找出总旅行次数和总旅行时间。这将返回一个多索引数据帧

player_total = df.groupby(by = ['team', 'player']).agg({'time' : 'sum', 'trips' : 'count'})

player_total
Out[4]:
                 trips  time
team    player      
Team1   Player1     2   19526
Team2   Player293   1   17656
        Player333   1   18373
Team3   Player20    1   1883
        Player22    1   17338899

所需输出： 我想打印输出，这样团队中的所有球员都在同一条线上

Team1   Player1 : 2 trips : 19526;
Team2   Player293 : 1 : 17656; Player333 : 1 : 18373;
Team3   Player22 : 1 trip : 17338899; Player20 : 1 trip : 1883

这被认为过于宽泛，因此我冒昧地将pandas数据帧创建/聚合从输出打印中分离出来

使用

groupby（）

迭代第0级（团队）

例如，在第二次迭代中，它将为

Team2

返回一个数据帧：

                trips   time
team  player              
Team2 Player293     1  17656
      Player333     1  18373

使用

reset_index（）

删除团队索引列，并将球员索引列作为数据框的一部分

>>>team_df = df2.reset_index(level = 0, drop = True).reset_index()
>>>team_df
      player  trips   time
0  Player293     1  17656
1  Player333     1  18373

将该数据帧转换为列表列表，以便我们可以遍历每个播放器

team_df.values.tolist()
>>>[['Player293', 1, 17656], ['Player333', 1, 18373]]

打印时，我们必须将整数映射到字符串，并使用print函数的end参数打印分号，而不是在末尾打印新行

>>>for player in team_df.values.tolist():
       print(': '.join(map(str, player)), end = '; ')
>>>Player293: 1: 17656; Player333: 1: 18373; 

完整解决方案：

from __future__ import print_function

#iterate through each team
for team, df2 in player_total.groupby(level = 0):
    print(team, end = '\t')
    #drop the 0th level (team) and move the first level (player) as the index
    team_df = df2.reset_index(level = 0, drop = True).reset_index()
    #iterate through each player on the team and print player, trip, and time
    for player in team_df.values.tolist():
        print(': '.join(map(str, player)), end = '; ')
    #After printing all players insert a new line
    print()

Player1: 2: 19526; 
Player293: 1: 17656; Player333: 1: 18373; 
Player20: 1: 1883; Player22: 1: 17338899; 

输出：

from __future__ import print_function

#iterate through each team
for team, df2 in player_total.groupby(level = 0):
    print(team, end = '\t')
    #drop the 0th level (team) and move the first level (player) as the index
    team_df = df2.reset_index(level = 0, drop = True).reset_index()
    #iterate through each player on the team and print player, trip, and time
    for player in team_df.values.tolist():
        print(': '.join(map(str, player)), end = '; ')
    #After printing all players insert a new line
    print()

Player1: 2: 19526; 
Player293: 1: 17656; Player333: 1: 18373; 
Player20: 1: 1883; Player22: 1: 17338899;