在一行

Question

我有一个数据框，在执行groupby()和aggregation后转换为multiIndex数据框。

In[1]:

mydata = [['Team1', 'Player1', 'idTrip13', 133], ['Team2', 'Player333', 'idTrip10', 18373],
['Team3', 'Player22', 'idTrip12', 17338899], ['Team2', 'Player293','idTrip02', 17656], 
['Team3', 'Player20', 'idTrip11', 1883], ['Team1', 'Player1', 'idTrip19', 19393]]

df = pd.DataFrame(mydata, columns = ['team', 'player', 'trips', 'time'])
df
Out[1]:
     team    player       trips      time
0   Team1   Player1     idTrip13    133
1   Team2   Player333   idTrip10    18373
2   Team3   Player22    idTrip12    17338899
3   Team2   Player293   idTrip02    17656
4   Team3   Player20    idTrip11    1883
5   Team1   Player1     idTrip19    19393

对于团队中的每位玩家，查找旅行总次数和旅行总时间。这将返回一个multiIndex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

期望的输出： 我想打印输出，以便团队中的所有玩家都在同一条线上。

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

此question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。

Answer 1

1. 使用groupby()遍历第0级（团队）。

   for team, df2 in player_total.groupby(level = 0):
   

   例如，在第二次迭代中，它将返回Team2的数据帧：

                   trips   time
   team  player              
   Team2 Player293     1  17656
         Player333     1  18373
   

2. 使用reset_index()删除团队索引列，并将播放器索引列作为数据帧的一部分。

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

3. 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]]
   

4. 打印时我们必须将整数映射到字符串，然后使用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; 
   

完整的解决方案：

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()

<强>输出：

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