df.head()
Player Tourn Score
Tom a 65
Henry a 72
Johno a 69
Ingram a 79
Ben a 76
Harry a 66
Nick b 70
Ingram b 79
Johno b 69
我有各种锦标赛(‘a’到’m’)的球员得分数据框.有些球员参加过多场比赛,有些球员只参加过一场比赛.我希望为每个玩家创建一个额外的列,如果玩家参加该锦标赛,则为1,如果没有,则为0(所以基本上是虚拟变量).
看起来像这样(为每个玩家重复):
Player Tourn Score Tom(Dummy)
Tom a 65 1
Henry a 72 1
Johno a 69 1
Ingram a 79 1
Ben a 76 1
Harry a 66 1
Nick b 70 0
Ingram b 79 0
Johno b 69 0
在代码中实现此目的的最佳方法是什么? (理想情况下,我需要能够在大型数据帧中很好地扩展的东西!)
有兴趣听取您的回复.
最佳答案 首先使用
get_dummies
然后
groupby
按列Tourn与
transform
的
any
,转换为int并将
join
转换为原始:
df1 = pd.get_dummies(df['Player'])
df2 = df.join(df1.groupby(df['Tourn']).transform('any').astype(int))
另一个更快的解决方案(每场锦标赛只有一次比赛):
df.join(df.groupby(['Tourn','Player']).size().unstack(fill_value=0), on='Tourn')
print (df2)
Player Tourn Score Ben Harry Henry Ingram Johno Nick Tom
0 Tom a 65 1 1 1 1 1 0 1
1 Henry a 72 1 1 1 1 1 0 1
2 Johno a 69 1 1 1 1 1 0 1
3 Ingram a 79 1 1 1 1 1 0 1
4 Ben a 76 1 1 1 1 1 0 1
5 Harry a 66 1 1 1 1 1 0 1
6 Nick b 70 0 0 0 1 1 1 0
7 Ingram b 79 0 0 0 1 1 1 0
8 Johno b 69 0 0 0 1 1 1 0
时序:
N = 10000
a = ['Tom', 'Henry', 'Johno', 'Ingram', 'Ben', 'Harry', 'Nick', 'Ingram', 'Johno']
a = ['{}{}'.format(i, j) for i in range(5) for j in a]
df = pd.DataFrame({'Player':np.random.choice(a, size=N),
'Tourn':np.random.randint(1000, size=N).astype(str)})
df = df.sort_values('Tourn')
#print (df.head())
In [486]: %%timeit
...: df.join(df.groupby(['Tourn','Player']).size().unstack(fill_value=0), on='Tourn')
...:
100 loops, best of 3: 12.6 ms per loop
In [487]: %%timeit
...: df.join(pd.crosstab(df.Tourn, df.Player), on='Tourn')
10 loops, best of 3: 60.9 ms per loop
In [488]: %%timeit
...: df1 = pd.get_dummies(df['Player'])
...: df2 = df.join(df1.groupby(df['Tourn']).transform('any').astype(int))
...:
10 loops, best of 3: 120 ms per loop
In [489]: %%timeit
...: df.join(pd.get_dummies(df.Tourn).T.dot(pd.get_dummies(df.Player)), on='Tourn')
...:
1 loop, best of 3: 895 ms per loop
In [490]: %%timeit
...: dd = df.Tourn.str.get_dummies()
...: df.assign(**{x.Player: dd[x.Tourn] for x in df.itertuples()})
...:
1 loop, best of 3: 7.02 s per loop
In [491]: %%timeit
...: df.assign(**{x.Player:df.Tourn.eq(x.Tourn).astype(int) for x in df.itertuples()})
...:
1 loop, best of 3: 13.7 s per loop
警告
考虑到DataFrame的组数和长度,结果不能解决性能问题,这将影响其中一些解决方案的时序.