我以前问过一个问题(这里回答正确):
简而言之,我有以下数据帧:
| winner | loser | tournament |
+--------+---------+------------+
| John | Steve | A |
+--------+---------+------------+
| Steve | John | B |
+--------+---------+------------+
| John | Michael | A |
+--------+---------+------------+
| Steve | John | A |
+--------+---------+------------+
我想基本上以这个结束:
+--------+---------+------------+-------------+------------+---------------+--------------+--------------+-------------+
| winner | loser | tournament | winner wins | loser wins | winner losses | loser losses | winner win % | loser win % |
+--------+---------+------------+-------------+------------+---------------+--------------+--------------+-------------+
| John | Steve | A | 0 | 0 | 0 | 0 | 0/(0+0) | 0/(0+0) |
+--------+---------+------------+-------------+------------+---------------+--------------+--------------+-------------+
| Steve | John | B | 0 | 0 | 0 | 0 | 0/(0+0) | 0/(0+0) |
+--------+---------+------------+-------------+------------+---------------+--------------+--------------+-------------+
| John | Michael | A | 1 | 0 | 0 | 0 | 1/(1+0) | 0/(0+0) |
+--------+---------+------------+-------------+------------+---------------+--------------+--------------+-------------+
| Steve | John | A | 0 | 2 | 1 | 0 | 0/(0+1) | 2/(2+0) |
+--------+---------+------------+-------------+------------+---------------+--------------+--------------+-------------
提议的解决方案之一是这段代码:
def win_los_percent(sdf):
sdf['winner wins'] = sdf.groupby('winner').cumcount()
sdf['winner losses'] = [(sdf.loc[0:i, 'loser'] == sdf.loc[i, 'winner']).sum() for i in sdf.index]
sdf['loser losses'] = sdf.groupby('loser').cumcount()
sdf['loser wins'] = [(sdf.loc[0:i, 'winner'] == sdf.loc[i, 'loser']).sum() for i in sdf.index]
sdf['winner win %'] = sdf['winner wins'] / (sdf['winner wins'] + sdf['winner losses'])
sdf['loser win %'] = sdf['loser wins'] / (sdf['loser wins'] + sdf['loser losses'])
return sdf
ddf = df.groupby('tournament').apply(win_los_percent)
这确实给出了正确的计算和答案。但是,我有一个很大的数据帧,运行它需要很长时间(>;10分钟)
有人能提出一个加速这个功能的方法吗?一般来说,我对熊猫和numpy还不熟悉,但我读到的一个解决方案是使用矢量化
我找不到一种方法来矢量化这样的函数。有人能给我指出正确的方向吗?我不介意为中间计算创建更多的列,只要答案是正确的,并且做得相当快
谢谢
目前没有回答
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