pandas 大熊猫在群体中的百分位排名
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percentile rank in pandas in groups
提问by itjcms18
I can't quite figure out how to write function to accomplish a grouped percentile. I have all teams from years 1985-2012 in a data frame; the first 10 are shown below: it's currently sorted by year. I was looking to give a percentile for LgRnkgrouped by Year. So for instance, 23 LgRank (worst team) for 1985 would be a 100 percentile and a 1 LgRank (best team) for 1985 would be a 1 percentile. 30 LgRank (worst team) for 2010 would be 100 percentile, etc. It needs to be grouped by year b/c of the differing number of LgRnks.
我不太清楚如何编写函数来完成分组的百分位数。我将 1985 年至 2012 年的所有团队放在一个数据框中；前 10 个如下所示：目前按年份排序。我想给LgRnk分组的百分位数Year。因此，例如，1985 年 23 LgRank（最差球队）将是 100 个百分点，而 1985 年的 1 LgRank（最佳球队）将是 1 个百分点。2010 年的 30 LgRank（最差团队）将是 100 个百分位，等等。它需要按不同数量的LgRnks 的b/c 年分组。
Team WLPer Year LgRnk W L
19 Sacramento Kings 0.378 1985 18 31 51
0 Atlanta Hawks 0.415 1985 17 34 48
17 Phoenix Suns 0.439 1985 16 36 46
4 Cleveland Cavaliers 0.439 1985 15 36 46
13 Milwaukee Bucks 0.720 1985 3 59 23
3 Chicago Bulls 0.463 1985 14 38 44
16 Philadelphia 76ers 0.707 1985 4 58 24
22 Washington Wizards 0.488 1985 13 40 42
20 San Antonio Spurs 0.500 1985 12 41 41
21 Utah Jazz 0.500 1985 11 41 41
I've tried creating a function using: scipy.stats.percentileofscoreand I can't quite get it.
我试过使用：创建一个函数scipy.stats.percentileofscore，但我不太明白。
回答by Andy Hayden
You can do an apply on the LgRnk column:
您可以对 LgRnk 列进行申请：
# just for me to normalize this, so my numbers will go from 0 to 1 in this example
In [11]: df['LgRnk'] = g.LgRnk.rank()
In [12]: g = df.groupby('Year')
In [13]: g.LgRnk.apply(lambda x: x / len(x))
Out[13]:
19 1.0
0 0.9
17 0.8
4 0.7
13 0.1
3 0.6
16 0.2
22 0.5
20 0.4
21 0.3
Name: 1985, dtype: float64
The Series groupby rank (which just applies Series.rank) take a pct argument to do just this:
系列 groupby 排名（仅适用Series.rank）采用 pct 参数来执行此操作：
In [21]: g.LgRnk.rank(pct=True)
Out[21]:
19 1.0
0 0.9
17 0.8
4 0.7
13 0.1
3 0.6
16 0.2
22 0.5
20 0.4
21 0.3
Name: 1985, dtype: float64
and directly on the WLPercolumn (although this is slightly different due to draws):
并直接在WLPer列上（尽管由于抽签而略有不同）：
In [22]: g.WLPer.rank(pct=True, ascending=False)
Out[22]:
19 1.00
0 0.90
17 0.75
4 0.75
13 0.10
3 0.60
16 0.20
22 0.50
20 0.35
21 0.35
Name: 1985, dtype: float64
Note: I've changed the numbers on the first line, so you'll get different scores on your completeframe.
注意：我已经更改了第一行的数字，因此您将在整个框架上获得不同的分数。
回答by user636224
You need to calculate rank within the group before normalizing within the group. The other answers will result in percentiles over 100%. I suggest:
在组内归一化之前，您需要计算组内的排名。其他答案将导致百分位数超过 100%。我建议：
df['percentile'] = df.groupby('year')['LgRnk'].rank(pct=True)
