按分位数对 Pandas 数据框进行排名

问题描述

我有一个 Pandas 数据框，其中每一列代表一个单独的属性，每一行都保存特定日期的属性值:

I have a Pandas dataframe in which each column represents a separate property, and each row holds the properties' value on a specific date:

import pandas as pd

dfstr = \
'''         AC        BO         C       CCM        CL       CRD        CT        DA        GC        GF
2010-01-19  0.844135 -0.194530 -0.231046  0.245615 -0.581238 -0.593562  0.057288  0.655903  0.823997  0.221920
2010-01-20 -0.204845 -0.225876  0.835611 -0.594950 -0.607364  0.042603  0.639168  0.816524  0.210653  0.237833
2010-01-21  0.824852 -0.216449 -0.220136  0.234343 -0.611756 -0.624060  0.028295  0.622516  0.811741  0.201083'''
df = pd.read_csv(pd.compat.StringIO(dfstr), sep='\s+')

使用 rank 方法，我可以找到每个属性相对于特定日期的百分位排名:

Using the rank method, I can find the percentile rank of each property with respect to a specific date:

df.rank(axis=1, pct=True)

输出:

             AC   BO    C  CCM   CL  CRD   CT   DA   GC   GF
2010-01-19  1.0  0.4  0.3  0.7  0.2  0.1  0.5  0.8  0.9  0.6
2010-01-20  0.4  0.3  1.0  0.2  0.1  0.5  0.8  0.9  0.6  0.7
2010-01-21  1.0  0.4  0.3  0.7  0.2  0.1  0.5  0.8  0.9  0.6

我想得到的是每个属性的分位数(例如四分位数、五分位数、十分位数等)等级.例如，对于五分位数，我想要的输出是:

What I'd like to get instead is the quantile (eg quartile, quintile, decile, etc) rank of each property. For example, for quintile rank my desired output would be:

             AC   BO    C  CCM   CL  CRD   CT   DA   GC   GF
2010-01-19   5    2     2  4     1   1     3    4    5    3
2010-01-20   2    2     5  1     1   3     4    5    3    4
2010-01-21   5    2     2  4     1   1     3    4    5    3

我可能遗漏了一些东西，但似乎没有内置的方法可以使用 Pandas 进行这种分位数排名.获得所需输出的最简单方法是什么?

I might be missing something, but there doesn't seem to a built-in way to do this kind of quantile ranking with Pandas. What's the simplest way to get my desired output?

推荐答案

Method 1 mul &np.ceil

你的排名非常接近.只需将 .mul 乘以 5 即可获得所需的分位数，也可以使用 np.ceil 进行四舍五入:

Method 1 mul & np.ceil

You were quite close with the rank. Just multiplying by 5 with .mul to get the desired quantile, also rounding up with np.ceil:

np.ceil(df.rank(axis=1, pct=True).mul(5))

输出

             AC   BO    C  CCM   CL  CRD   CT   DA   GC   GF
2010-01-19  5.0  2.0  2.0  4.0  1.0  1.0  3.0  4.0  5.0  3.0
2010-01-20  2.0  2.0  5.0  1.0  1.0  3.0  4.0  5.0  3.0  4.0
2010-01-21  5.0  2.0  2.0  4.0  1.0  1.0  3.0  4.0  5.0  3.0

如果你想要整数使用 astype:

If you want integers use astype:

np.ceil(df.rank(axis=1, pct=True).mul(5)).astype(int)

甚至更好从熊猫版本 0.24.0 开始，我们有 可空整数类型:Int64.
所以我们可以使用:

Or even betterSince pandas version 0.24.0 we have nullable integer type: Int64.
So we can use :

np.ceil(df.rank(axis=1, pct=True).mul(5)).astype('Int64')

输出

            AC  BO  C  CCM  CL  CRD  CT  DA  GC  GF
2010-01-19   5   2  2    4   1    1   3   4   5   3
2010-01-20   2   2  5    1   1    3   4   5   3   4
2010-01-21   5   2  2    4   1    1   3   4   5   3

方法二 scipy.stats.percentileofscore

d = df.apply(lambda x: [np.ceil(stats.percentileofscore(x, a, 'rank')*0.05) for a in x], axis=1).values

pd.DataFrame(data=np.concatenate(d).reshape(d.shape[0], len(d[0])),
             columns=df.columns,
             dtype='int',
             index=df.index)

输出

            AC  BO  C  CCM  CL  CRD  CT  DA  GC  GF
2010-01-19   5   2  2    4   1    1   3   4   5   3
2010-01-20   2   2  5    1   1    3   4   5   3   4
2010-01-21   5   2  2    4   1    1   3   4   5   3

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