pandas Groupby 给定所选 DataFrame 列值的百分位数
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Groupby given percentiles of the values of the chosen DataFrame column
提问by pms
Imagine that I have a DataFramewith columns that contain only real values.
想象一下，我有一个DataFrame只包含真实值的列。
>> df
col1 col2 col3
0 0.907609 82 4.207991
1 3.743659 1523 6.488842
2 2.358696 324 5.092592
3 0.006793 0 0.000000
4 19.319746 11969 7.405685
I want to group it by quartiles (or any other percentiles specified by me) of the chosen column (e.g., col1), to perform some operations on these groups. Ideally, I would like to do something like:
我想按所选列（例如col1）的四分位数（或我指定的任何其他百分位数）对其进行分组，以对这些组执行一些操作。理想情况下，我想做类似的事情：
df.groupy( quartiles_of_col1 ).mean() # not working, how to code quartiles_of_col1?
The output should give the mean of each of the columns for four groups corresponding to the quartiles of col1. Is this possible with the groupbycommand? What's the simplest way of achieving it?
输出应给出对应于 的四分位数的四组的每一列的平均值col1。这可以通过groupby命令实现吗？实现它的最简单方法是什么？
回答by CT Zhu
I don't have a computer to test it right now, but I think you can do it by: df.groupby(pd.cut(df.col0, np.percentile(df.col0, [0, 25, 75, 90, 100]), include_lowest=True)).mean(). Will update after 150mins.
我现在没有电脑来测试它，但我认为你可以通过：df.groupby(pd.cut(df.col0, np.percentile(df.col0, [0, 25, 75, 90, 100]), include_lowest=True)).mean(). 150分钟后更新。
Some explanations:
一些解释：
In [42]:
#use np.percentile to get the bin edges of any percentile you want
np.percentile(df.col0, [0, 25, 75, 90, 100])
Out[42]:
[0.0067930000000000004,
0.907609,
3.7436589999999996,
13.089311200000001,
19.319745999999999]
In [43]:
#Need to use include_lowest=True
print df.groupby(pd.cut(df.col0, np.percentile(df.col0, [0, 25, 75, 90, 100]), include_lowest=True)).mean()
col0 col1 col2
col0
[0.00679, 0.908] 0.457201 41.0 2.103996
(0.908, 3.744] 3.051177 923.5 5.790717
(3.744, 13.0893] NaN NaN NaN
(13.0893, 19.32] 19.319746 11969.0 7.405685
In [44]:
#Or the smallest values will be skiped
print df.groupby(pd.cut(df.col0, np.percentile(df.col0, [0, 25, 75, 90, 100]))).mean()
col0 col1 col2
col0
(0.00679, 0.908] 0.907609 82.0 4.207991
(0.908, 3.744] 3.051177 923.5 5.790717
(3.744, 13.0893] NaN NaN NaN
(13.0893, 19.32] 19.319746 11969.0 7.405685
回答by biobirdman
I hope this will solve your problem. It is not pretty but I hope it will work for you
我希望这能解决你的问题。它不漂亮，但我希望它对你有用
import pandas as pd
import random
import numpy as np
## create a mock df as example. with column A, B, C and D
df = pd.DataFrame(np.random.randn(100, 4), columns=list('ABCD'))
## select dataframe based on the quantile of column A, using the quantile method.
df[df['A'] < df['A'].quantile(0.3)].mean()
this will print
这将打印
A -1.157615
B 0.205529
C -0.108263
D 0.346752
dtype: float64
回答by RickB
Pandas has a native solution, pandas.qcut, to this as well:
Pandas 也有一个原生的解决方案pandas.qcut：
http://pandas.pydata.org/pandas-docs/stable/generated/pandas.qcut.html
http://pandas.pydata.org/pandas-docs/stable/generated/pandas.qcut.html
