pandas 熊猫系列的分位数函数的倒数是多少？
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what's the inverse of the quantile function on a pandas Series?
提问by Mannaggia
The quantile functions gives us the quantile of a given pandas series s,
分位数函数为我们提供给定Pandas系列s的分位数，
E.g.
例如
s.quantile(0.9) is 4.2
s.quantile(0.9) 是 4.2
Is there the inverse function (i.e. cumulative distribution) which finds the value x such that
是否有反函数（即累积分布）找到值 x 使得
s.quantile(x)=4
s.quantile(x)=4
Thanks
谢谢
回答by fernandosjp
I had the same question as you did! I found an easy way of getting the inverse of quantile using scipy.
我和你有同样的问题！我找到了一种使用 scipy 获取分位数倒数的简单方法。
#libs required
from scipy import stats
import pandas as pd
import numpy as np
#generate ramdom data with same seed (to be reproducible)
np.random.seed(seed=1)
df = pd.DataFrame(np.random.uniform(0,1,(10)), columns=['a'])
#quantile function
x = df.quantile(0.5)[0]
#inverse of quantile
stats.percentileofscore(df['a'],x)
回答by ILoveCoding
Sorting can be expensive, if you look for a single value I'd guess you'd be better of computing it with:
排序可能很昂贵，如果您寻找单个值，我猜您最好使用以下方法计算它：
s = pd.Series(np.random.uniform(size=1000))
( s < 0.7 ).astype(int).mean() # =0.7ish
There's probably a way to avoid the int(bool) shenanigan.
可能有一种方法可以避免 int(bool) 恶作剧。
回答by Mike
There's no 1-liner that I know of, but you can achieve this with scipy:
我所知道的没有 1-liner，但是您可以使用 scipy 实现这一点：
import pandas as pd
import numpy as np
from scipy.interpolate import interp1d
# set up a sample dataframe
df = pd.DataFrame(np.random.uniform(0,1,(11)), columns=['a'])
# sort it by the desired series and caculate the percentile
sdf = df.sort('a').reset_index()
sdf['b'] = sdf.index / float(len(sdf) - 1)
# setup the interpolator using the value as the index
interp = interp1d(sdf['a'], sdf['b'])
# a is the value, b is the percentile
>>> sdf
index a b
0 10 0.030469 0.0
1 3 0.144445 0.1
2 4 0.304763 0.2
3 1 0.359589 0.3
4 7 0.385524 0.4
5 5 0.538959 0.5
6 8 0.642845 0.6
7 6 0.667710 0.7
8 9 0.733504 0.8
9 2 0.905646 0.9
10 0 0.961936 1.0
Now we can see that the two functions are inverses of each other.
现在我们可以看到这两个函数是互逆的。
>>> df['a'].quantile(0.57)
0.61167933268395969
>>> interp(0.61167933268395969)
array(0.57)
>>> interp(df['a'].quantile(0.43))
array(0.43)
interp can also take in list, a numpy array, or a pandas data series, any iterator really!
interp 还可以接收列表、numpy 数组或 Pandas 数据系列，真的是任何迭代器！
回答by Calvin Ku
Just came across the same problem. Here's my two cents.
刚刚遇到同样的问题。这是我的两分钱。
def inverse_percentile(arr, num):
arr = sorted(arr)
i_arr = [i for i, x in enumerate(arr) if x > num]
return i_arr[0] / len(arr) if len(i_arr) > 0 else 1
