Question

我有一个pandas数据框my_df，在那里我可以找到给定列的mean（），median（），mode（）：

my_df['field_A'].mean()
my_df['field_A'].median()
my_df['field_A'].mode()

我想知道是否有可能找到更详细的统计数据，如90％？谢谢！

Answer 1

您可以使用pandas.DataFrame.quantile()功能，如下所示。

import pandas as pd
import random

A = [ random.randint(0,100) for i in range(10) ]
B = [ random.randint(0,100) for i in range(10) ]

df = pd.DataFrame({ 'field_A': A, 'field_B': B })
df
#    field_A  field_B
# 0       90       72
# 1       63       84
# 2       11       74
# 3       61       66
# 4       78       80
# 5       67       75
# 6       89       47
# 7       12       22
# 8       43        5
# 9       30       64

df.field_A.mean()   # Same as df['field_A'].mean()
# 54.399999999999999

df.field_A.median() 
# 62.0

# You can call `quantile(i)` to get the i'th quantile,
# where `i` should be a fractional number.

df.field_A.quantile(0.1) # 10th percentile
# 11.9

df.field_A.quantile(0.5) # same as median
# 62.0

df.field_A.quantile(0.9) # 90th percentile
# 89.10000000000001

Answer 2

假设系列s

s = pd.Series(np.arange(100))

获取[.1, .2, .3, .4, .5, .6, .7, .8, .9]

的分位数

s.quantile(np.linspace(.1, 1, 9, 0))

0.1     9.9
0.2    19.8
0.3    29.7
0.4    39.6
0.5    49.5
0.6    59.4
0.7    69.3
0.8    79.2
0.9    89.1
dtype: float64

OR

s.quantile(np.linspace(.1, 1, 9, 0), 'lower')

0.1     9
0.2    19
0.3    29
0.4    39
0.5    49
0.6    59
0.7    69
0.8    79
0.9    89
dtype: int32

Answer 3

我发现下面会有效：

grep

Answer 4

您甚至可以为多列提供空值并获得多个分位数（我将95％用于离群值处理）

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pandas：找到给定列的百分位统计数据

4 个答案: