Pandas统计重复的列里面的值方法

pandas

代码如下:

import pandas as pd
import numpy as np

salaries = pd.dataframe({
 'name': ['boss', 'lilei', 'lilei', 'han', 'boss', 'boss', 'han', 'boss'],
 'year': [2016, 2016, 2016, 2016, 2017, 2017, 2017, 2017],
 'salary': [1, 2, 3, 4, 5, 6, 7, 8],
 'bonus': [2, 2, 2, 2, 3, 4, 5, 6]
})
print(salaries)
print(salaries['bonus'].duplicated(keep='first'))
print(salaries[salaries['bonus'].duplicated(keep='first')].index)
print(salaries[salaries['bonus'].duplicated(keep='first')])
print(salaries['bonus'].duplicated(keep='last'))
print(salaries[salaries['bonus'].duplicated(keep='last')].index)
print(salaries[salaries['bonus'].duplicated(keep='last')])

输出如下：

 bonus salary year name
0  2  1 2016 boss
1  2  2 2016 lilei
2  2  3 2016 lilei
3  2  4 2016 han
4  3  5 2017 boss
5  4  6 2017 boss
6  5  7 2017 han
7  6  8 2017 boss
0 false
1  true
2  true
3  true
4 false
5 false
6 false
7 false
name: bonus, dtype: bool
int64index([1, 2, 3], dtype='int64')
 bonus salary year name
1  2  2 2016 lilei
2  2  3 2016 lilei
3  2  4 2016 han
0  true
1  true
2  true
3 false
4 false
5 false
6 false
7 false
name: bonus, dtype: bool
int64index([0, 1, 2], dtype='int64')
 bonus salary year name
0  2  1 2016 boss
1  2  2 2016 lilei
2  2  3 2016 lilei

非pandas

对于如nunpy中的这些操作主要如下:

假设有数组

a = np.array([1, 2, 1, 3, 3, 3, 0])

想找出 [1 3]

则有

方法1

m = np.zeros_like(a, dtype=bool)
m[np.unique(a, return_index=true)[1]] = true
a[~m]

方法2

a[~np.in1d(np.arange(len(a)), np.unique(a, return_index=true)[1], assume_unique=true)]

方法3

np.setxor1d(a, np.unique(a), assume_unique=true)

方法4

u, i = np.unique(a, return_inverse=true)
u[np.bincount(i) > 1]

方法5

s = np.sort(a, axis=none)
s[:-1][s[1:] == s[:-1]]

参考：https://*.com/questions/11528078/determining-duplicate-values-in-an-array

以上这篇pandas统计重复的列里面的值方法就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持。