pandas

代码如下:

import pandas as pdimport numpy as npsalaries = 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 name0 2 1 2016 BOSS1 2 2 2016 Lilei2 2 3 2016 Lilei3 2 4 2016 Han4 3 5 2017 BOSS5 4 6 2017 BOSS6 5 7 2017 Han7 6 8 2017 BOSS0 False1 True2 True3 True4 False5 False6 False7 FalseName: Bonus, dtype: boolInt64Index([1, 2, 3], dtype='int64') Bonus Salary Year name1 2 2 2016 Lilei2 2 3 2016 Lilei3 2 4 2016 Han0 True1 True2 True3 False4 False5 False6 False7 FalseName: Bonus, dtype: boolInt64Index([0, 1, 2], dtype='int64') Bonus Salary Year name0 2 1 2016 BOSS1 2 2 2016 Lilei2 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1m = np.zeros_like(a, dtype=bool)m[np.unique(a, return_index=True)[1]] = Truea[~m]方法2a[~np.in1d(np.arange(len(a)), np.unique(a, return_index=True)[1], assume_unique=True)]方法3np.setxor1d(a, np.unique(a), assume_unique=True)方法4u, i = np.unique(a, return_inverse=True)u[np.bincount(i) > 1]方法5s = np.sort(a, axis=None)s[:-1][s[1:] == s[:-1]]

参考：https://stackoverflow.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。