pandas

代碼如下:

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')])

輸出如下：

 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

對于如nunpy中的這些操作主要如下:

假設有數組

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

想找出 [1 3]

則有

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]]

參考：https://stackoverflow.com/questions/11528078/determining-duplicate-values-in-an-array

以上這篇Pandas統計重復的列里面的值方法就是小編分享給大家的全部內容了，希望能給大家一個參考，也希望大家多多支持武林站長站。