(编辑：jimmy 日期: 2025/9/18 浏览：2)
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://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。