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pandas
代碼如下:
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')])
輸出如下:
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
對(duì)于如nunpy中的這些操作主要如下:
假設(shè)有數(shù)組
a = np.array([1, 2, 1, 3, 3, 3, 0])
想找出 [1 3]
則有
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]]
參考:https://stackoverflow.com/questions/11528078/determining-duplicate-values-in-an-array
以上這篇Pandas統(tǒng)計(jì)重復(fù)的列里面的值方法就是小編分享給大家的全部內(nèi)容了,希望能給大家一個(gè)參考,也希望大家多多支持億速云。
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