Pandas统计重复的列里面的值方法

时间:2022-02-03 10:02:52

pandas

代码如下:

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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:

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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

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方法1
 
m = np.zeros_like(a, dtype=bool)
m[np.unique(a, return_index=True)[1]] = True
a[~m]
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方法2
 
a[~np.in1d(np.arange(len(a)), np.unique(a, return_index=True)[1], assume_unique=True)]
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方法3
 
np.setxor1d(a, np.unique(a), assume_unique=True)
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方法4
 
u, i = np.unique(a, return_inverse=True)
u[np.bincount(i) > 1]
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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服务器之家。

原文链接:https://blog.csdn.net/hguo11/article/details/82556171