如何在pandas中定义用户定义函数

2024-03-27 22:33:26 发布

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我有一个csv文件,其中包含如下信息

name    salary  department
a        2500      x
b        5000      y
c        10000      y
d        20000      x 

我需要用熊猫把它转换成

dept    name    position
x        a       Normal Employee
x        b       Normal Employee
y        c       Experienced Employee
y        d       Experienced Employee

如果薪资<;=8000职位是普通员工

如果薪资8000<;=25000是有经验的员工

分组依据的默认代码

import csv
import pandas
pandas.set_option('display.max_rows', 999)
data_df = pandas.read_csv('employeedetails.csv')
#print(data_df.columns)
t = data_df.groupby(['dept'])
print t

为了得到上面提到的输出,我需要在这段代码中做哪些更改


Tags: 文件csv代码nameimportltpandasdf
3条回答

一个有用的函数是apply

data_df['position'] = data_df['salary'].apply(lambda salary: 'Normal Employee' if salary <= 8000 else 'Experienced Employee', axis=1)

这将lambda函数应用于salary列中的每个元素。

我会使用一个简单的函数,比如:

def f(x):
    if x <= 8000:
        x = 'Normal Employee'
    elif 8000 < x <= 25000:
        x = 'Experienced Employee'
    return x

然后将其应用于df

df['position'] = df['salary'].apply(f)

您可以定义两个掩码并将它们传递给np.where

In [91]:
normal = df['salary'] <= 8000
experienced = (df['salary'] > 8000) & (df['salary'] <= 25000)
df['position'] = np.where(normal, 'normal emplyee', np.where(experienced, 'experienced employee', 'unknown'))
df

Out[91]:
  name  salary department              position
0    a    2500          x        normal emplyee
1    b    5000          y        normal emplyee
2    c   10000          y  experienced employee
3    d   20000          x  experienced employee

或者更具可读性的是将它们传递给loc

In [92]:
df.loc[normal, 'position'] = 'normal employee'
df.loc[experienced,'position'] = 'experienced employee'
df

Out[92]:
  name  salary department              position
0    a    2500          x       normal employee
1    b    5000          y       normal employee
2    c   10000          y  experienced employee
3    d   20000          x  experienced employee

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