Python：出现“列表索引超出范围”错误；我知道为什么但不知道如何解决

Question

我目前正在从事一个数据科学项目。想法是从“glassdoor_jobs.csv”中清理数据，并以更易于理解的方式呈现。

import pandas as pd

df = pd.read_csv('glassdoor_jobs.csv')

#salary parsing
#Removing "-1" Ratings
#Clean up "Founded"
#state field
#Parse out job description

df['hourly'] = df['Salary Estimate'].apply(lambda x: 1 if 'per hour' in x.lower() else 0)
df['employer_provided'] = df['Salary Estimate'].apply(lambda x: 1 if 'employer provided salary' in x.lower() else 0)
df = df[df['Salary Estimate'] != '-1']
Salary = df['Salary Estimate'].apply(lambda x: x.split('(')[0])
minus_Kd = Salary.apply(lambda x: x.replace('K', '').replace('$',''))

minus_hr = minus_Kd.apply(lambda x: x.lower().replace('per hour', '').replace('employer provided salary:', ''))

df['min_salary'] = minus_hr.apply(lambda x: int(x.split('-')[0]))
df['max_salary'] = minus_hr.apply(lambda x: int(x.split('-')[1]))

我在最后一行收到错误。经过一番挖掘，我在 minus_hr 中发现，一些“Salary Estimate”只有一个数字而不是范围：

index	Salary Estimate
0	150
1	58
2	130
3	125-150
4	110-140
5	200
6	67- 77

等等。现在我想弄清楚如何解决“列表索引超出范围”的问题，并使 max_salary 与只有一个值的单元格的 min_salary 相同。

我也在尝试获取最低和最高工资之间的平均值，如果单元格只有一个值，则将该值设为平均值

所以最后，像索引 0 这样的东西看起来像：

index	min	max	average
0	150	150	150

Answer 1

您必须在某处添加条件语句。

df['min_salary'] = minus_hr.apply(lambda x: int(x.split('-')[0]) if '-' in x else x)

上面可能会做，或者你可以定义一个函数。

def max_salary(cell_value):
    if '-' in cell_value:
        max_salary = split(cell_value, '-')[1]
    else:
        max_salary = cell_value
return max_salary

df['max_salary'] = minus_hr.apply(lambda x: max_salary(x))


def avg_salary(cell_value):
    if '-' in cell_value:
        salaries = split(cell_value,'-')
        avg = sum(salaries)/len(salaries)
    else:
        avg = cell_value
return avg

df['avg_salary'] = minus_hr.apply(lambda x: avg_salary(x))

交换 min_salary 并重复

Answer 2

在访问元素之前测试x.split('-') 的长度。

salaries = x.split('-')
if len(salaries) == 1:
    # only one salary number is given, so assign the same value to min and max 
    df['min_salary'] = df['max_salary'] = minus_hr.apply(lambda x: int(salaries[0]))
else:
    # two salary numbers are given
    df['min_salary'] = minus_hr.apply(lambda x: int(salaries[0]))
    df['max_salary'] = minus_hr.apply(lambda x: int(salaries[1]))

Answer 3

如果you want to avoid.apply()...

试试：

import numpy as np

# extract the two numbers (if there are two numbers) from the 'Salary Estimate' column
sals =  df['Salary Estimate'].str.extractall(r'(?P<min_salary>\d+)[^0-9]*(?P<max_salary>\d*)?')

# reset the new frame's index
sals = sals.reset_index()

# join the extracted min/max salary columns to the original dataframe and fill any blanks with nan
df = df.join(sals[['min_salary', 'max_salary']].fillna(np.nan))

# fill any nan values in the 'max_salary' column with values from the 'min_salary' column
df['max_salary'] = df['max_salary'].fillna(df['min_salary'])

# set the type of the columns to int
df['min_salary'] = df['min_salary'].astype(int)
df['max_salary'] = df['max_salary'].astype(int)

# calculate the average
df['average_salary'] = df.loc[:,['min_salary', 'max_salary']].mean(axis=1).astype(int)

# see what you've got
print(df)

或者不使用正则表达式：

import numpy as np

# extract the two numbers (if there are two numbers) from the 'Salary Estimate' column
df['sals'] =  df['Salary Estimate'].str.split('-')

# expand the list in sals to two columns filling with nan
df[['min_salary', 'max_salary']] = pd.DataFrame(df.sals.tolist()).fillna(np.nan)

# delete the sals column
del df['sals']

# # fill any nan values in the 'max_salary' column with values from the 'min_salary' column
df['max_salary'] = df['max_salary'].fillna(df['min_salary'])

# # set the type of the columns to int
df['min_salary'] = df['min_salary'].astype(int)
df['max_salary'] = df['max_salary'].astype(int)

# # calculate the average
df['average_salary'] = df.loc[:,['min_salary', 'max_salary']].mean(axis=1).astype(int)

# see you've got
print(df)

输出：

  Salary Estimate  min_salary  max_salary  average_salary
0             150         150         150             150
1              58          58          58              58
2             130         130         130             130
3         125-150         125         150             137
4         110-140         110         140             125
5             200         200         200             200
6          67- 77          67          77              72