使用字符串列输入python将年月转换为月

Question

数据集示例：

experience

5 month
nan
1 months
8 month
17 months
8 year
11 years
1.7 year
3.1 years
15.7 months
18 year
2017.2 years
98.3 years
68 year

我有一个专栏，讲述申请人多年的经历。它非常混乱，我试图通过它并创建一个样本。我有数字后跟（月或月）和（年）。

有很多 nan 条目，应该忽略它。

目标是在几个月内创造专栏体验：

if nan
  copy nan to the corresponding column
if the row has month or months 
  copy the number to the corresponding column
if year or years in the row and the number <55 
  the number shall be multiplied by 12 and copied to the corresponding column
else copy nan to the corresponding column

如何做到这一点？

Answer 1

使用 reg 表达式的简单解决方案，保持透明度。

import numpy as np
df = pd.read_csv(io.StringIO("""experience

5 month
nan
1 months
8 month
17 months
8 year
11 years
1.7 year
3.1 years
15.7 months
18 year
2017.2 years
98.3 years
68 year"""))

df = df.assign(unit=lambda dfa: dfa["experience"].str.extract("([a-z]+)+"),
         val=lambda dfa: dfa["experience"].str.extract("([0-9,\.]+)").astype(float),
         months=lambda dfa: np.where(dfa["unit"].isin(["month","months"]), dfa["val"],
                                    np.where(dfa["unit"].isin(["year","years"])
                                             &dfa["val"].lt(55), dfa["val"]*12, np.nan)))

print(df.to_string(index=False))

输出

   experience    unit     val  months
      5 month   month     5.0     5.0
          NaN     NaN     NaN     NaN
     1 months  months     1.0     1.0
      8 month   month     8.0     8.0
    17 months  months    17.0    17.0
       8 year    year     8.0    96.0
     11 years   years    11.0   132.0
     1.7 year    year     1.7    20.4
    3.1 years   years     3.1    37.2
  15.7 months  months    15.7    15.7
      18 year    year    18.0   216.0
 2017.2 years   years  2017.2     NaN
   98.3 years   years    98.3     NaN
      68 year    year    68.0     NaN

Answer 2

使用熊猫数据框可能有更好的方法来做到这一点，但这是您想要实现的目标吗？如果没有别的，您可能可以使用正则表达式。我没有添加条件

import re
applicants = []

applicant1 = {'name': 'Lisa', 'experience': 'nan'}
applicant2 = {'name': 'Bill', 'experience': '3.1 months'}
applicant3 = {'name': 'Mandy', 'experience': '1 month'}
applicant4 = {'name': 'Geoff', 'experience': '6.7 years'}
applicant5 = {'name': 'Patricia', 'experience': '1 year'}
applicant6 = {'name': 'Kirsty', 'experience': '2017.2 years'}

applicants.append(applicant1)
applicants.append(applicant2)
applicants.append(applicant3)
applicants.append(applicant4)
applicants.append(applicant5)
applicants.append(applicant6)

print(applicants)

month_pattern = '^([\d]+[\.\d]*) month(s*)'
year_pattern = '^([\d]+[\.\d]*) year(s*)'

applicant_output = []

for applicant in applicants:
    if applicant['experience'] == 'nan':
        applicant_output.append(applicant)
    else:
        month = re.search(month_pattern, applicant['experience'])
        if month is not None:
            applicant_output.append(
                {
                    'name': applicant['name'],
                    "exprience_months": month.group(1)
                })
        else:
            year = re.search(year_pattern, applicant['experience'])
            if year is not None:
                months = str(float(year.group(1)) * 12)
                applicant_output.append(
                    {
                        'name': applicant['name'],
                        "exprience_months": months
                    })

print(applicant_output)

这给出了输出：

[{'name': 'Lisa', 'experience': 'nan'}, {'name': 'Bill', 'experience': '3.1 months'}, {'name': 'Mandy', 'experience': '1 month'}, {'name': 'Geoff', 'experience': '6.7 years'}, {'name': 'Patricia', 'experience': '1 year'}, {'name': 'Kirsty', 'experience': '2017. years'}]

结果：

[{'name': 'Lisa', 'experience': 'nan'}, {'name': 'Bill', 'exprience_months': '3.1'}, {'name': 'Mandy', 'exprience_months': '1'}, {'name': 'Geoff', 'exprience_months': '80.4'}, {'name': 'Patricia', 'exprience_months': '12.0'}, {'name': 'Kirsty', 'exprience_months': '24206.4'}]

Answer 3

这假设格式是一致的（值、空间、时间段）。您可以使用 split 来获取这两个部分。

df = pd.DataFrame({'experience': ['5 month', np.nan, '1 months', '8 month', '17 months', '8 year', '11 years']})

def get_values(x):
    if pd.notnull(x):
        val = int(x.split(' ')[0])
        prd = x.split(' ')[1]
        if prd in ['month', 'months']:
            return val
        elif prd in ['year', 'years'] and val < 55: 
            return val * 12
    else:
        return x

df['months'] = df.apply(lambda x: get_values(x.experience), axis=1)  

输出：

  experience  months
0    5 month     5.0
1        NaN     NaN
2   1 months     1.0
3    8 month     8.0
4  17 months    17.0
5     8 year    96.0
6   11 years   132.0

如果 NaN 的百分比很高，可以先过滤再运行 lambda 函数

df[df.experience.notnull()].apply(lambda x: get_values(x.experience), axis=1)

Answer 4

temp_df 分离月/年部分

temp_df = df['experience'].str.split('([A-Za-z]+)', expand=True)
temp_df = temp_df.loc[:, ~(temp_df == "").any(axis=0)]  # deleting the extra column coming upon split
temp_df[0] = temp_df[0].astype(float)
temp_df

获得经验值的乘数

multiplier = pd.Series([1] * len(temp_df), index=temp_df.index)
year_rows = temp_df[1].str.contains('year', case=False).fillna(False)  # getting the rows which has year
temp_df.loc[(year_rows) & (temp_df[0]>=55), 0] = np.nan  # converting exp value to nan where value is >= 55 and unit is year
multiplier[year_rows] = 12
df['experience_in_months'] = temp_df[0] * multiplier
df

Answer 5

my_dict = {'Experience': ['5 month', 'nan', '1 months', '8 month','17 months','8 year',
                          '11 years','1.7 year', '3.1 years', '15.7 months','18 year',
                          '2017.2 years', '98.3 years', '68 year']}

df = pd.DataFrame(my_dict)

# Create filter for month/months
month_filt = df['Experience'].str.contains('month')

# Filter DataFrame for rows that contain month/months 
df['Months'] = df.loc[month_filt, 'Experience'].str.strip('month|months')

# Create filter for year/years
year_filt = df['Experience'].str.contains('year')

# Filter DataFrame for rows that contain year/years
df['Years'] = df.loc[year_filt, 'Experience'].str.strip('year|years')

# Fill NaN in Years column
df.loc[df['Years'].isna(),'Years'] = np.nan

# Convert Years to months
df.loc[df['Months'].isna(),'Months'] = df['Years'].astype('float') * 12

# Set years greater than 55 to NaN
df.loc[df['Years'].astype('float') > 55, 'Months'] = np.nan

    Experience  Months  Years
0   5 month     5       NaN
1   nan         NaN     NaN
2   1 months    1       NaN
3   8 month     8       NaN
4   17 months   17      NaN
5   8 year      96      8
6   11 years    132     11
7   1.7 year    20.4    1.7
8   3.1 years   37.2    3.1
9   15.7 months 15.7    NaN
10  18 year     216     18
11  2017.2 yearsNaN 2017.2
12  98.3 years  NaN     98.3
13  68 year     NaN     68