大熊猫从约会开始(例如:出生日期)

Question

如何计算人的年龄(基于dob列)并使用新值向数据框添加列？

dataframe如下所示:

    lname      fname     dob
0    DOE       LAURIE    03011979
1    BOURNE    JASON     06111978
2    GRINCH    XMAS      12131988
3    DOE       JOHN      11121986

我尝试过以下操作:

now = datetime.now()
df1['age'] = now - df1['dob']

但是,收到以下错误:

TypeError:不支持的操作数类型 - :'datetime.datetime'和'str'

Answer 1

import datetime as DT
import io
import numpy as np
import pandas as pd

pd.options.mode.chained_assignment = 'warn'

content = '''     ssno        lname         fname    pos_title             ser  gender  dob 
0    23456789    PLILEY     JODY        BUDG ANAL             0560  F      031871 
1    987654321   NOEL       HEATHER     PRTG SRVCS SPECLST    1654  F      120852
2    234567891   SONJU      LAURIE      SUPVY CONTR SPECLST   1102  F      010999
3    345678912   MANNING    CYNTHIA     SOC SCNTST            0101  F      081692
4    456789123   NAUERTZ    ELIZABETH   OFF AUTOMATION ASST   0326  F      031387'''

df = pd.read_csv(io.StringIO(content), sep='\s{2,}')
df['dob'] = df['dob'].apply('{:06}'.format)

now = pd.Timestamp('now')
df['dob'] = pd.to_datetime(df['dob'], format='%m%d%y')    # 1
df['dob'] = df['dob'].where(df['dob'] < now, df['dob'] -  np.timedelta64(100, 'Y'))   # 2
df['age'] = (now - df['dob']).astype('<m8[Y]')    # 3
print(df)

产量

        ssno    lname      fname            pos_title   ser gender  \
0   23456789   PLILEY       JODY            BUDG ANAL   560      F   
1  987654321     NOEL    HEATHER   PRTG SRVCS SPECLST  1654      F   
2  234567891    SONJU     LAURIE  SUPVY CONTR SPECLST  1102      F   
3  345678912  MANNING    CYNTHIA           SOC SCNTST   101      F   
4  456789123  NAUERTZ  ELIZABETH  OFF AUTOMATION ASST   326      F   

                  dob  age  
0 1971-03-18 00:00:00   43  
1 1952-12-08 18:00:00   61  
2 1999-01-09 00:00:00   15  
3 1992-08-16 00:00:00   22  
4 1987-03-13 00:00:00   27  

1. 看起来您的dob列目前是字符串.首先,将它们转换为Timestamps使用pd.to_datetime.
2. 该格式'%m%d%y'的最后两个数字转换为多年,但不幸的是假定52手段2052自那可能不是希瑟诺埃尔的birthyear,让我们减去100年dob 只要dob是大于now.您可能希望now在这种情况下减去几年,df['dob'] < now因为与一名1岁的工人相比,他可能稍微有一个101岁的工人......
3. 您可以减去dob从now获得timedelta64 [NS] .要将其转换为年,请使用astype('<m8[Y]')或astype('timedelta64[Y]').

Answer 2

# Data setup
df

    lname   fname        dob
0     DOE  LAURIE 1979-03-01
1  BOURNE   JASON 1978-06-11
2  GRINCH    XMAS 1988-12-13
3     DOE    JOHN 1986-11-12

# Make sure to parse all datetime columns in advance
df['dob'] = pd.to_datetime(df['dob'], errors='coerce')

如果您只想要年龄的年份部分，请使用@unutbu 的解决方案。. .

now = pd.to_datetime('now')
now
# Timestamp('2019-04-14 00:00:43.105892')

(now - df['dob']).astype('<m8[Y]') 

0    40.0
1    40.0
2    30.0
3    32.0
Name: dob, dtype: float64

另一种选择是减去年份部分并使用

(now.year - df['dob'].dt.year) - ((now.month - df['dob'].dt.month) < 0)

0    40
1    40
2    30
3    32
Name: dob, dtype: int64

如果您想要（几乎）精确的年龄（包括小数部分），请查询total_seconds并除以。

(now - df['dob']).dt.total_seconds() / (60*60*24*365.25)

0    40.120446
1    40.840501
2    30.332630
3    32.418872
Name: dob, dtype: float64

Answer 3

我找到了更简单的解决方案：

import pandas as pd
from datetime import datetime
from datetime import date

d = {'col0': [1, 2, 6], 'col1': [3, 8, 3], 'col2': ['17.02.1979',
          '11.11.1993',
          '01.08.1961']}

df = pd.DataFrame(data=d)

def calculate_age(born):
    born = datetime.strptime(born, "%d.%m.%Y").date()
    today = date.today()
    return today.year - born.year - ((today.month, today.day) < (born.month, born.day))

df['age'] = df['col6'].apply(calculate_age)
print(df)

输出：

     col0  col1  col3        age
0       1     3  17.02.1979   39
1       2     8  11.11.1993   24
2       6     3  01.08.1961   57