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？

数据框如下所示：

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

我尝试了以下操作：

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 - ：＆＃39; datetime.datetime＆＃39;和＆＃39; str＆＃39;

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

您的dob列目前似乎是字符串。第一，使用Timestamps将其转换为pd.to_datetime。
格式'%m%d%y'将最后两位数字转换为年份，但是不幸的是假设52意味着2052年。因为那可能不是希瑟诺埃尔的出生年份，让我们从dob减去100年只要dob大于now。您可能希望在条件now中减去几年至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......
您可以从dob中减去now以获取timedelta64[ns]。至将其转换为年份，请使用astype('<m8[Y]')或astype('timedelta64[Y]')。

Answer 2

我找到了更简单的解决方案：

// In code Use share GIF and Video for WhatsApp....

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savePath = [[NSBundle mainBundle] pathForResource:@"Movie" ofType:@"m4v"];
_documentInteractionController = [UIDocumentInteractionController interactionControllerWithURL:_videourl];
_documentInteractionController.UTI = @"net.whatsapp.movie";
_documentInteractionController.delegate = (id)self;
[_documentInteractionController presentOpenInMenuFromRect:CGRectMake(0, 0, 0, 0) inView:self.view animated: YES];

输出：

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)

Answer 3

首先想到的是你的年龄是两位数，这在这个时代并不是一个不错的选择。无论如何，我会假设像05这样的所有年份实际上都是1905。这可能不正确（！）但是提出正确的规则将在很大程度上依赖于您的数据。

from datetime import date

def age(date1, date2):
    naive_yrs = date2.year - date1.year
    if date1.replace(year=date2.year) > date2:
        correction = -1
    else:
        correction = 0
    return naive_yrs + correction

df1['age'] = df1['dob'].map(lambda x: age(date(int('19' + x[-2:]), int(x[:2]), int(x[2:-2])), date.today()))

Answer 4

# 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')

如果只希望年龄中的年份部分，请使用@unutbu's solution。。。

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来计算月份差异

(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

如果要（几乎）精确的年龄（包括小数部分），请查询total_seconds并除。

(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 5

以下解决方案呢？

import datetime as dt
import numpy as np
import pandas as pd
from dateutil.relativedelta import relativedelta

df1['age'] = [relativedelta(pd.to_datetime('now'), d).years for d in df1['dob']]

Answer 6

当您尝试从出生日期列中查找当前年份的年龄时，请使用此衬纸

import pandas as pd

df["dob"] = pd.to_datetime(data["dob"])

df["age"] = df["dob"].apply(lambda x : (pd.datetime.now().year - x.year))

大熊猫从约会开始（例如：出生日期）

6 个答案: