我有一个可复制的数据框架:
df <- data.frame(ID =c(864,121,582,300,765,56,571,819,923,789,438,987,30,446,369,445),
city=c("del","mum","nav","pun","bang","chen","triv","vish","del","mum","bang","vish","bhop","kol","noi","gurg"),
name= c("xab","Lan","mun","mmc","aaf","nnhu","njam","jiha","ntha","gydbt","hytb","kula","huta","vcge","bhsue","nudj"),
DOJ = c("9/5/2005","8/23/2006","3/30/2006","5/29/2009","12/29/2009","6/20/2005","10/2/2010","11/15/2003","3/3/2004","4/23/2004","7/28/2003","8/27/2004","6/14/2007","3/24/2007","9/29/2009","9/4/2007"))我正在尝试创建一个函数,它将询问一个DOJ的列名,比如雇用日期,== "DOJ“,然后变异一个新的列并计算该列中的任期。
此外,如果给出出生日期,则新的雇用日期列大于出生日期+ 20。
我尝试过使用mondf和difftime,但无法为此创建函数。
产出应如下:

发布于 2020-09-08 18:51:51
试试这个解决方案:
#Format date
df$DOJ <- as.Date(df$DOJ,'%m/%d/%Y')
#Compute variable
df$Tenure <- round(as.numeric(difftime(Sys.Date(),df$DOJ,units = 'weeks')/52.25),0)输出:
ID city name DOJ Tenure
1 864 del xab 2005-09-05 15
2 121 mum Lan 2006-08-23 14
3 582 nav mun 2006-03-30 14
4 300 pun mmc 2009-05-29 11
5 765 bang aaf 2009-12-29 11
6 56 chen nnhu 2005-06-20 15
7 571 triv njam 2010-10-02 10
8 819 vish jiha 2003-11-15 17
9 923 del ntha 2004-03-03 16
10 789 mum gydbt 2004-04-23 16
11 438 bang hytb 2003-07-28 17
12 987 vish kula 2004-08-27 16
13 30 bhop huta 2007-06-14 13
14 446 kol vcge 2007-03-24 13
15 369 noi bhsue 2009-09-29 11
16 445 gurg nudj 2007-09-04 13您的数据集没有出生日期,但是新变量中的df$NewVar <- df$Date1>Date2+20这样的情况应该会产生您想要的结果。
https://stackoverflow.com/questions/63799796
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