R 计算经验年数

我有一个可复制的数据框：

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"))

df尝试以下解决方案：

#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)

您的数据集没有出生日期，但新变量中类似于
df$NewVar Date2+20
的条件应该会产生您想要的结果。
我现在收到错误，以前它是as.POSIXlt.character（x，tz，…）中的工作错误：字符串不是标准的无歧义格式format@rjunkie2请检查您的数据结构。可能某些日期是因子或具有其他格式。df$NewVar df$DOB+20 ok将进行检查，但这也会在df$DOB+20中给出错误：二进制运算符>类（雇用日期）[1]“日期”的非数字参数，但它表示只有日期格式您的方法应该有效。我在虚拟数据上进行了测试，它是有效的。任何变量都是字符！
    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