I am looking for a shorter function to group similar datasets from a list

I have scattered databases in a list. I would like to group databases of the same year;

mydata12<-data.frame(Age=c(12,13),Sex=c("F","H"), Weight=c(70,75),year=c(2012))
mydata13<-data.frame(Age=c(14,15),Sex=c("F","H"), Weight=c(70,75),year=c(2013))
mydata14<-data.frame(Age=c(16,17),Sex=c("F","H"), Weight=c(70,75),year=c(2014))
mydata2012<-data.frame(Age=c(18,19),Sex=c("F","H"), Weight=c(70,75),year=c(2012))
mydata2013<-data.frame(Age=c(20,13),Sex=c("H","H"), Weight=c(70,75),year=c(2013))
mydata2014<-data.frame(Age=c(22,13),Sex=c("F","F"), Weight=c(70,75),year=c(2014))



 mydatalist<-list(

  `12`=mydata12,
  `13`=mydata13,
  `14`=mydata14,
  `2013`=mydata2012,
  `2014`=mydata2013,
  `2015`=mydata2014
)

I can do it with this function

list(`2012`=rbind(mydatalist$`12`,mydata2012),
     `2013`=rbind(mydatalist$`13`,mydata2013),
     `2014`=rbind(mydatalist$`14`,mydata2014))

but I would like to make it shorter (without a line of code for each year), since we already have patterns 2012:2021,

12:21

We could use bind_rows with group_split:

library(dplyr)

bind_rows(mydatalist) %>% 
  split(f = as.factor(.$year))

$`2012`
  Age Sex Weight year
1  12   F     70 2012
2  13   H     75 2012
7  18   F     70 2012
8  19   H     75 2012

$`2013`
   Age Sex Weight year
3   14   F     70 2013
4   15   H     75 2013
9   20   H     70 2013
10  13   H     75 2013

$`2014`
   Age Sex Weight year
5   16   F     70 2014
6   17   H     75 2014
11  22   F     70 2014
12  13   F     75 2014

In base R, use either the substring of the list names i.e. the last 2 digits or add 20 as prefix to those have only 2 characters in the list name, then split the list and rbind

out <- lapply(split(mydatalist, sub("^(\\d{2})$", "20\", 
  names(mydatalist))), \(x) `row.names<-`(do.call(rbind, x), NULL))

-output

> out
$`2012`
  Age Sex Weight year
1  12   F     70 2012
2  13   H     75 2012
3  18   F     70 2012
4  19   H     75 2012

$`2013`
  Age Sex Weight year
1  14   F     70 2013
2  15   H     75 2013
3  20   H     70 2013
4  13   H     75 2013

$`2014`
  Age Sex Weight year
1  16   F     70 2014
2  17   H     75 2014
3  22   F     70 2014
4  13   F     75 2014


</details>



# 答案3
**得分**: 3

以下是翻译好的内容：

> split(row.names<-(do.call(rbind, mydatalist), NULL), ~year)
$2012
Age Sex Weight year
1 12 F 70 2012
2 13 H 75 2012
7 18 F 70 2012
8 19 H 75 2012

$2013
Age Sex Weight year
3 14 F 70 2013
4 15 H 75 2013
9 20 H 70 2013
10 13 H 75 2013

$2014
Age Sex Weight year
5 16 F 70 2014
6 17 H 75 2014
11 22 F 70 2014
12 13 F 75 2014

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<summary>英文:</summary>

I am not sure how &quot;short&quot; you are happy with, but below might be an option

> split(row.names<-(do.call(rbind, mydatalist), NULL), ~year)
$2012
Age Sex Weight year
1 12 F 70 2012
2 13 H 75 2012
7 18 F 70 2012
8 19 H 75 2012

$2013
Age Sex Weight year
3 14 F 70 2013
4 15 H 75 2013
9 20 H 70 2013
10 13 H 75 2013

$2014
Age Sex Weight year
5 16 F 70 2014
6 17 H 75 2014
11 22 F 70 2014
12 13 F 75 2014

</details>