英文:
I am looking for a shorter function to group similar datasets from a list
问题
我有一个分散的数据库列表。我想要将相同年份的数据库分组；
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)
你可以使用以下函数来完成：
list(`2012`=rbind(mydatalist$`12`,mydata2012),`2013`=rbind(mydatalist$`13`,mydata2013),`2014`=rbind(mydatalist$`14`,mydata2014))
但我想要让它更简洁（不需要为每一年编写一行代码），因为我们已经有模式 2012:2021，可以这样做：
12:21
英文:
I have scattered databases in a list. I would like to group databases of the same year;
mydata12<-data.frame(Age=c(12,13),Sex=c("F","H"), Weight=c(70,75),year=c(2012))mydata13<-data.frame(Age=c(14,15),Sex=c("F","H"), Weight=c(70,75),year=c(2013))mydata14<-data.frame(Age=c(16,17),Sex=c("F","H"), Weight=c(70,75),year=c(2014))mydata2012<-data.frame(Age=c(18,19),Sex=c("F","H"), Weight=c(70,75),year=c(2012))mydata2013<-data.frame(Age=c(20,13),Sex=c("H","H"), Weight=c(70,75),year=c(2013))mydata2014<-data.frame(Age=c(22,13),Sex=c("F","F"), Weight=c(70,75),year=c(2014))mydatalist<-list(`12`=mydata12,`13`=mydata13,`14`=mydata14,`2013`=mydata2012,`2014`=mydata2013,`2015`=mydata2014)
I can do it with this function
list(`2012`=rbind(mydatalist$`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
答案1
得分: 3
We could use bind_rows with group_split:
library(dplyr)bind_rows(mydatalist) %>%split(f = as.factor(.$year))
$`2012`Age Sex Weight year1 12 F 70 20122 13 H 75 20127 18 F 70 20128 19 H 75 2012$`2013`Age Sex Weight year3 14 F 70 20134 15 H 75 20139 20 H 70 201310 13 H 75 2013$`2014`Age Sex Weight year5 16 F 70 20146 17 H 75 201411 22 F 70 201412 13 F 75 2014
英文:
We could use bind_rows with group_split:
library(dplyr)bind_rows(mydatalist) %>%split(f = as.factor(.$year))
$`2012`Age Sex Weight year1 12 F 70 20122 13 H 75 20127 18 F 70 20128 19 H 75 2012$`2013`Age Sex Weight year3 14 F 70 20134 15 H 75 20139 20 H 70 201310 13 H 75 2013$`2014`Age Sex Weight year5 16 F 70 20146 17 H 75 201411 22 F 70 201412 13 F 75 2014
答案2
得分: 3
In base R, 使用以下方式之一来处理列表名称的子串，即最后两位数字，或者在列表名称仅包含2个字符的情况下添加20作为前缀，然后执行split和rbind：
out <- lapply(split(mydatalist, sub("^(\\d{2})$", "20\", names(mydatalist))), \(x) `row.names<-`(do.call(rbind, x), NULL))
-output
> out$`2012`Age Sex Weight year1 12 F 70 20122 13 H 75 20123 18 F 70 20124 19 H 75 2012$`2013`Age Sex Weight year1 14 F 70 20132 15 H 75 20133 20 H 70 20134 13 H 75 2013$`2014`Age Sex Weight year1 16 F 70 20142 17 H 75 20143 22 F 70 20144 13 F 75 2014
英文:
In base R, use either the substring of the list names i.e. the last 2 digits or add 20 as prefix to those have only 2 characters in the list name, then split the list and rbind
out <- lapply(split(mydatalist, sub("^(\\d{2})$", "20\",names(mydatalist))), \(x) `row.names<-`(do.call(rbind, x), NULL))
-output
> out$`2012`Age Sex Weight year1 12 F 70 20122 13 H 75 20123 18 F 70 20124 19 H 75 2012$`2013`Age Sex Weight year1 14 F 70 20132 15 H 75 20133 20 H 70 20134 13 H 75 2013$`2014`Age Sex Weight year1 16 F 70 20142 17 H 75 20143 22 F 70 20144 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
<details><summary>英文:</summary>I am not sure how "short" 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>
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