在R中按顺序合并数据帧对

时间:2016-03-31 14:46:18

标签: r

我有一个数据框,其中包含多个采样间隔的多个站点的标记个体。见下面的例子:

> df
   Tag   Site Interval Ind_ID
1  507 Golden        7      1
2  507 Golden        8      1
3  552 Golden        2      1
4  552 Golden        1      1
5  847 Golden        4      1
6  847 Golden        6      1
8  847 Golden        5      1
9  847 Golden        3      1
31 541 Golden        1      1
33 541 Golden        3      1
34 541 Golden        4      1
35 541 Golden        7      1
36 541 Golden        6      1
37 541 Golden        5      1
39 810 Golden        7      1
40 810 Golden        8      1
41 840 Golden        7      1
42 840 Golden        8      1
43 840 Golden        3      1
44 840 Golden        2      1

我尝试做的是按时间间隔标记被标记的个体,我已经使用此for循环完成:

for (i in 1:nlevels(factor(df$Interval))){
  I<-subset(df,Interval==levels(factor(df$Interval))[i])
  assign(paste("Interval_", i, sep = ""), I)}

然后按顺序合并数据帧,我目前正在使用此代码进行合并:

IPl2<-merge(Interval_1, Interval_2, by=c("Tag", "Site", "Ind_ID"))
IPl3<-merge(Interval_2, Interval_3, by=c("Tag", "Site", "Ind_ID"))
IPl4<-merge(Interval_3, Interval_4, by=c("Tag", "Site", "Ind_ID"))
IPl5<-merge(Interval_4, Interval_5, by=c("Tag", "Site", "Ind_ID"))
IPl6<-merge(Interval_5, Interval_6, by=c("Tag", "Site", "Ind_ID"))
IPl7<-merge(Interval_6, Interval_7, by=c("Tag", "Site", "Ind_ID"))
IPl8<-merge(Interval_7, Interval_8, by=c("Tag", "Site", "Ind_ID"))

我确信这是一种更有效的方式。此外,我不断向数据集添加数据(即更多间隔),我希望每次添加新数据时都不必编辑代码。有什么想法吗?

2 个答案:

答案 0 :(得分:0)

也许是这样的:

dfs <- split(df,df$Interval)
n <- nlevels(factor(df$Interval))-1
results <- setNames(vector("list",length = n),paste0("IPl",2:(n+1)))
for (i in seq_len(n)){
    results[[i]] <- merge(dfs[[i]],dfs[[i+1]],by = c('Tag','Site','Ind_ID'))
}

> head(results)

$IPl2
  Tag   Site Ind_ID Interval.x Interval.y
1 552 Golden      1          1          2

$IPl3
  Tag   Site Ind_ID Interval.x Interval.y
1 840 Golden      1          2          3

$IPl4
  Tag   Site Ind_ID Interval.x Interval.y
1 541 Golden      1          3          4
2 847 Golden      1          3          4

$IPl5
  Tag   Site Ind_ID Interval.x Interval.y
1 541 Golden      1          4          5
2 847 Golden      1          4          5

$IPl6
  Tag   Site Ind_ID Interval.x Interval.y
1 541 Golden      1          5          6
2 847 Golden      1          5          6

$IPl7
  Tag   Site Ind_ID Interval.x Interval.y
1 541 Golden      1          6          7

答案 1 :(得分:0)

下面是一个dplyr解决方案,它将数据框与自身连接起来,并将结果放入数据框中。

library(dplyr)
## Join the 'df' to itself based on the intervals to compare; this is done by
## creating a key to indicate which intervals to join on.
resultdf <-
    ## Create match_interval to next sequential value
    df %>% mutate(match_interval = paste0('IPl', as.numeric(Interval)+1)) %>% arrange(Interval, Site) %>%
    ## Join to self by match_interval and other columns.
    inner_join(df %>% mutate(match_interval = paste0('IPl', as.numeric(Interval))),
               by = c('Tag', 'Site', 'Ind_ID', 'match_interval')) %>%
    ## Order columns
    select(match_interval, Tag, Site, Ind_ID, Interval.x, Interval.y)


resultsdf

##    match_interval Tag   Site Ind_ID Interval.x Interval.y
## 1            IPl2 552 Golden      1          1          2
## 2            IPl3 840 Golden      1          2          3
## 3            IPl4 847 Golden      1          3          4
## 4            IPl4 541 Golden      1          3          4
## 5            IPl5 847 Golden      1          4          5
## 6            IPl5 541 Golden      1          4          5
## 7            IPl6 847 Golden      1          5          6
## 8            IPl6 541 Golden      1          5          6
## 9            IPl7 541 Golden      1          6          7
## 10           IPl8 507 Golden      1          7          8
## 11           IPl8 810 Golden      1          7          8
## 12           IPl8 840 Golden      1          7          8