R 将多id重复变量读数从长改宽
这就是我所拥有的:R 将多id重复变量读数从长改宽,r,reshape,data-science,reshape2,R,Reshape,Data Science,Reshape2,这就是我所拥有的: id<-c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2) measure<-c("speed","weight","time","speed","weight","time","speed","weight","time", "speed","weight","time","speed","weight","time","speed","weight","time") value<-c(1.23,10.3,3
id<-c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2)
measure<-c("speed","weight","time","speed","weight","time","speed","weight","time",
"speed","weight","time","speed","weight","time","speed","weight","time")
value<-c(1.23,10.3,33,1.44,10.4,31,1.21,10.1,33,4.25,12.5,38,1.74,10.8,31,3.21,10.3,33)
testdf<-data.frame(id,measure,value)
id如果您想走tidyverse路线:
library(tidyr)
library(dplyr)
testdf %>%
# add unique id for rows to be able to use spread
group_by(measure) %>% mutate(unique_id = row_number()) %>%
spread(measure, value) %>% select(-unique_id )
R Cookbook是这类问题的绝佳资源:使用data.table中的rowid
(与@Kelli Jean的答案非常相似):
这是我的解决办法
library(plyr)
a=daply(testdf, .(id, measure), function(x) x$value)
listdf=apply(a, c(3), function(x) rbind(data.frame(x,id=row.names(x))))
df <- ldply(listdf, data.frame)
df$.id=NULL
df <- df[order(df$id),]
df
speed time weight id
1 1.23 33 10.3 1
3 1.44 31 10.4 1
5 1.21 33 10.1 1
2 4.25 38 12.5 2
4 1.74 31 10.8 2
6 3.21 33 10.3 2
库(plyr)
a=daply(testdf,.(id,measure),函数(x)x$value)
listdf=apply(a,c(3),函数(x)rbind(data.frame(x,id=row.names(x)))
df首先安装重塑2以帮助重新格式化数据
然后创建另一个标识符,以帮助按所需数据集中的三行顺序一次组织数据
A<-c("A","A","A")
B<-c("B","B","B")
C<-c("C","C","C")
D<-c("D","D","D")
E<-c("E","E","E")
F<-c("F","F","F")
A <- as.data.frame(A)
colnames(A) <- "id2"
B <- as.data.frame(B)
colnames(B) <- "id2"
C <- as.data.frame(C)
colnames(C) <- "id2"
D <- as.data.frame(D)
colnames(D) <- "id2"
E <- as.data.frame(E)
colnames(E) <- "id2"
F <- as.data.frame(F)
colnames(F) <- "id2"
如果有必要,可以使用删除id2变量
testdf$id2 <- NULL
testdf$id2我被之前发布的答案弄糊涂了,不过你的答案很有道理~补充我的:),谢谢你的回答,澄清这个问题你为什么要使用Reforme2?您是否试图将其保持为data.frame格式?为什么不发布完整的data.table
solution:library(data.table);dcast(如.data.table(testdf)[,rn:=rowid(measure)],id+rn~measure)[,-“rn”]
或dcast(setDT(testdf),粘贴(id,rowid(measure),sep=“”)~measure)
?
library(plyr)
a=daply(testdf, .(id, measure), function(x) x$value)
listdf=apply(a, c(3), function(x) rbind(data.frame(x,id=row.names(x))))
df <- ldply(listdf, data.frame)
df$.id=NULL
df <- df[order(df$id),]
df
speed time weight id
1 1.23 33 10.3 1
3 1.44 31 10.4 1
5 1.21 33 10.1 1
2 4.25 38 12.5 2
4 1.74 31 10.8 2
6 3.21 33 10.3 2
A<-c("A","A","A")
B<-c("B","B","B")
C<-c("C","C","C")
D<-c("D","D","D")
E<-c("E","E","E")
F<-c("F","F","F")
A <- as.data.frame(A)
colnames(A) <- "id2"
B <- as.data.frame(B)
colnames(B) <- "id2"
C <- as.data.frame(C)
colnames(C) <- "id2"
D <- as.data.frame(D)
colnames(D) <- "id2"
E <- as.data.frame(E)
colnames(E) <- "id2"
F <- as.data.frame(F)
colnames(F) <- "id2"
x<-rbind(A,B,C,D,E,F)
testdf <- cbind(testdf, x)
x2<-dcast(testdf, id + id2 ~ measure, value.var="value")
id id2 speed time weight
1 1 A 1.23 33 10.3
2 1 B 1.44 31 10.4
3 1 C 1.21 33 10.1
4 2 D 4.25 38 12.5
5 2 E 1.74 31 10.8
6 2 F 3.21 33 10.3
testdf$id2 <- NULL