我是R的新手并且有一个数据格式问题。 我需要改变这个:
Poly Tran Strat Surv MALLP MALLS MALLG MALLF GADWP GADWS GADWG GADWF
AL 1 M y 1 2 0 0 1 4 0 0
ARL 1 M y 0 0 0 0 0 0 20 0
AM 1 M y 0 0 0 0 0 0 0 0
AM 2 M y 1 0 0 0 0 0 0 5
到此:
Poly Tran Strat Surv Spp Num Status
AL 1 M y mall 1 p
AL 1 M y mall 2 s
AL 1 M y gadw 1 p
AL 1 M y gadw 4 s
ARL 1 M y gadw 20 g
AM 2 M y mall 1 p
AM 2 M y gadw 5 f
我需要一些帮助! 三江源。
答案 0 :(得分:3)
使用dplyr
和tidyr
即可:
library(tidyr)
library(dplyr)
df %>% gather(Spp, Num, -Poly, -Tran, -Strat, -Surv) %>%
mutate(Status = tolower(substr(Spp, 5, 5)),
Spp = tolower(substr(Spp, 1, 4))) %>%
filter(!Num == 0) %>%
arrange(Tran)
给出了:
# Poly Tran Strat Surv Spp Num Status
#1 AL 1 M y mall 1 p
#2 AL 1 M y mall 2 s
#3 AL 1 M y gadw 1 p
#4 AL 1 M y gadw 4 s
#5 ARL 1 M y gadw 20 g
#6 AM 2 M y mall 1 p
#7 AM 2 M y gadw 5 f
答案 1 :(得分:2)
使用基础R的解决方案(reshape2::melt
除外):
dat2 <- reshape2::melt(dat, id.vars=c("Poly","Tran","Strat","Surv"))
dat2 <- subset(dat2, subset = value > 0)
dat2$variable <- as.character(dat2$variable)
dat2$Status <- with(dat2, substr(tolower(variable), nchar(variable), nchar(variable)))
dat2$variable <- with(dat2, substr(tolower(variable), 1, nchar(variable)-1))
dat2 <- dat2[order(dat2$Tran),]
colnames(dat2) <- c("Poly", "Tran", "Strat", "Surv", "Spp", "Num", "Status")
rownames(dat2) <- NULL
<强>结果
> dat2
Poly Tran Strat Surv Spp Num Status
1 AL 1 M y mall 1 p
2 AL 1 M y mall 2 s
3 AL 1 M y gadw 1 p
4 AL 1 M y gadw 4 s
5 ARL 1 M y gadw 20 g
6 AM 2 M y mall 1 p
7 AM 2 M y gadw 5 f
数据强>
dat <- read.csv(text = "Poly,Tran,Strat,Surv,MALLP,MALLS,MALLG,MALLF,GADWP,GADWS,GADWG,GADWF
AL,1,M,y,1,2,0,0,1,4,0,0
ARL,1,M,y,0,0,0,0,0,0,20,0
AM,1,M,y,0,0,0,0,0,0,0,0
AM,2,M,y,1,0,0,0,0,0,0,5")
答案 2 :(得分:2)
为了好玩,这里有一个基础R方法。我已经取代了&#34; 0&#34;使用NA
,以便na.omit
可用于&#34;收缩&#34;在stack
值之后关闭数据集。从那里开始,有一些基本的gsub
bing来获得&#34; spp&#34;和&#34;状态&#34;值。
within(na.omit(
cbind(dat[1:4],
stack(replace(dat[-c(1:4)], dat[-c(1:4)] == 0, NA)))), {
ind <- tolower(ind)
spp <- gsub("(mall|gadw).*", "\\1", ind)
status <- gsub("mall|gadw", "", ind)
rm(ind)
})
# Poly Tran Strat Surv values status spp
# 1 AL 1 M y 1 p mall
# 4 AM 2 M y 1 p mall
# 5 AL 1 M y 2 s mall
# 17 AL 1 M y 1 p gadw
# 21 AL 1 M y 4 s gadw
# 26 ARL 1 M y 20 g gadw
# 32 AM 2 M y 5 f gadw
正如多米尼克和史蒂文的答案所证明的那样,订购非常简单。
答案 3 :(得分:0)
这是使用data.table
:
dt.m = melt(dt, id=1:4, variable.name="Spp", value.name="Num")[Num != 0L]
dt.m[, c("Spp", "Status") := list(substring(Spp, 1, 4), substring(Spp, 5, 5))]
或者使用read.fwf
:
dt.m = melt(dt, id=1:4, variable.name="Spp", value.name="Num", variable.factor = FALSE)
dt.m[Num != 0L][, c("Spp", "Status") := read.fwf(textConnection(Spp), widths=c(4,1))]