R中如何将列转换为行？

发布于 2024-10-21 06:11:15 字数 322 浏览 9 评论 0原文

我也有同样的问题。我有这种顺序的数据： ;=column

D1 ;hurs

1  ;0.12

1  ;0.23

1  ;0.34

1  ;0.01

2  ;0.24

2  ;0.67

2  ;0.78

2  ;0.98

我喜欢这样：

D1; X; X; X; X    
1;0.12; 0.23; 0.34; 0.01; 
2;0.24; 0.67; 0.78; 0.98;

我想根据 D1 对它进行排序并想重塑它？有人有想法吗？我需要对 D1 的 7603 个值执行此操作。

原文

I kind of have the same problem. I have data in this kind of order: ;=column

D1 ;hurs

1  ;0.12

1  ;0.23

1  ;0.34

1  ;0.01

2  ;0.24

2  ;0.67

2  ;0.78

2  ;0.98

and I like to have it like this:

D1; X; X; X; X    
1;0.12; 0.23; 0.34; 0.01; 
2;0.24; 0.67; 0.78; 0.98;

I would like to sort it with respect to D1 and like to reshape it? Does anyone have an idea? I need to do this for 7603 values of D1.

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请远离我 2024-10-28 06:11:15

我会研究 Hadley 的 reshape 包。它可以做各种很棒的事情。下面的代码将适用于您的玩具示例，但可能有一种更优雅的方法来执行此操作。简而言之，您的数据已经呈现为 ?melt 形式，因此您可以简单地 ?cast 它。

另外，请查看这些链接

http://www.statmethods.net/management/reshape.html

http://had.co.nz/reshape/

library(reshape)

help(package=reshape)
?melt

D1 <- c(1,1,1,1,2,2,2,2)
hurs <- c(.12, .23, .34, .01, .24, .67, .78, .98)
var <- rep(paste("X", 1:4, sep=""), 2)

foo <- data.frame(D1, var, hurs)
foo

cast(foo, D1~var)

I would look into Hadley's reshape package. It does all sorts of great stuff. The code below will work with your toy example, but there is probably a more elegant way of doing this. Simply, your data already appear to be in the ?melt form, so you can simply ?cast it.

Also, check out these links

http://www.statmethods.net/management/reshape.html

http://had.co.nz/reshape/

library(reshape)

help(package=reshape)
?melt

D1 <- c(1,1,1,1,2,2,2,2)
hurs <- c(.12, .23, .34, .01, .24, .67, .78, .98)
var <- rep(paste("X", 1:4, sep=""), 2)

foo <- data.frame(D1, var, hurs)
foo

cast(foo, D1~var)

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情深如许 2024-10-28 06:11:15

挖掘不可能被认领的骷髅，为什么不使用aggregate()呢？

dat = read.table(header = TRUE, sep = ";", text = "D1 ;hurs
1  ;0.12
1  ;0.23
1  ;0.34
1  ;0.01
2  ;0.24
2  ;0.67
2  ;0.78
2  ;0.98")
aggregate(hurs ~ D1, dat, c)
#   D1 hurs.1 hurs.2 hurs.3 hurs.4
# 1  1   0.12   0.23   0.34   0.01
# 2  2   0.24   0.67   0.78   0.98

如果 D1 中每个 id 的长度不相同，您还可以在首先创建“时间”变量后使用基本 R reshape()：

dat2 <- dat[-8, ]
dat2$timeSeq <- ave(dat2$D1, dat2$D1, FUN = seq_along)
reshape(dat2, direction="wide", idvar="D1", timevar="timeSeq")
#   D1 hurs.1 hurs.2 hurs.3 hurs.4
# 1  1   0.12   0.23   0.34   0.01
# 5  2   0.24   0.67   0.78     NA

Digging up skeletons not likely to ever be claimed, why not use aggregate()?

dat = read.table(header = TRUE, sep = ";", text = "D1 ;hurs
1  ;0.12
1  ;0.23
1  ;0.34
1  ;0.01
2  ;0.24
2  ;0.67
2  ;0.78
2  ;0.98")
aggregate(hurs ~ D1, dat, c)
#   D1 hurs.1 hurs.2 hurs.3 hurs.4
# 1  1   0.12   0.23   0.34   0.01
# 2  2   0.24   0.67   0.78   0.98

If the lengths of each id in D1 are not the same, you can also use base R reshape() after first creating a "time" variable:

dat2 <- dat[-8, ]
dat2$timeSeq <- ave(dat2$D1, dat2$D1, FUN = seq_along)
reshape(dat2, direction="wide", idvar="D1", timevar="timeSeq")
#   D1 hurs.1 hurs.2 hurs.3 hurs.4
# 1  1   0.12   0.23   0.34   0.01
# 5  2   0.24   0.67   0.78     NA

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伤痕我心 2024-10-28 06:11:15

我假设每个 D1 的小时数不相等（7603 值）

txt = 'D1 ;hurs
 1 ;0.12
 1 ;0.23
 1 ;0.34
 1 ;0.01
 2 ;0.24
 2 ;0.67
 2 ;0.78
 2 ;0.98'

dat <- read.table(textConnection(txt),header=T,sep=";")
dat$Lp <- 1:nrow(dat)
dat <- dat[order(dat$D1,dat$Lp),]
out <- split(dat$hurs,dat$D1)
out <- sapply(names(out),function(x) paste(paste(c(x,out[[x]]),collapse=";"),";",sep="",collapse=""))

I have assumed that there are unequal number of hurs per D1 (7603 values)

txt = 'D1 ;hurs
 1 ;0.12
 1 ;0.23
 1 ;0.34
 1 ;0.01
 2 ;0.24
 2 ;0.67
 2 ;0.78
 2 ;0.98'

dat <- read.table(textConnection(txt),header=T,sep=";")
dat$Lp <- 1:nrow(dat)
dat <- dat[order(dat$D1,dat$Lp),]
out <- split(dat$hurs,dat$D1)
out <- sapply(names(out),function(x) paste(paste(c(x,out[[x]]),collapse=";"),";",sep="",collapse=""))

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