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This question already has answers here:

Reshaping multiple sets of measurement columns (wide format) into single columns (long format) (8 answers)

Closed 5 years ago.

我拥有如此庞大的数据框架:

id    x1    x2    x3    y1    y2    y3    z1    z2    z3     v 
 1     2     4     5    10    20    15   200   150   170   2.5
 2     3     7     6    25    35    40   300   350   400   4.2

我需要建立这样一个数据框架:

id   xsource   xvalue   yvalue   zvalue       v 
 1        x1        2       10      200     2.5
 1        x2        4       20      150     2.5
 1        x3        5       15      170     2.5
 2        x1        3       25      300     4.2
 2        x2        7       35      350     4.2
 2        x3        6       40      400     4.2

I m quite sure I have to do it with the reshape package, but I m not able to get what I want.

你们能否帮助我?

增编

Answer 1

此处为<代码>reshape() 解决办法。

关键比值是,<代码>varying= 论点可以采用与长期单一变量相对应的广泛格式的栏目矢量清单。在这种情况下,原始数据框架中的<代码>x1”、“x2”、“x3”>栏将发送至长期数据框架中的一个栏目<代码>y1, y2, y3”。

# Read in the original data, x, from Andrie s answer

res <- reshape(x, direction = "long", idvar = "id",
               varying = list(c("x1","x2", "x3"), 
                              c("y1", "y2", "y3"), 
                              c("z1", "z2", "z3")),
               v.names = c("xvalue", "yvalue", "zvalue"), 
               timevar = "xsource", times = c("x1", "x2", "x3"))
#      id   v xsource xvalue yvalue zvalue
# 1.x1  1 2.5      x1      2     10    200
# 2.x1  2 4.2      x1      3     25    300
# 1.x2  1 2.5      x2      4     20    150
# 2.x2  2 4.2      x2      7     35    350
# 1.x3  1 2.5      x3      5     15    170
# 2.x3  2 4.2      x3      6     40    400

最后,如你的问题所示,要取得以下成果,就必须采取两步纯共同的步骤:

res <- res[order(res$id, res$xsource), c(1,3,4,5,6,2)]
row.names(res) <- NULL
res
#   id xsource xvalue yvalue zvalue   v
# 1  1      x1      2     10    200 2.5
# 2  1      x2      4     20    150 2.5
# 3  1      x3      5     15    170 2.5
# 4  2      x1      3     25    300 4.2
# 5  2      x2      7     35    350 4.2
# 6  2      x3      6     40    400 4.2

Answer 2

在此,我关于reshape2的方法。

步骤1:确定已经列入各栏的变量。在本案中:id,和 v. 这些变量是我们带的变量。

library(reshape2)
xm <- melt(x, c("id", "v"))

步骤2:将目前合并为一个栏的变量分开。在这种情况下,来源(特性部分)和复制(分类部分):

有许多方法可以做到这一点,Im将使用带载体的提取法stringr<>>。一揽子计划



library(stringr)
xm$source <- str_sub(xm$variable, 1, 1)
xm$rep <- str_sub(xm$variable, 2, 2)
xm$variable <- NULL


步骤3:重新列出目前各行各行各行各业但我们在各栏目中希望的变数:

dcast(xm, ... ~ source)

#   id   v rep x  y   z
# 1  1 2.5     1 2 10 200
# 2  1 2.5     2 4 20 150
# 3  1 2.5     3 5 15 170
# 4  2 4.2     1 3 25 300
# 5  2 4.2     2 7 35 350
# 6  2 4.2     3 6 40 400

Answer 3

有些人要证明我错了,但我认为,利用<代码>reshape<>/code>的包裹或基底<代码>reshape<<>/code>功能,解决这一问题并不容易。

However, it s easy enough using lapply and do.call:

Replicate the data:

x <- read.table(text="
id    x1    x2    x3    y1    y2    y3    z1    z2    z3     v 
1     2     4     5    10    20    15   200   150   170   2.5
2     3     7     6    25    35    40   300   350   400   4.2
", header=TRUE)

分析意见

chunks <- lapply(1:nrow(x), 
    function(i)cbind(x[i, 1], 1:3, matrix(x[i, 2:10], ncol=3), x[i, 11]))
res <- do.call(rbind, chunks)
colnames(res) <- c("id", "source", "x", "y", "z", "v")
res

     id source x y  z   v  
[1,] 1  1      2 10 200 2.5
[2,] 1  2      4 20 150 2.5
[3,] 1  3      5 15 170 2.5
[4,] 2  1      3 25 300 4.2
[5,] 2  2      7 35 350 4.2
[6,] 2  3      6 40 400 4.2

Answer 4

采用重塑古阿姆集团一揽子计划。它利用了纸浆包和沙皮书包,为你提供了便于使用的接口,使你能够在你执行之前审查你的重塑。该文件还给你以重塑法,使你能够把它传入你的可再生文字,从而你能够学会使用斜线,并在改塑中投下指挥。它对复杂的数据操纵,例如对背叛国的人的操纵,是一种冰.。

Answer 5

这里是了解这一问题的人可能感兴趣的两个新办法:

备选办法1:反对

library(tidyverse)
x %>% 
  gather(var, val, -id, -v) %>% 
  extract(var, into = c("header", "source"), regex = "([a-z])([0-9])") %>% 
  spread(header, val)
#   id   v source x  y   z
# 1  1 2.5      1 2 10 200
# 2  1 2.5      2 4 20 150
# 3  1 2.5      3 5 15 170
# 4  2 4.2      1 3 25 300
# 5  2 4.2      2 7 35 350
# 6  2 4.2      3 6 40 400

备选办法2:数据。

library(data.table)
setDT(x)
melt(x, measure.vars = patterns("x", "y", "z"), 
     value.name = c("x", "y", "z"), 
     variable.name = "source")
#    id   v source x  y   z
# 1:  1 2.5      1 2 10 200
# 2:  2 4.2      1 3 25 300
# 3:  1 2.5      2 4 20 150
# 4:  2 4.2      2 7 35 350
# 5:  1 2.5      3 5 15 170
# 6:  2 4.2      3 6 40 400

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