简单多了!

devtools::install_github("yikeshu0611/onetree") #install onetree package

library(onetree)
widedata=reshape_toWide(data = dat1,id = "name",j = "numbers",value.var.prefix = "value")
widedata

        name     value1     value2     value3     value4
   firstName  0.3407997 -0.7033403 -0.3795377 -0.7460474
  secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357

如果你想从宽返回到长，只改变宽为长，不改变对象。

reshape_toLong(data = widedata,id = "name",j = "numbers",value.var.prefix = "value")

        name numbers      value
   firstName       1  0.3407997
  secondName       1 -0.8981073
   firstName       2 -0.7033403
  secondName       2 -0.3347941
   firstName       3 -0.3795377
  secondName       3 -0.5013782
   firstName       4 -0.7460474
  secondName       4 -0.1745357

即使你有丢失的对，它也能工作，而且它不需要排序(as.matrix(dat1)[，1:2]可以用cbind(dat1[，1]，dat1[，2])替换):

> set.seed(45);dat1=data.frame(name=rep(c("firstName","secondName"),each=4),numbers=rep(1:4,2),value=rnorm(8))
> u1=unique(dat1[,1]);u2=unique(dat1[,2])
> m=matrix(nrow=length(u1),ncol=length(u2),dimnames=list(u1,u2))
> m[as.matrix(dat1)[,1:2]]=dat1[,3]
> m
                    1          2          3          4
firstName   0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357

如果你有缺少的对并且需要排序，这是行不通的，但如果对已经排序了，它会更短一些:

> u1=unique(dat1[,1]);u2=unique(dat1[,2])
> dat1=dat1[order(dat1[,1],dat1[,2]),] # not actually needed in this case
> matrix(dat1[,3],length(u1),,T,list(u1,u2))
                    1          2          3          4
firstName   0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357

下面是第一个方法的函数版本(添加as.data.frame使其与tibbles一起工作):

l2w=function(x,row=1,col=2,val=3,sort=F){
  u1=unique(x[,row])
  u2=unique(x[,col])
  if(sort){u1=sort(u1);u2=sort(u2)}
  out=matrix(nrow=length(u1),ncol=length(u2),dimnames=list(u1,u2))
  out[cbind(x[,row],x[,col])]=x[,val]
  out
}

或者如果你只有下三角形的值，你可以这样做:

> euro=as.matrix(eurodist)[1:3,1:3]
> lower=data.frame(V1=rownames(euro)[row(euro)[lower.tri(euro)]],V2=colnames(euro)[col(euro)[lower.tri(euro)]],V3=euro[lower.tri(euro)])
> lower
         V1        V2   V3
1 Barcelona    Athens 3313
2  Brussels    Athens 2963
3  Brussels Barcelona 1318
> n=unique(c(lower[,1],lower[,2]))
> full=rbind(lower,setNames(lower[,c(2,1,3)],names(lower)),data.frame(V1=n,V2=n,V3=0))
> full
         V1        V2   V3
1 Barcelona    Athens 3313
2  Brussels    Athens 2963
3  Brussels Barcelona 1318
4    Athens Barcelona 3313
5    Athens  Brussels 2963
6 Barcelona  Brussels 1318
7    Athens    Athens    0
8 Barcelona Barcelona    0
9  Brussels  Brussels    0
> l2w(full,sort=T)
          Athens Barcelona Brussels
Athens         0      3313     2963
Barcelona   3313         0     1318
Brussels    2963      1318        0

或者还有另一种方法:

> rc=as.matrix(lower[-3])
> n=sort(unique(c(rc)))
> m=matrix(0,length(n),length(n),,list(n,n))
> m[rc]=lower[,3]
> m[rc[,2:1]]=lower[,3]
> m
          Athens Barcelona Brussels
Athens         0      3313     2963
Barcelona   3313         0     1318
Brussels    2963      1318        0

base R中的另一个简单方法是使用xtabs。xtabs的结果基本上只是一个带有花哨类名的矩阵，但你可以让它看起来像一个普通的矩阵，class(x)=NULL;attr(x，"call")=NULL;dimnames(x)=unname(dimnames(x)):

> x=xtabs(value~name+numbers,dat1);x
            numbers
name                  1          2          3          4
  firstName   0.3407997 -0.7033403 -0.3795377 -0.7460474
  secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357
> str(x)
 'xtabs' num [1:2, 1:4] 0.341 -0.898 -0.703 -0.335 -0.38 ...
 - attr(*, "dimnames")=List of 2
  ..$ name   : chr [1:2] "firstName" "secondName"
  ..$ numbers: chr [1:4] "1" "2" "3" "4"
 - attr(*, "call")= language xtabs(formula = value ~ name + numbers, data = dat1)
> class(x)
[1] "xtabs" "table"
> class(as.matrix(x)) # `as.matrix` has no effect because `x` is already a matrix
[1] "xtabs" "table"
> class(x)=NULL;class(x)
[1] "matrix" "array"
> attr(x,"call")=NULL;dimnames(x)=unname(dimnames(x))
> x # now it looks like a regular matrix
                    1          2          3          4
firstName   0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357
> str(x)
 num [1:2, 1:4] 0.341 -0.898 -0.703 -0.335 -0.38 ...
 - attr(*, "dimnames")=List of 2
  ..$ : chr [1:2] "firstName" "secondName"
  ..$ : chr [1:4] "1" "2" "3" "4"

通常as.data.frame(x)将xtab的结果转换回长格式，但你可以使用class(x)=NULL来避免这种情况:

> x=xtabs(value~name+numbers,dat1);as.data.frame(x)
        name numbers       Freq
1  firstName       1  0.3407997
2 secondName       1 -0.8981073
3  firstName       2 -0.7033403
4 secondName       2 -0.3347941
5  firstName       3 -0.3795377
6 secondName       3 -0.5013782
7  firstName       4 -0.7460474
8 secondName       4 -0.1745357
> class(x)=NULL;as.data.frame(x)
                    1          2          3          4
firstName   0.3407997 -0.7033403 -0.3795377 -0.7460474
secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357

这将宽格式的数据转换为长格式(unlist将数据帧转换为向量，c将矩阵转换为向量):

w2l=function(x)data.frame(V1=rownames(x)[row(x)],V2=colnames(x)[col(x)],V3=unname(c(unlist(x))))

如果考虑性能，另一个选择是使用数据。表格对reshape2的melt和dcast函数的扩展

(参考:使用data.tables进行高效重塑)

library(data.table)

setDT(dat1)
dcast(dat1, name ~ numbers, value.var = "value")

#          name          1          2         3         4
# 1:  firstName  0.1836433 -0.8356286 1.5952808 0.3295078
# 2: secondName -0.8204684  0.4874291 0.7383247 0.5757814

至于数据。表v1.9.6可以对多个列进行强制转换

## add an extra column
dat1[, value2 := value * 2]

## cast multiple value columns
dcast(dat1, name ~ numbers, value.var = c("value", "value2"))

#          name    value_1    value_2   value_3   value_4   value2_1   value2_2 value2_3  value2_4
# 1:  firstName  0.1836433 -0.8356286 1.5952808 0.3295078  0.3672866 -1.6712572 3.190562 0.6590155
# 2: secondName -0.8204684  0.4874291 0.7383247 0.5757814 -1.6409368  0.9748581 1.476649 1.1515627

将三列数据框架重塑为矩阵(“长”到“宽”格式)。这个问题已经结束了，所以我在这里写了一个替代解。

我找到了另一种解决方案，可能对寻找将三列转换为矩阵的人有用。我指的是去耦(2.3.2)包。以下摘自他们的网站

生成一种表，其中行来自id_cols，列来自names_from，值来自values_from。

使用

pivot_wider_profile(
data,
id_cols,
names_from,
values_from,
values_fill = NA,
to_matrix = FALSE,
to_sparse = FALSE,
...
)

基本重塑功能工作得非常好:

df <- data.frame(
  year   = c(rep(2000, 12), rep(2001, 12)),
  month  = rep(1:12, 2),
  values = rnorm(24)
)
df_wide <- reshape(df, idvar="year", timevar="month", v.names="values", direction="wide", sep="_")
df_wide

在哪里

Idvar是分隔行的类列 Timevar是要宽转换的类列 V.names是包含数值的列 方向指定宽或长格式 可选的sep参数是输出data.frame中timevar类名和v.names之间的分隔符。

如果不存在idvar，在使用重塑()函数之前创建一个:

df$id   <- c(rep("year1", 12), rep("year2", 12))
df_wide <- reshape(df, idvar="id", timevar="month", v.names="values", direction="wide", sep="_")
df_wide

只需要记住idvar是必需的!timevar和v.names部分很简单。这个函数的输出比其他一些函数更可预测，因为所有内容都是显式定义的。

只使用dplyr和map。

library(dplyr)
library(purrr)
set.seed(45)
dat1 <- data.frame(
  name = rep(c("firstName", "secondName"), each=4),
  numbers = rep(1:4, 2), value = rnorm(8)
)
longer_to_wider <- function(data, name_from, value_from){
  group <- colnames(data)[!(colnames(data) %in% c(name_from,value_from))]
  data %>% group_by(.data[[group]]) %>%
    summarise( name = list(.data[[name_from]]), 
               value = list(.data[[value_from]])) %>%
    {
      d <- data.frame(
        name = .[[name_from]] %>% unlist() %>% unique()
      )
      e <- map_dfc(.[[group]],function(x){
          y <- data_frame(
            x = data %>% filter(.data[[group]] == x) %>% pull(value_from)
          )
          colnames(y) <- x
          y
      })
      cbind(d,e)
    }
}
longer_to_wider(dat1, "name", "value")
#    name          1          2          3          4
# 1  firstName  0.3407997 -0.7033403 -0.3795377 -0.7460474
# 2 secondName -0.8981073 -0.3347941 -0.5013782 -0.1745357