4

鉴于下表：

X = 

        col1    col2    col3
row1    "A"      "A"     "1.0"
row2    "A"      "B"     "0.9"
row3    "A"      "C"     "0.4"
row4    "B"      "A"     "0.9"
row5    "B"      "B"     "1.0"
row6    "B"      "C"     "0.2"
row7    "C"      "A"     "0.4"
row8    "C"      "B"     "0.2"
row9    "C"      "C"     "1.0"

其中 col3 是 col1 和 col2 中实体对之间的相关性度量。

如何构造一个列名称为 col1、行名称为 col2、矩阵单元格中的值由 col3 填充的矩阵？

4

2 回答 2

3 
df <- read.table(textConnection('col1    col2    col3
row1    "A"      "A"     "1.0"
row2    "A"      "B"     "0.9"
row3    "A"      "C"     "0.4"
row4    "B"      "A"     "0.9"
row5    "B"      "B"     "1.0"
row6    "B"      "C"     "0.2"
row7    "C"      "A"     "0.4"
row8    "C"      "B"     "0.2"
row9    "C"      "C"     "1.0"'), header=T)

## fetch row/column indices
rows <- match(df$col1, LETTERS)
cols <- match(df$col2, LETTERS)

## create matrix
m <- matrix(0, nrow=max(rows), ncol=max(cols))

## fill matrix
m[cbind(rows, cols)] <- df$col3

m
#     [,1] [,2] [,3]
#[1,]  1.0  0.9  0.4
#[2,]  0.9  1.0  0.2
#[3,]  0.4  0.2  1.0
于 2013-09-01T17:32:07.203 回答
3

需要一些数据来处理，所以我会弥补一些。

# Make fake data
x <- c('A','B','C')
dat <- expand.grid(x, x)
dat$Var3 <- rnorm(9)

我们可以使用base R来做到这一点。我对“重塑”功能不是很好，但你可以这样做。列名之后需要清理

> reshape(dat, idvar = "Var1", timevar = "Var2", direction = "wide")
  Var1     Var3.A      Var3.B     Var3.C
1    A -1.2442937 -0.01132871 -0.5693153
2    B -1.6044295 -1.34907504  1.6778866
3    C  0.5393472 -1.00637345 -0.7694940

或者，您可以使用dcastreshape2 包中的函数。我认为输出更干净一些。

> library(reshape2)
> dcast(dat, Var1 ~ Var2, value.var = "Var3")
  Var1          A           B          C
1    A -1.2442937 -0.01132871 -0.5693153
2    B -1.6044295 -1.34907504  1.6778866
3    C  0.5393472 -1.00637345 -0.7694940
于 2013-09-01T17:41:20.463 回答