Question

我有一个相当大的表，这是一个示例：

dput(data)
structure(list(ID = 1:5, area = c(6L, 6L, 6L, 6L, 6L), ERC = c(1L, 
1L, 1L, 1L, 1L), a = c(33, 34, 35, 38, 39), b = c(38, 41, 45, 
8, NA), c = c(53, 35, 38, 39, 53), d = c(32, 33, 65, 36, 34)), .Names = c("ID", 
"area", "ERC", "a", "b", "c", "d"), row.names = c(NA, -5L), class = "data.frame")

我想创建一个观察缺席/存在的矩阵。结果应如下所示：

dput(result)
structure(list(ID = c(1, 2), `32` = c(1, 0), `33` = c(1, 1), 
    `34` = c(0, 1), `35` = c(0, 1), `37` = c(0, 0), `38` = c(0, 
    0), `41` = c(1, 0), `53` = c(0, 1), `54` = c(1, 0)), .Names = c("ID", 
"32", "33", "34", "35", "37", "38", "41", "53", "54"), row.names = c(NA, 
-2L), class = "data.frame")

由于第I列有因子（观察），1表示存在，0表示缺席。

有没有办法在不为每列生成简单的矩阵列的情况下进行此操作？

Answer 1

library(reshape2)
dcast(melt(DF, id.vars = c("ID", "area", "ERC")), 
  ID ~ value, fill = 0, fun.aggregate = length)
#  ID 8 32 33 34 35 36 38 39 41 45 53 65 NA
#1  1 0  1  1  0  0  0  1  0  0  0  1  0  0
#2  2 0  0  1  1  1  0  0  0  1  0  0  0  0
#3  3 0  0  0  0  1  0  1  0  0  1  0  1  0
#4  4 1  0  0  0  0  1  1  1  0  0  0  0  0
#5  5 0  0  0  1  0  0  0  1  0  0  1  0  1

Answer 2

这里有一种使用基本R的可能方法。

输入数据：

input
  ID area ERC  a  b  c  d
1  1    6   1 33 38 53 32
2  2    6   1 34 41 35 33
3  3    6   1 35 45 38 65
4  4    6   1 38  8 39 36
5  5    6   1 39 NA 53 34

使用f：

为每行应用简单的函数apply

unique_n<-unique(as.numeric(unlist(input[,c(4:7)])))
f<-function(input_1,unique_n)
{
return(as.numeric(unique_n  %in% input_1[c(4:7)]))
}
count_n<-t(apply(input,1,f,unique_n=unique_n))
colnames(count_n)<-unique_n
cbind(ID=input[,1],count_n)
     ID 33 34 35 38 39 41 45 8 <NA> 53 32 65 36
[1,]  1  1  0  0  1  0  0  0 0    0  1  1  0  0
[2,]  2  1  1  1  0  0  1  0 0    0  0  0  0  0
[3,]  3  0  0  1  1  0  0  1 0    0  0  0  1  0
[4,]  4  0  0  0  1  1  0  0 1    0  0  0  0  1
[5,]  5  0  1  0  0  1  0  0 0    1  1  0  0  0

R：来自大表的矩阵

2 个答案: