Question

我试图从这个数据帧中获取频率表：

tmp2 <- structure(list(a1 = c(1L, 0L, 0L), a2 = c(1L, 0L, 1L),
                       a3 = c(0L, 1L, 0L), b1 = c(1L, 0L, 1L),
                       b2 = c(1L, 0L, 0L), b3 = c(0L, 1L, 1L)),
                       .Names = c("a1", "a2", "a3", "b1", "b2", "b3"),
                       class = "data.frame", row.names = c(NA, -3L))


tmp2 <- read.csv("tmp2.csv", sep=";")
tmp2
> tmp2
  a1 a2 a3 b1 b2 b3
1  1  1  0  1  1  0
2  0  0  1  0  0  1
3  0  1  0  1  0  1

我尝试获得频率表如下：

table(tmp2[,1:3], tmp2[,4:6])

但我明白了：

sort.list（y）出错：'x'必须是'sort.list'原子的原子你有没有在名单上打电话给'排序'？

预期产出：

信息：没有必要使用方形矩阵我应该能够添加b4 b5并保持a1 a2 a3

Answer 1

选项：

如果您的b和acols<-1:3 #state the indices of the a columns bcols<-4:6 #same for b; if you add a column this should be 4:7 matrix(colSums(tmp2[,rep(acols,length(bcols))] & tmp2[,rep(bcols,each=length(acols))]), ncol=length(bcols),nrow=length(acols), dimnames=list(colnames(tmp2)[acols],colnames(tmp2)[bcols]))列数不同，可以尝试：

controls

Answer 2

这是一个可能的解决方案：

aIdxs <- 1:3
bIdxs <- 4:7

# init matrix
m <- matrix(0,
            nrow = length(aIdxs), ncol=length(bIdxs),
            dimnames = list(colnames(tmp2)[aIdxs],colnames(tmp2)[bIdxs]))

# create all combinations of a's and b's column indexes
idxs <- expand.grid(aIdxs,bIdxs)

# for each line and for each combination we add 1
# to the matrix if both a and b column are 1 
for(r in 1:nrow(tmp2)){
  m <- m + matrix(apply(idxs,1,function(x){ all(tmp2[r,x]==1) }),
                  nrow=length(aIdxs), byrow=FALSE)
}
> m
   b1 b2 b3
a1  1  1  0
a2  2  1  1
a3  0  0  1

Answer 3

另一种可能的解决方案。你的输入对于'table'来说有点棘手，因为你固有地有两个'a'和'b'在每一行中都有二进制指示符，表示只在'a'和'b'之间的成对实例，你想要循环它们。下面是一个广义的（但可能不那么优雅）函数，可以使用不同长度的'a'和'b'：

tmp2 <- structure(list(a1 = c(1L, 0L, 0L), a2 = c(1L, 0L, 1L), a3 = c(0L, 
                                                              1L, 0L), b1 = c(1L, 0L, 1L), b2 = c(1L, 0L, 0L), b3 = c(0L, 1L, 
                                                                                                                      1L)), .Names = c("a1", "a2", "a3", "b1", "b2", "b3"), class = "data.frame", row.names = c(NA, 
                                                                                                                                                                                                                -3L))                                                                                                                                                                                                               
fun = function(x) t(do.call("cbind", lapply(x[,grep("a", colnames(x))], 
    function(p) rowSums(do.call("rbind", lapply(x[,grep("b", colnames(x))], 
    function(q) q*p ))))))
fun(tmp2)
#> fun(tmp2)
#   b1 b2 b3
#a1  1  1  0
#a2  2  1  1
#a3  0  0  1

# let's do a bigger example
set.seed(1)
m = matrix(rbinom(size=1, n=50, prob=0.75), ncol=10, dimnames=list(paste("instance_", 1:5, sep=""), c(paste("a",1:4,sep=""), paste("b",1:6,sep=""))))

# Notice that the count of possible a and b elements are not equal
#> m
#           a1 a2 a3 a4 b1 b2 b3 b4 b5 b6
#instance_1  1  0  1  1  0  1  1  1  0  0
#instance_2  1  0  1  1  1  1  1  0  1  1
#instance_3  1  1  1  0  1  1  1  1  0  1
#instance_4  0  1  1  1  1  0  1  1  1  1
#instance_5  1  1  0  0  1  1  0  1  1  1

fun(as.data.frame(m))
#> fun(as.data.frame(m))
#   b1 b2 b3 b4 b5 b6
#a1  3  4  3  3  2  3
#a2  3  2  2  3  2  3
#a3  3  3  4  3  2  3
#a4  2  2  3  2  2  2

来自R中多个col和多行的表频率

3 个答案: