从R中的combn函数中排除某些组合
从R中的combn函数中排除某些组合
提问于 2017-08-03 18:42:56
我正在尝试排除由包含"var4“和"var5”的combn函数生成的所有组合。下面是目前不起作用的代码：
mod_headers <- c("var1", "var2", "var3", "var4", "var5", "var6")
f <- function(){
for(i in 1:length(mod_headers)){
tab <- combn(mod_headers,i,function(mod_headers){
if (combn(mod_headers,i) %in% c("var4","var5")) {return()}
})
for(j in 1:ncol(tab)){
tab_new <- c(tab[,j])
mod_tab_new <- c(tab_new, "newcol")
print(mod_tab_new)
}
}
}
f()谢谢你的帮忙!
回答 5
Stack Overflow用户
发布于 2017-08-03 19:03:12
我真的不确定您希望如何格式化结果，所以我在获取排除两个值出现在一起的组合时停了下来。它依赖于这样一个事实：combn返回一个矩阵，其中每一列都是一个组合。
mod_headers <- c("var1", "var2", "var3", "var4", "var5", "var6")
combn_with_exclusion <- function(x, n, exclude){
full <- combn(x, n)
# remove any columns that have all elements of `exclude`
full[, !apply(full, 2, function(y) all(exclude %in% y))]
}
combn_with_exclusion(mod_headers, 2, c("var4", "var5"))Stack Overflow用户
发布于 2017-08-03 19:12:02
这是另一种方法，生成所有组合的列表，然后排除同时包含var4和var5的组合...
lapply(
lapply(1:length(mod_headers),
function(i) combn(mod_headers, i)),
function(x) x[,apply(x, 2, function(y) !all(c("var4", "var5") %in% y))])
[[1]]
[1] "var1" "var2" "var3" "var4" "var5" "var6"
[[2]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,] "var1" "var1" "var1" "var1" "var1" "var2" "var2" "var2" "var2" "var3" "var3" "var3" "var4" "var5"
[2,] "var2" "var3" "var4" "var5" "var6" "var3" "var4" "var5" "var6" "var4" "var5" "var6" "var6" "var6"
[[3]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16]
[1,] "var1" "var1" "var1" "var1" "var1" "var1" "var1" "var1" "var1" "var2" "var2" "var2" "var2" "var2" "var3" "var3"
[2,] "var2" "var2" "var2" "var2" "var3" "var3" "var3" "var4" "var5" "var3" "var3" "var3" "var4" "var5" "var4" "var5"
[3,] "var3" "var4" "var5" "var6" "var4" "var5" "var6" "var6" "var6" "var4" "var5" "var6" "var6" "var6" "var6" "var6"
...etcStack Overflow用户
发布于 2017-08-03 19:12:58
我只在TIO上尝试过，所以没有基准测试，但我打赌这个版本在大型集合上会更快，如果这应该是重要的话。
m <- c("var2", "var3", "var4", "var5", "var6")
comb <- combn(m, 3)
csums <- colSums((comb == "var4") + (comb == "var5"))
comb[, csums < 2]
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
# [1,] "var2" "var2" "var2" "var2" "var2" "var3" "var3"
# [2,] "var3" "var3" "var3" "var4" "var5" "var4" "var5"
# [3,] "var4" "var5" "var6" "var6" "var6" "var6" "var6"或者等同于OP的f()
f2 <- function(m=mod_headers) {
lapply(1:length(m), function(x) {
comb <- combn(m, x)
csums <- colSums((comb == "var4") + (comb == "var5"))
comb[, csums < 2]
})
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/45481931
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