如何绕过r中缺少的参数

Question

我编写了一个以多个向量作为参数的程序。如果输入单个向量作为输入，则该函数提供以下错误消息：

Error in combn(letters[1:length(mylist)], 2) : n < m
Called from: combn(letters[1:length(mylist)], 2)

我知道我为什么会犯这个错误。我使用了一个组合函数组合，它选择了两个向量。我该怎么解决这个问题？我想得到一个输出，即使我的输入是一个单一的向量。在这种情况下，我希望我的代码跳过那些需要两个向量的行，然后运行其余的来获得输出。为了方便起见，我在这里提供了我的全部代码。

### Function to check whether two sets a,b are incomparable

incomparable <- function(a,b){
  
  cond1 <- sum(a>b) > 0 # Returns true is if there is at least one case such that a_i is 1 and b_i is 0
  cond2 <- sum(b>a) > 0 # True if at least one case such that one case with b_i is 1 and a_i is 0
  ifelse (length(a)== length(b), cond1*cond2, "Strings are not of the same length")
}



incomparable(a,b)

### Function to get subsets of a binary vector

binary_subset<-function(a){
  a_seq = lapply(a, seq, 0)   # keep 0s as 0, make 1s c(1, 0)
  subset=do.call(expand.grid, a_seq)
  colnames(subset)=(1:length(a))
  return(subset)
} 

#### Function to generate all lower-order interaction terms correspond to interaction terms a,b,c...upto any arbitrary number L (including terms a, b, c...L)

all_lower_order_interactions<-function(...){
  mylist <- list(...)

  combination<-combn(letters[1:length(mylist)], 2) #choose 2 out of L vectors
  
  check_incomparable<-0
  for (j in 1:ncol(combination)){
    check_incomparable[j]<- (incomparable(get(combination[1,j]), get(combination[2,j])))
  }
  check_incomparable
  
  if(all(check_incomparable>0)==FALSE) {stop( "at least one of the interaction terms is a special case (or a subset) of another term.")}
  
  interactions_abc <- do.call("rbind", lapply(mylist, binary_subset))
  interactions_no_duplicate <- unique(interactions_abc[1:length(mylist[[1]])])
  rownames(interactions_no_duplicate) <- 1:nrow(interactions_no_duplicate)
  
  interactions_no_duplicate
}



a<-c(0,1,0)
b<-c(0,0,1)
all_lower_order_interactions(a)

Answer 1

在没有至少两个向量的情况下，可以使用if语句跳过行，这取决于combn：

all_lower_order_interactions <- function(...) {
  mylist <- list(...)
  if (length(mylist) >= 2) {
    {
      combination <-
        combn(letters[1:length(mylist)], 2) #choose 2 out of L vectors
    }
    
    check_incomparable <- 0
    for (j in 1:ncol(combination)) {
      check_incomparable[j] <-
        (incomparable(get(combination[1, j]), get(combination[2, j])))
    }
    check_incomparable
    
    if (all(check_incomparable > 0) == FALSE) {
      stop("at least one of the interaction terms is a special case (or a subset) of another term.")
    }
  }
  interactions_abc <-
    do.call("rbind", lapply(mylist, binary_subset))
  interactions_no_duplicate <-
    unique(interactions_abc[1:length(mylist[[1]])])
  rownames(interactions_no_duplicate) <-
    1:nrow(interactions_no_duplicate)
  
  interactions_no_duplicate
}


a <- c(0, 1, 0)
b <- c(0, 0, 1)
all_lower_order_interactions(a)

  1 2 3
1 0 1 0
2 0 0 0

Answer 2

您可以根据列表的长度尝试设置combn的第二个参数。

替换

combination<-combn(letters[1:length(mylist)], 2) #choose 2 out of L vectors

出自：

if (length(mylist) < 2) {
  n_elements <- 1
} else {
  n_elements <- 2
}

combination<-combn(letters[1:length(mylist)], n_elements)

编辑

若要使用提供的变量名，请将最后一行更改为：

combination<-combn(names(mylist)[1:length(mylist)], n_elements)