通过输入数字

Question

我想为下面的代码编写一个函数，以避免在长格式代码中重复它们。例如，这些是我的代码：

l = nrow(p)
for (i in 1:l) {
      node<- apply(p[c((i + 1) : l), ], 1, function(x){
        a = c(as.character(p[i,1]) , as.character(x[1]))

        cond1 = ifelse(nrow(Network1[((Network1$from == a[1]) & (Network1$to) == a[2]), ]) != 0 |
                         nrow(Network1[((Network1$from == a[2]) & (Network1$to) == a[1]), ]) != 0, TRUE, FALSE)

        if (((cond1   == 1) & (zsd== 1)) {
          return(c(as.character(p[i, 5]), as.character(x[5])))
        }

        if ((zsd== FALSE)) {
          t1 = as.character(p[i, 1]) == as.character(x[1])
          mycond = cond1  
          TT = t1  
          if (TT + mycond >= z){
            return(c(as.character(p[i, 2]), as.character(x[2])))
          }

        }
      })

此代码中已编写1：2数字。但是，我喜欢在函数中收集它们，在这个函数中输入的数字是多少，它生成更多数字的行，如下所示，代码以1：2的形式更改为输入：

l = nrow(p)
for (i in 1:l) {
      node<- apply(p[c((i + 1) : l), ], 1, function(x){
        a = c(as.character(p[i,1]) , as.character(x[1]))
        b = c(as.character(p[i,2]) , as.character(x[2]))

        cond1 = ifelse(nrow(Network1[((Network1$from == a[1]) & (Network1$to) == a[2]), ]) != 0 |
                         nrow(Network1[((Network1$from == a[2]) & (Network1$to) == a[1]), ]) != 0, TRUE, FALSE)
        cond2 = ifelse(nrow(Network2[((Network2$from == b[1]) & (Network2$to) == b[2]), ]) != 0 |
                         nrow(Network2[((Network2$from == b[2]) & (Network2$to) == b[1]), ]) != 0, TRUE, FALSE)

        if (((cond1 + cond2  == 1) & (zsd== 1)) |
            ((cond1 + cond2  == 2) & (zsd== 2)) {
          return(c(as.character(p[i, 5]), as.character(x[5])))
        }

        if ((zsd== FALSE)) {
          t1 = as.character(p[i, 1]) == as.character(x[1])
          t2 = as.character(p[i, 2]) == as.character(x[2])
          mycond = cond1 + cond2 
          TT = t1 + t2 
          if (TT + mycond >= z){
            return(c(as.character(p[i, 3]), as.character(x[3])))
          }

        }
      })

任何人都知道我该怎么做？或者通过其他方式可以实现这一点吗？我想在我的代码中重复这些行，而不是这些数字。