递归申请

Question

我尝试执行以下递归：y [i] = y [i-1] + a [i]我试过了：

   apply(c(2:10),2,function(x) y[x]=y[x-1]+a[x])

但它没有用。是否有其他方法可以在没有循环的情况下执行此操作？

Answer 1

这更快：

c(0,cumsum(a[c(-1,-length(a))])) + 1

或等效地：

c(0,cumsum(a[seq(2,10)])) + 1

Answer 2

如果您想在sapply内为<<-分配值，则此时需要y，还需要sapply运算符。以下作品：

示例数据：

y    <- vector() #initiate a vector (can also simply do y <- 1 )
y[1] <- 1        #I suppose the first value will be provided
a    <- 20:30    #some random data for a

解决方案：

#sapply works for vectors i.e. for your 2:10, c() is unnecessary btw
#<<- operator modifies the y vector outside the sapply according to the formula 
> sapply(c(2:10), function(x) {y[x] <<- y[x-1] + a[x]} )
[1]  22  44  67  91 116 142 169 197 226

#y looks like this after the assignments 
> y
 [1]   1  22  44  67  91 116 142 169 197 226