在 R 中创建没有循环或递归的特定向量 [英] Creation of a specific vector without loop or recursion in R

问题描述

我有第一个向量，假设 x 只包含 1 和 -1.然后，我有第二个向量 y，它由 1、-1 和零组成.现在，如果 x[i] 等于 1 并且 n 前导元素 (y[(in):i])...

I've got a first vector, let's say x that consists only of 1's and -1's. Then, I have a second vector y that consists of 1's, -1's, and zeros. Now, I'd like to create a vector z that contains in index i a 1 if x[i] equals 1 and a 1 exists within the vector y between the n precedent elements (y[(i-n):i])...

更正式的:z <- ifelse(x == 1 && 1 %in% y[(index(y)-n):index(y)],1,0)

我希望在 R 中创建这样一个向量而无需循环或递归.上面的命题不起作用，因为它不能识别以一个元素一个元素地使用表达式 y[(index(y)-n):index(y)].

I'm looking to create such a vector in R without looping or recursion. The proposition above does not work since it does not recognize to take the expression y[(index(y)-n):index(y)] element by element.

非常感谢您的支持

推荐答案

这里有一种方法，它使用 cumsum 函数来测试到目前为止已经看到的数量.如果i位置的1个数大于i-n位置的1个数，则满足右边的条件.

Here's an approach that uses the cumsum function to test for the number of ones that have been seen so far. If the number of ones at position i is larger than the number of ones at position i-n, then the condition on the right will be satisfied.

## Generate some random y's.
> y <- sample(-1:1, 25, replace=T)
> y
 [1]  0  1 -1 -1 -1 -1 -1  1 -1 -1 -1 -1  0  0 -1 -1 -1  1 -1  1  1  0  0  0  1
> n <- 3
## Compute number of ones seen at each position.
> cs <- cumsum(ifelse(y == 1, 1, 0))
> lagged.cs <- c(rep(0, n), cs[1:(length(cs)-n)])
> (cs - lagged.cs) > 0
 [1] FALSE  TRUE  TRUE  TRUE FALSE FALSE FALSE  TRUE  TRUE  TRUE FALSE FALSE
[13] FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE
[25]  TRUE

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