本文介绍了按矩阵计算在上限范围内的N次出现的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我想每次在矩阵行中将值在给定范围内时都进行计数，然后对这些逻辑结果求和以得出每一行的一致性度量"./p>

可复制的示例:

  m1<-matrix(c(1,2,1,6,3,7,4,2,6,8,11,15)，ncol = 4，byrow = TRUE)#预期结果，每边均在+/- 1范围内exp.outcome< -matrix(c(TRUE，TRUE，TRUE，FALSE，TRUE，FALSE，TRUE，TRUE，FALSE，FALSE，FALSE，FALSE)，ncol = 4，byrow = TRUE)

在每个值都位于该行中任何其他值的+/- 1范围内的情况下，我已经指出了预期的结果.

在 m1 的第一行中，第一个值(1)在该行中任何其他值的+/- 1之内，因此等于 TRUE ，依此类推.

相比之下， m1 的第4行中的任何值都不在彼此的单个数字值之内，因此每个值都被分配为 FALSE .

任何指针将不胜感激?

更新:

借助所提供的帮助，我现在可以计算任意任意大矩阵(使用二项式系数，k从n中抽取，无需替换)满足上限条件的唯一值对.

解决方案

在回答之前，我只是想澄清一下，在您的问题中您已经说过:

但是

 >>m1 [1,4][1] 6

6不在1的+/- 1范围内，并且您的答案中有 FALSE 值是正确的结果.

解决方案

此解决方案应使您获得所需的结果:

  t(apply(X = m1，#从矩阵中取出每一行保证金= 1，FUN =函数(x){sapply(X = x，#现在遍历该行的每个元素FUN =函数(y){#您的条件y％in％c(x-1)|y％in％c(x + 1)})}))

结果

  [，1] [，2] [，3] [，4][1，]是是是否[2，]是否是是[3，] FALSE FALSE FALSE FALSE

检查

用于存储为 res 的结果.

 >>相同(res，exp.outcome)[1]是

I would like to tally each time a value lies within a given range in a matrix by-row, and then sum these logical outcomes to derive a "measure of consistency" for each row.

Reproducible example:

m1 <- matrix(c(1,2,1,6,3,7,4,2,6,8,11,15), ncol=4, byrow = TRUE)


# expected outcome, given a range of +/-1 either side

exp.outcome<-matrix(c(TRUE,TRUE,TRUE,FALSE,
                          TRUE,FALSE,TRUE,TRUE,
                          FALSE,FALSE,FALSE,FALSE),
                          ncol=4, byrow=TRUE)

Above I've indicated the the expected outcome, in the case where each value lies within +/- 1 range of any other values within that row.

Within the first row of m1 the first value (1) is within +/-1 of any other value in that row hence equals TRUE, and so on.

By contrast, none of the values in row 4 of m1 are within a single digit value of each other and hence each is assigned FALSE.

Any pointers would be much appreciated?

Update:

Thanks to the help provided I can now count the unique pairs of values which meet the ceiling criteria for any arbitrarily large matrix (using the binomial coefficient, k draws from n, without replacement).

解决方案

Before progressing with the answer I just wanted to clarify that in your question you have said:

However,

>> m1[1,4]
[1] 6

6 is not within the +/- 1 from 1, and there is FALSE value as a correct result in your answer.

Solution

This solution should get you to the desired results:

t(apply(
    X = m1,
    # Take each row from the matrix
    MARGIN = 1,
    FUN = function(x) {
        sapply(
            X = x,
            # Now go through each element of that row
            FUN = function(y) {
                # Your conditions
                y %in% c(x - 1) | y %in% c(x + 1)
            }
        )
    }
))

Results

      [,1]  [,2]  [,3]  [,4]
[1,]  TRUE  TRUE  TRUE FALSE
[2,]  TRUE FALSE  TRUE  TRUE
[3,] FALSE FALSE FALSE FALSE

Check

For results stored as res.

>> identical(res, exp.outcome)
[1] TRUE

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