R要替换的项目数不是替换长度/结果的倍数，但是是正确的 [英] R number of items to replace is not a multiple of replacement length / results however correct

问题描述

我知道这个问题上已经有了一些线索，但是经过这些，我还是无法弄清楚问题所在 - 请原谅我。

我试图运行代码

  for（i in 1：a）{
 matrix $ new_column [矩阵[i，1：b-1] ==矩阵$ col_b [i]）
} 
  

我正在尝试的是：
a 的矩阵 找到包含与列 b 中的值相同的值的列（在每行的列中）。 总是有这样一个值），并将相应的列号写入* new_column *

My Code不断抛出错误

在矩阵$ new_column [i]中的警告< - 其中（矩阵[i，：要替换的
项的数目不是替换长度的倍数

然而，结果是完全正确的。已经尝试了

• 首先创建* new_column *填充0
  
• a 或 a + 1 如上所述，结果是正确的，但是我觉得如果我做的一切都正确，我不应该得到警告信息，所以我非常感谢任何关于如何解决这个问题的建议。
  
  
  

  最后，不要问我为什么选择1：b-1，当我想从2跳到b-1时，我刚刚看到， 2：b-1，它会在第3列开始执行。
  

  解决方案

  / code>可以返回一个向量，如果有多个匹配的话。例如：
  
  

   哪（（1:12）%% 2 == 0）＃哪个都是偶数？ 
    

  
  
  矩阵$ col_b [i] unique ？结果可能仍然是正确的。注意在这种情况下会发生什么：
  
  

    x < -  1：2 
  x [1] < -  3： 4 
  x 
    

  另外， 1：b-1 不会给你从 1 到 b - 1 的数字，但是从 1 至 b ，全部减去 1 ：
  
  

    b < -  10 
   1：b-1 
    

  你需要括号先强制减法： 1：（b-1）。

  
  

  I know there have been already some threads on this, however going through those I was not able to figure out what the problem might be - please forgive me for that..

  I am trying to run the code

    for (i in 1:a){
      matrix$new_column[i]<-which(matrix[i,1:b-1]==matrix$col_b[i])
    }
  

  What I am attempting is: For the matrix of a lines and b columns, in each line´s columns 2 to b-1, find the one that contains the same value as the one in column b (there is always such a value) and write the according column number into the *new_column*

  My Code keeps throwing the error

  Warning in matrix$new_column[i] <- which(matrix[i, : number of items to replace is not a multiple of replacement length

  However, the result is completely correct. I have tried

  • creating the *new_column* filled with 0s first
  • changing the end indices from a to a-1 or a+1

  As said, the outcome is correct, however I feel I should not be getting the warning message if I did everything correctly, so I´m really grateful for any advice on how to fix this.

  Finally, don´t ask me why I chose 1:b-1 when I wanted to go from 2 to b-1, I just saw that when I would use 2:b-1, it would acutally begin in column 3..

  解决方案

  which() can return a vector if there are multiple matches. For example:

  which((1:12)%%2 == 0) # which are even?
  

  Is matrix$col_b[i] unique? The results may still look correct. Notice what happens in this case:

  x <- 1:2
  x[1] <- 3:4
  x
  

  Also, 1:b-1 does not give you the numbers from 1 to b - 1 but the number from 1 to b, all minus 1:

  b <- 10
  1:b-1
  

  You need parentheses to force the subtraction first: 1:(b - 1).

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