R 编码中的 If-else 语句不起作用

Question

I did coding in R as below,

  data1<-c(25,35,60,79,50)
  data2<-c(100,150,170,200,1000)
  
  g1=sort(data1)
  g2=sort(data2)
  
  ybar1<-mean(g1)
  ybar2<-mean(g2)
  
  #BIWEIGHT
  
  med1=median(g1)
  med2=median(g2)
  
  mad1=1.4826*(median(abs(g1-med1)))
  mad2=1.4826*(median(abs(g2-med2)))

  u1=(g1-med1)/(9*mad1)
  u2=(g2-med2)/(9*mad2)
  
  #cat("\nu1:",u1)
  u=rbind(u1,u2)
  print(u)
  
  abs=abs(u)
  print(abs)
  
    for(j in abs){
      if(j < 1){
        num1 = ((g1-med1)^2)*((1-(u1^2))^4)
        den1 = ((1-(u1^2))*(1-5*(u1^2)))
        
        num2 = ((g2-med2)^2)*((1-(u2^2))^4)
        den2 = ((1-(u2^2))*(1-5*(u2^2)))
      }
    }
  
  cat("\num2:",num2)

but when i calculate manually, for data1, every value u1 is less than 1, thus the coding is right with my manual calculation, but for data2, it include value of u2 that is more than 1. can anyone help me figure out why it happen and how to fix it? thankyou in advance.

Answer 1

The problem seems to me that within the if condition (and therefore also within the for loop), you always take the entire vectors ( g1 , g2 , u1 , u2 ). You do not tell R to only use those values in g1 and g2 for which j<1. Instead you use the entire vector as many times as there are values in abs that are <1.

Furthermore, you combine u1 and u2 as a 2-row matrix, but leave g1 and g2 separate. This is confusing and gets you more problems. I suggest you do separate loops for the two vectors.

You can do that as follows:

idx1 <- which(abs(u1) < 1) #Gives you the indexes/positions in vector u1, where abs(u1)<1 is TRUE
idx2 <- which(abs(u2) < 1) #same for u2

for(i in idx1){
               num1[i] = ((g1[i]-med1)^2)*((1-(u1[i]^2))^4)
               den1[i] = ((1-(u1[i]^2))*(1-5*(u1[i]^2)))
               }
for(j in idx2){
               num2[j] = ((g2[j]-med2)^2)*((1-(u2[j]^2))^4)
               den2[j] = ((1-(u2[j]^2))*(1-5*(u2[j]^2)))
               }

This way, num1 and den1 have 5 elements because all absolute values of u1 are <1.
In contrast, num2 and den2 have 4 elements, because the last value of u2 is ca. 2.07.

Hopefully, this helps you.

EDIT: Forgot to add the indices ( [i] and [j] ) to num1 / 2 and den1 / 2 .