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if else condition - help

7 messages · Adrian Johnson, WRAY NICHOLAS, Dylan Keenan +3 more

Adrian Johnson
# 
Hi group:
I am having difficulty with if else condition. I kindly request some help.

I have a matrix k
C1         C2         C3         C4
A  0.09902175 -0.1083887  0.2018689 -0.3546167
B  1.60623838 -1.4167034  0.9076373 -0.3161138
C -0.10433133 -1.7060911 -0.4030050  1.0153297
D -2.91485614  2.9201895 -2.4771802 -2.6991517

I want to convert values > 1.5 to 1, < -1.5 to -1 and rest to 0;
C1         C2         C3         C4
A          0           0            0           0
B           1          0            0           0
C           0        -1             0           0
D           -1        1            -1           -1


I am trying with if else but cannot do it. I could only define one
condition.  Could someone help how I can do this. I dont mean only if
else, but any other way.

k =
structure(c(0.0990217544905328, 1.60623837694539, -0.104331330281166,
-2.91485614212114, -0.108388742328104, -1.41670341534772, -1.70609114096417,
2.92018951284015, 0.201868946570178, 0.907637296638577, -0.403004972105994,
-2.47718015803221, -0.354616729237253, -0.316113789733413, 1.01532974064126,
-2.69915170731852), .Dim = c(4L, 4L), .Dimnames = list(c("A",
"B", "C", "D"), c("C1", "C2", "C3", "C4")))



k1 <- t(apply(k, 1, function(x) ifelse(x > 1.5,1,-1)))
C1 C2 C3 C4
A -1 -1 -1 -1
B  1 -1 -1 -1
C -1 -1 -1 -1
D -1  1 -1 -1



Thanks
Adrian
WRAY NICHOLAS
# 
Hi Adrian I'm not sure that you need to use the ifelse here.  You can simply
assign values ina vector or matrix using a simple condition -- here is a simple
example:

v<-c(4,5,6,7)
v1<-v
v1[]<-0
v1[v<5]<--1
v1[v>6]<-1
v1

Nick
Dylan Keenan
# 
Try this:
C1 C2 C3 C4
A  0  0  0  0
B  1  0  0  0
C  0 -1  0  0
D -1  1 -1 -1

On Sun, May 22, 2016 at 2:00 PM Adrian Johnson <oriolebaltimore at gmail.com>
wrote:

  
  


  


      

    

    

      
      

        


  

        

      Adrian Johnson
    

    

      
      #
    

  


  

      
Thank you both Dylan and Wray.

since my matrix is quite large and for simplicity in downstream
operation, i will use sign function. thanks a lot.

        
On Sun, May 22, 2016 at 2:12 PM, Dylan Keenan <dylan.keenan at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      David Winsemius
    

    

      
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If the problems were somewhat less symmetric or  more complex this would be a method that could be easily generalized to a larger number of less "absolutely" symmetric intervals:

 k2 <- k
 k2[] <- findInterval(k2, c(-Inf, -1.5, 1.5, Inf) ) -2  
                                             # shifts the 1-3 values to -1 to 1
 k2
  C1 C2 C3 C4
A  0  0  0  0
B  1  0  0  0
C  0 -1  0  0
D -1  1 -1 -1

Using k2[] <- ... preserves the matrix structure

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

        
David Winsemius
Alameda, CA, USA

  
  


  


      

    

    

      
      

        


  

        

      jim holtman
    

    

      
      #
    

  


  

      
if you want to use 'ifelse', here is a way:

            

      
      
      
    

        
+ structure(c(0.0990217544905328, 1.60623837694539, -0.104331330281166,
+ -2.91485614212114, -0.108388742328104, -1.41670341534772,
-1.70609114096417,
+ 2.92018951284015, 0.201868946570178, 0.907637296638577,
-0.403004972105994,
+ -2.47718015803221, -0.354616729237253, -0.316113789733413,
1.01532974064126,
+ -2.69915170731852), .Dim = c(4L, 4L), .Dimnames = list(c("A",
+ "B", "C", "D"), c("C1", "C2", "C3", "C4")))

            

      
      
      
    

        
+             , 1
+             , ifelse(k < -1.5
+                 , -1
+                 , 0
+                 )
+             )

            

      
      
      
    

        
num [1:4, 1:4] 0 1 0 -1 0 0 -1 1 0 0 ...
 - attr(*, "dimnames")=List of 2
  ..$ : chr [1:4] "A" "B" "C" "D"
  ..$ : chr [1:4] "C1" "C2" "C3" "C4"

            

      
      
      
    

        
C1 C2 C3 C4
A  0  0  0  0
B  1  0  0  0
C  0 -1  0  0
D -1  1 -1 -1

            

      
      
      
    

        
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

On Sun, May 22, 2016 at 1:58 PM, Adrian Johnson <oriolebaltimore at gmail.com>
wrote:

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Martin Maechler
    

    

      
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> if you want to use 'ifelse', here is a way:

hmm, why should he want that ?
The OP did mention that it's about somewhat large objects, so
efficiency is one of the considerations :

ifelse() is often convenient and nicely self-explaining, but it 
is (because of its generality, but also by its definition)
much *less efficient* than the (sometimes slightly less
convenient) ways you were shown previously in this thread :

- For the generalized case findInterval() is order of magnitudes
  better, and
- for the simple case you were shown to use logical indexing,
  i.e., calls ? la    x[x > k] <- ..


In summary:
   Use  ifelse()  much less -- notably if writing
   functions/code which should scale !


Martin Maechler
ETH Zurich