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:
Hi group:
I am having difficulty with if else condition. I kindly request some help.
I have a matrix k
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;
k1 - desired output
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)))
k1
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