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.kee...@gmail.com> wrote:
> Try this:
>
>> sign(ifelse(abs(k)<=1.5, 0, k))
>
>
> 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 <oriolebaltim...@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
>>
>> ______________________________________________
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>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
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