# [R] if else condition - help

David Winsemius dwinsemius at comcast.net
Sun May 22 20:37:09 CEST 2016

```> On May 22, 2016, at 11:23 AM, Adrian Johnson <oriolebaltimore at gmail.com> wrote:
>
> 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:
>> 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
>>

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

>> On Sun, May 22, 2016 at 2:00 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
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> and provide commented, minimal, self-contained, reproducible code.

David Winsemius
Alameda, CA, USA

```