Or perhaps you wanted:
W <- z
W[z>4|z<2] <- 0
Sent from my iPhone
> On Feb 1, 2021, at 9:41 PM, David Winsemius <dwinsem...@comcast.net> wrote:
>
> Just drop the “+” if you want logical.
>
> Sent from my iPhone
>
>>> On Feb 1, 2021, at 9:36 PM, Shaami <nzsh...@gmail.com> wrote:
>>>
>>
>> Hi Prof. David
>>
>> Thank you. I will always follow your advice. The suggested code worked. It
>> gives either 1 or 0 depending on the condition to be true. I want index of z
>> for which the condition is true (instead of 1) else zero. Could you please
>> suggest?
>>
>> Thank you
>>
>> Shaami
>>
>>> On Tue, Feb 2, 2021 at 10:16 AM David Winsemius <dwinsem...@comcast.net>
>>> wrote:
>>> Cc’ed the list as should always be your practice.
>>>
>>> Here’s one way (untested):
>>>
>>> W <- +(z>4| z<2) # assume z is of length 20
>>>
>>> —
>>> David
>>>
>>> Sent from my iPhone
>>>
>>>>> On Feb 1, 2021, at 7:08 PM, Shaami <nzsh...@gmail.com> wrote:
>>>>>
>>>>
>>>> Hi Prof. David
>>>>
>>>> In the following state
>>>>
>>>> W = (1:2000)[z >4|z<2)
>>>>
>>>> Could you please guide how I can assign zero if condition is not
>>>> satisfied?
>>>>
>>>> Best Regards
>>>>
>>>> Shaami
>>>>
>>>>> On Mon, 1 Feb 2021, 11:01 am David Winsemius, <dwinsem...@comcast.net>
>>>>> wrote:
>>>>>
>>>>> On 1/31/21 1:26 PM, Berry, Charles wrote:
>>>>> >
>>>>> >> On Jan 30, 2021, at 9:32 PM, Shaami <nzsh...@gmail.com> wrote:
>>>>> >>
>>>>> >> Hi
>>>>> >> I have made the sample code again. Could you please guide how to use
>>>>> >> vectorization for variables whose next value depends on the previous
>>>>> >> one?
>>>>> >>
>>>>>
>>>>> I agree with Charles that I suspect your results are not what you
>>>>> expect. You should try using cat or print to output intermediate results
>>>>> to the console. I would suggest you limit your examination to a more
>>>>> manageable length, say the first 10 results while you are working out
>>>>> your logic. After you have the logic debugged, you can move on to long
>>>>> sequences.
>>>>>
>>>>>
>>>>> This is my suggestion for a more compact solution (at least for the
>>>>> inner loop calculation):
>>>>>
>>>>> set.seed(123)
>>>>>
>>>>> x <- rnorm(2000)
>>>>>
>>>>> z <- Reduce( function(x,y) { sum(y+5*x) }, x, accumulate=TRUE)
>>>>>
>>>>> w<- numeric(2000)
>>>>>
>>>>> w <- (1:2000)[ z >4 | z < 1 ] # In your version the w values get
>>>>> overwritten and end up all being 2000
>>>>>
>>>>>
>>>>> I would also advise making a natural language statement of the problem
>>>>> and goals. I'm thinking that you may be missing certain aspects of the
>>>>> underying problem.
>>>>>
>>>>> --
>>>>>
>>>>> David.
>>>>>
>>>>> >
>>>>> > Glad to help.
>>>>> >
>>>>> > First, it could help you to trace your code. I suspect that the
>>>>> > results are not at all what you want and tracing would help you see
>>>>> > that.
>>>>> >
>>>>> > I suggest running this revision and printing out x, z, and w.
>>>>> >
>>>>> > #+begin_src R
>>>>> > w = NULL
>>>>> > for(j in 1:2)
>>>>> > {
>>>>> > z = NULL
>>>>> > x = rnorm(10)
>>>>> > z[1] = x[1]
>>>>> > for(i in 2:10)
>>>>> > {
>>>>> > z[i] = x[i]+5*z[i-1]
>>>>> > if(z[i]>4 | z[i]<1) {
>>>>> > w[j]=i
>>>>> > } else {
>>>>> > w[j] = 0
>>>>> > }
>>>>> > }
>>>>> > }
>>>>> > #+end_src
>>>>> >
>>>>> >
>>>>> > You should be able to see that the value of w can easily be obtained
>>>>> > outside of the `i' loop.
>>>>> >
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