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
Sent from my iPhone
> On Feb 1, 2021, at 10:16 PM, David Winsemius wrote:
>
> Or perhaps:
>
> W <- 1:2000
> W[z>4|z<2] <- 0
Another way:
W <- (1:2000)*(z>4|z<2)
As I said earlier you really should study logical class vectors and the
operators that use them and how to apply them as
Or perhaps:
W <- 1:2000
W[z>4|z<2] <- 0
Sent from my iPhone
> On Feb 1, 2021, at 9:56 PM, David Winsemius wrote:
>
> 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 wrote:
>>>
>> Just drop the “+” if you
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 wrote:
>
> Just drop the “+” if you want logical.
>
> Sent from my iPhone
>
>>> On Feb 1, 2021, at 9:36 PM, Shaami wrote:
>>>
>>
>> Hi Prof. David
>>
>> Thank you. I will
Just drop the “+” if you want logical.
Sent from my iPhone
> On Feb 1, 2021, at 9:36 PM, Shaami 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
IMTS length 2000
Sent from my iPhone
> On Feb 1, 2021, at 9:16 PM, David Winsemius 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
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 wrote:
>
>
> Hi Prof. David
>
> In the following state
>
> W = (1:2000)[z >4|z<2)
>
> Could you
Hi Duncan
It worked. Thank you.
Best Regards
On Mon, Feb 1, 2021 at 6:26 PM Duncan Murdoch
wrote:
> On 01/02/2021 7:03 a.m., Shaami wrote:
> > z = NULL
> > p = 0.25
> > x = rnorm(100)
> > z[1] = p*x[1] + (1-p)*5
> > for(i in 2:100)
> > {
> >z[i] = p*x[i]+(1-p)*z[i-1]
> > }
>
> That's the
On 01/02/2021 7:03 a.m., Shaami wrote:
z = NULL
p = 0.25
x = rnorm(100)
z[1] = p*x[1] + (1-p)*5
for(i in 2:100)
{
z[i] = p*x[i]+(1-p)*z[i-1]
}
That's the same as
p <- 0.25
x <- rnorm(100)
z <- stats::filter(p*x, 1-p, init = 5, method="recursive")
This leaves z as a time series; if that
Hi David and Charles
Your suggestions helped me a lot. Could you please suggest how I could
vectorize the following for loop?
z = NULL
p = 0.25
x = rnorm(100)
z[1] = p*x[1] + (1-p)*5
for(i in 2:100)
{
z[i] = p*x[i]+(1-p)*z[i-1]
}
Thank you
Regards
On Mon, Feb 1, 2021 at 11:01 AM David
On 1/31/21 1:26 PM, Berry, Charles wrote:
On Jan 30, 2021, at 9:32 PM, Shaami 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
> On Jan 30, 2021, at 9:32 PM, Shaami 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?
>
Glad to help.
First, it could help you to trace your code. I suspect that the results
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?
w = NULL
for(j in 1:1000)
{
z = NULL
x = rnorm(2000)
z[1] = x[1]
for(i in 2:2000)
{
z[i] = x[i]+5*z[i-1]
if(z[i]>4 | z[i]<1)
On 1/30/21 8:26 PM, Shaami wrote:
Hi
I have very large dependent nested for loops that are quite expensive
computationally. It takes weeks and does not end to give me results. Could
anyone please guide how could I use apply function or any other suggestion
for such big and dependent loops in
Hi
I have very large dependent nested for loops that are quite expensive
computationally. It takes weeks and does not end to give me results. Could
anyone please guide how could I use apply function or any other suggestion
for such big and dependent loops in R? A sample code is as follows.
w =
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