So this works... Thanks. My mistake was leaving out seq_len()
for (i in seq_len(nrow(Volumes))){
plot(as.numeric(deltas[i,]))
Sys.sleep(0.5); }
Ken Spriggs wrote:
>
> I appreciate the response but if this is what you have in mind I didn't
> get anything different.
> Any ideas why?
>
> for (i in nrow(Volumes)){
> plot(as.numeric(deltas[i,]))
> Sys.sleep(0.5); }
>
> Thanks
>
>
> Jan T. Kim wrote:
>>
>> On Mon, Jun 09, 2008 at 02:02:01PM -0700, Ken Spriggs wrote:
>>>
>>> Hello,
>>>
>>> This code works fine but is so fast I can't see anything but the last
>>> plot.
>>>
>>> for (i in nrow(X)){
>>> plot(as.numeric(d[i,])) }
>>>
>>> I'd like to view a plot every 500 milliseconds, nrow(X) = 400. How?
>>
>> Adding the line
>>
>> Sys.sleep(0.5);
>>
>> to the loop should do this trick.
>>
>> Best regards, Jan
>> --
>> +- Jan T. Kim -------------------------------------------------------+
>> | email: [EMAIL PROTECTED] |
>> | WWW: http://www.cmp.uea.ac.uk/people/jtk |
>> *-----=< hierarchical systems are for files, not for humans >=-----*
>>
>> ______________________________________________
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>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
>
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