

Hi all,
I have a time series which contain data collected weekly from week 26
to week 25 the following year. How do I plot this data, so that the
xaxis is displaying the week numbers, ordered as in the data?
Thanks in advance,
Gustaf

x<c(26:52,1:25)
y<rnorm(52)+1:52
plot(x,y) ## How do I get the x axis to be ordered by the current
ordering of x?

Gustaf Rydevik, M.Sci.
tel: +46(0)703 051 451
address:Essingetorget 40,112 66 Stockholm, SE
skype:gustaf_rydevik
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This should do what you want.
x<c(26:52,1:25)
y<rnorm(52)+1:52
plot(seq_along(x), y, xaxt='n')
axis(1, at=seq_along(x), labels=x)
On 9/14/07, Gustaf Rydevik < [hidden email]> wrote:
> Hi all,
>
> I have a time series which contain data collected weekly from week 26
> to week 25 the following year. How do I plot this data, so that the
> xaxis is displaying the week numbers, ordered as in the data?
>
> Thanks in advance,
>
> Gustaf
> 
> x<c(26:52,1:25)
> y<rnorm(52)+1:52
> plot(x,y) ## How do I get the x axis to be ordered by the current
> ordering of x?
>
>
>
> 
> Gustaf Rydevik, M.Sci.
> tel: +46(0)703 051 451
> address:Essingetorget 40,112 66 Stockholm, SE
> skype:gustaf_rydevik
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
>

Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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On Fri, 14Sep2007 at 02:18PM 0400, jim holtman wrote:
> This should do what you want.
>
> x<c(26:52,1:25)
> y<rnorm(52)+1:52
>
> plot(seq_along(x), y, xaxt='n')
> axis(1, at=seq_along(x), labels=x)
And if you have to economize on keystrokes,
seq(x) will achieve the same as seq_along(x)
HTH

~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___ Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Middle minds discuss events
(:_~*~_:) Small minds discuss people
(_)(_) ..... Anon
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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But not necessarily performance. Look at the code for 'seq' and
'seq_along'. Time difference may be small, but keystrokes are done
only once.
On 9/17/07, Patrick Connolly < [hidden email]> wrote:
> On Fri, 14Sep2007 at 02:18PM 0400, jim holtman wrote:
>
> > This should do what you want.
> >
> > x<c(26:52,1:25)
> > y<rnorm(52)+1:52
> >
> > plot(seq_along(x), y, xaxt='n')
> > axis(1, at=seq_along(x), labels=x)
>
> And if you have to economize on keystrokes,
> seq(x) will achieve the same as seq_along(x)
>
> HTH
>
> 
> ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
> ___ Patrick Connolly
> {~._.~} Great minds discuss ideas
> _( Y )_ Middle minds discuss events
> (:_~*~_:) Small minds discuss people
> (_)(_) ..... Anon
>
> ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
>

Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


I am calculating the median absolute deviation using mad function, and
it tends to ignore the parameter constant=1, when I am calculating it
for x=seq(1:5). Am I missing something here?
x<seq(1:5)
mad(x)# gives [1] 1.4826
mad(x, constant=1)# gives [1] 1
#Here is the long form
dev.from.median<abs((xmedian(x)))
dev.from.median # Gives [1] 2 1 0 1 2
sum(dev.from.median) # Gives [1] 6
sum(dev.from.median)/length(x) # Gives [1] 1.2
# The long form does not match the output from the function
# When x<seq(1:10) they match
x<seq(1:10)
dev.from.median<abs((xmedian(x)))
sum(dev.from.median)/length(x) # Gives 2.5
mad(x, constant=1) # Gives 2.5
#The long form matches the output from the function
Did I miss anything here?
Cheers../Murli
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On 9/17/07, Nair, Murlidharan T < [hidden email]> wrote:
>
> I am calculating the median absolute deviation using mad function, and
> it tends to ignore the parameter constant=1, when I am calculating it
> for x=seq(1:5). Am I missing something here?
>
> x<seq(1:5)
> mad(x)# gives [1] 1.4826
> mad(x, constant=1)# gives [1] 1
> #Here is the long form
> dev.from.median<abs((xmedian(x)))
> dev.from.median # Gives [1] 2 1 0 1 2
> sum(dev.from.median) # Gives [1] 6
> sum(dev.from.median)/length(x) # Gives [1] 1.2
> # The long form does not match the output from the function
>
> # When x<seq(1:10) they match
> x<seq(1:10)
> dev.from.median<abs((xmedian(x)))
> sum(dev.from.median)/length(x) # Gives 2.5
> mad(x, constant=1) # Gives 2.5
> #The long form matches the output from the function
>
> Did I miss anything here?
yes; mad := Median (not mean) absolute deviation (from the median, by default).
Deepayan
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Original Message
From: Deepayan Sarkar [mailto: [hidden email]]
Sent: Monday, September 17, 2007 5:10 PM
To: Nair, Murlidharan T
Cc: [hidden email]
Subject: Re: [R] MAD
On 9/17/07, Nair, Murlidharan T < [hidden email]> wrote:
>
> I am calculating the median absolute deviation using mad function, and
> it tends to ignore the parameter constant=1, when I am calculating it
> for x=seq(1:5). Am I missing something here?
>
> x<seq(1:5)
> mad(x)# gives [1] 1.4826
> mad(x, constant=1)# gives [1] 1
> #Here is the long form
> dev.from.median<abs((xmedian(x)))
> dev.from.median # Gives [1] 2 1 0 1 2
> sum(dev.from.median) # Gives [1] 6
> sum(dev.from.median)/length(x) # Gives [1] 1.2
> # The long form does not match the output from the function
>
> # When x<seq(1:10) they match
> x<seq(1:10)
> dev.from.median<abs((xmedian(x)))
> sum(dev.from.median)/length(x) # Gives 2.5
> mad(x, constant=1) # Gives 2.5
> #The long form matches the output from the function
>
> Did I miss anything here?
yes; mad := Median (not mean) absolute deviation (from the median, by
default).
Deepayan
Indeed, its median and that what I am calculating in the long form. So,
what is that you found I was doing differently? May be I missed your
point.
Thx../M
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If you read ?mad you will find this phrase:
"median of the absolute deviations from the median"
Note the first word. I think you're too focused on
the last word.
Peter Ehlers
Nair, Murlidharan T wrote:
>
> Original Message
> From: Deepayan Sarkar [mailto: [hidden email]]
> Sent: Monday, September 17, 2007 5:10 PM
> To: Nair, Murlidharan T
> Cc: [hidden email]
> Subject: Re: [R] MAD
>
> On 9/17/07, Nair, Murlidharan T < [hidden email]> wrote:
>> I am calculating the median absolute deviation using mad function, and
>> it tends to ignore the parameter constant=1, when I am calculating it
>> for x=seq(1:5). Am I missing something here?
>>
>> x<seq(1:5)
>> mad(x)# gives [1] 1.4826
>> mad(x, constant=1)# gives [1] 1
>> #Here is the long form
>> dev.from.median<abs((xmedian(x)))
>> dev.from.median # Gives [1] 2 1 0 1 2
>> sum(dev.from.median) # Gives [1] 6
>> sum(dev.from.median)/length(x) # Gives [1] 1.2
>> # The long form does not match the output from the function
>>
>> # When x<seq(1:10) they match
>> x<seq(1:10)
>> dev.from.median<abs((xmedian(x)))
>> sum(dev.from.median)/length(x) # Gives 2.5
>> mad(x, constant=1) # Gives 2.5
>> #The long form matches the output from the function
>>
>> Did I miss anything here?
>
> yes; mad := Median (not mean) absolute deviation (from the median, by
> default).
>
> Deepayan
>
> Indeed, its median and that what I am calculating in the long form. So,
> what is that you found I was doing differently? May be I missed your
> point.
> Thx../M
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
>
>
______________________________________________
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https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


On 9/18/07, Nair, Murlidharan T < [hidden email]> wrote:
>
>
> Original Message
> From: Deepayan Sarkar [mailto: [hidden email]]
> Sent: Monday, September 17, 2007 5:10 PM
> To: Nair, Murlidharan T
> Cc: [hidden email]
> Subject: Re: [R] MAD
>
> On 9/17/07, Nair, Murlidharan T < [hidden email]> wrote:
> >
> > I am calculating the median absolute deviation using mad function, and
> > it tends to ignore the parameter constant=1, when I am calculating it
> > for x=seq(1:5). Am I missing something here?
> >
> > x<seq(1:5)
> > mad(x)# gives [1] 1.4826
> > mad(x, constant=1)# gives [1] 1
> > #Here is the long form
> > dev.from.median<abs((xmedian(x)))
> > dev.from.median # Gives [1] 2 1 0 1 2
> > sum(dev.from.median) # Gives [1] 6
> > sum(dev.from.median)/length(x) # Gives [1] 1.2
I'm pretty sure that adding up a bunch of numbers and dividing the
total by the number of numbers qualifies as computing the mean, not
the median.
> > Did I miss anything here?
>
> yes; mad := Median (not mean) absolute deviation (from the median, by
> default).
>
> Deepayan
>
> Indeed, its median and that what I am calculating in the long form. So,
> what is that you found I was doing differently? May be I missed your
> point.
> Thx../M
>
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