x-axis order

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x-axis order

Gustaf Rydevik
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
x-axis 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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Re: x-axis order

jholtman
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
> x-axis 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/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


--
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

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Re: x-axis order

Patrick Connolly-2
On Fri, 14-Sep-2007 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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Re: x-axis order

jholtman
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, 14-Sep-2007 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?

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MAD

Nair, Murlidharan T

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((x-median(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((x-median(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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Re: MAD

Deepayan Sarkar
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((x-median(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((x-median(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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Re: MAD

Nair, Murlidharan T


-----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((x-median(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((x-median(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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Re: MAD

P Ehlers
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((x-median(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((x-median(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/r-help
> 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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Re: MAD

Deepayan Sarkar
In reply to this post by Nair, Murlidharan T
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((x-median(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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