Any assistance will be greatly appreciated and many thanks for the same.
Regards,
Krishna
Date: Sun, 19 Dec 2010 15:42:15 -0800
From: Dennis Murphy<djmu...@gmail.com>
To: HUXTERE<emilyhux...@gmail.com>
Cc: r-help@r-project.org
Subject: Re: [R] monthly median in a daily dataset
Message-ID:
<aanlktimxtjhbse1mq4o121fekxtf8d1psyeegzkkz...@mail.gmail.com>
Content-Type: text/plain
Hi:
There is a months() function associated with Date objects, so you should be
able to do something like
aggregate(value ~ months(date), data = data$flow$daily, FUN = median)
Here's a toy example because your data are not in a ready form:
df<- data.frame(date = seq(as.Date('2010-01-01'), by = 'days', length =
250),
val = rnorm(250))
aggregate(val ~ months(date), data = df, FUN = median)
months(date) val
1 April -0.18864817
2 August -0.16203705
3 February 0.03671700
4 January 0.04500988
5 July -0.12753151
6 June 0.09864811
7 March 0.23652105
8 May 0.25879994
9 September 0.53570764
HTH,
Dennis
On Sun, Dec 19, 2010 at 2:31 PM, HUXTERE<emilyhux...@gmail.com> wrote:
Hello,
I have a multi-year dataset (see below) with date, a data value and a flag
for the data value. I want to find the monthly median for each month in
this
dataset and then plot it. If anyone has suggestions they would be greatly
apperciated. It should be noted that there are some dates with no values
and
they should be removed.
Thanks
Emily
print ( str(data$flow$daily) )
'data.frame': 16071 obs. of 3 variables:
$ date :Class 'Date' num [1:16071] -1826 -1825 -1824 -1823 -1822 ...
$ value: num NA NA NA NA NA NA NA NA NA NA ...
$ flag : chr "" "" "" "" ...
NULL
520 2008-11-01 0.034
1041 2008-11-02 0.034
1562 2008-11-03 0.034
2083 2008-11-04 0.038
2604 2008-11-05 0.036
3125 2008-11-06 0.035
3646 2008-11-07 0.036
4167 2008-11-08 0.039
4688 2008-11-09 0.039
5209 2008-11-10 0.039
5730 2008-11-11 0.038
6251 2008-11-12 0.039
6772 2008-11-13 0.039
7293 2008-11-14 0.038
7814 2008-11-15 0.037
8335 2008-11-16 0.037
8855 2008-11-17 0.037
9375 2008-11-18 0.037
9895 2008-11-19 0.034 B
10415 2008-11-20 0.034 B
10935 2008-11-21 0.033 B
11455 2008-11-22 0.034 B
11975 2008-11-23 0.034 B
12495 2008-11-24 0.034 B
13016 2008-11-25 0.034 B
13537 2008-11-26 0.033 B
14058 2008-11-27 0.033 B
14579 2008-11-28 0.033 B
15068 2008-11-29 0.034 B
15546 2008-11-30 0.035 B
521 2008-12-01 0.035 B
1042 2008-12-02 0.034 B
1563 2008-12-03 0.033 B
2084 2008-12-04 0.031 B
2605 2008-12-05 0.031 B
3126 2008-12-06 0.031 B
3647 2008-12-07 0.032 B
4168 2008-12-08 0.032 B
4689 2008-12-09 0.032 B
5210 2008-12-10 0.033 B
5731 2008-12-11 0.033 B
6252 2008-12-12 0.032 B
6773 2008-12-13 0.031 B
7294 2008-12-14 0.030 B
7815 2008-12-15 0.030 B
8336 2008-12-16 0.029 B
8856 2008-12-17 0.028 B
9376 2008-12-18 0.028 B
9896 2008-12-19 0.028 B
10416 2008-12-20 0.027 B
10936 2008-12-21 0.027 B
11456 2008-12-22 0.028 B
11976 2008-12-23 0.028 B
12496 2008-12-24 0.029 B
13017 2008-12-25 0.029 B
13538 2008-12-26 0.029 B
14059 2008-12-27 0.030 B
14580 2008-12-28 0.030 B
15069 2008-12-29 0.030 B
15547 2008-12-30 0.031 B
15851 2008-12-31 0.031 B
--
View this message in context:
http://r.789695.n4.nabble.com/monthly-median-in-a-daily-dataset-tp3094917p30
94917.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help@r-project.org 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.
[[alternative HTML version deleted]]
------------------------------
Message: 35
Date: Mon, 20 Dec 2010 00:03:24 +0000
From: "Enrico R. Crema"<enryu_cr...@yahoo.it>
To: r-help@r-project.org
Subject: [R] Time Series of Histograms
Content-Type: text/plain; charset=us-ascii
Dear List,
I have a set of distributions recorded at an equal interval of time and I
would like to plot them as series of horizontal histograms (with the x-axis
representing time, and y-axis representing the bins) since the distribution
shifts from unimodal to multimodal in several occasions. What I would like
to see is something close to a violinplot, but I do not want a kernel
density estimate...
[[elided Yahoo spam]]
Thanks in Advance,
Enrico
------------------------------
Message: 36
Date: Mon, 20 Dec 2010 00:21:02 +0000
From: Paolo Rossi<statmailingli...@googlemail.com>
To: r-help@r-project.org
Subject: [R] Turning a Variable into String
Message-ID:
<aanlkti=fa+982znie+z-idwurvq8ee5zjoq7opxlh...@mail.gmail.com>
Content-Type: text/plain
I would like to know how to turn a variable into a string. I have tried
as.symbol and as.name but it doesnt work for what I'd like to do
Essentially, I'd like to feed the function below with two variables. This
works fine in the bit working out number of elements in each variable.
In the print(sprintf("OK with %s and %s\n", var1, var2)) line I would like
var1 and var2 to be magically substituted with a string containing the name
of var1 and name of var2.
Thanks in advance
Paolo
haveSameLength<- function(var1, var2) {
if (length(var1)==length(var2))
{
print(sprintf("OK with %s and %s\n", var1, var2))
} else {
print("Problems!!")
}
}
[[alternative HTML version deleted]]
------------------------------
Message: 37
Date: Sun, 19 Dec 2010 16:30:38 -0800 (PST)
From: Phil Spector<spec...@stat.berkeley.edu>
To: Paolo Rossi<statmailingli...@googlemail.com>
Cc: r-help@r-project.org
Subject: Re: [R] Turning a Variable into String
Message-ID:
<alpine.deb.2.00.1012191627170.26...@springer.berkeley.edu>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
Paolo -
One way to make the function do what you want is to replace
the line
print(sprintf("OK with %s and %s\n", var1, var2))
with
cat('OK with',substitute(var1),'and',substitute(var2),'\n')
With sprintf, you'd need
print(sprintf("OK with %s and %s\n", deparse(substitute(var1)),
deparse(substitute(var2))))
but since you're just printing the string returned by sprintf, I'd
go with cat.
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Mon, 20 Dec 2010, Paolo Rossi wrote:
I would like to know how to turn a variable into a string. I have tried
as.symbol and as.name but it doesnt work for what I'd like to do
Essentially, I'd like to feed the function below with two variables. This
works fine in the bit working out number of elements in each variable.
In the print(sprintf("OK with %s and %s\n", var1, var2)) line I would
like
var1 and var2 to be magically substituted with a string containing the
name
of var1 and name of var2.
Thanks in advance
Paolo
haveSameLength<- function(var1, var2) {
if (length(var1)==length(var2))
{
print(sprintf("OK with %s and %s\n", var1, var2))
} else {
print("Problems!!")
}
}
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org 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.
------------------------------
Message: 38
Date: Sun, 19 Dec 2010 19:35:28 -0500
From: Duncan Murdoch<murdoch.dun...@gmail.com>
To: Paolo Rossi<statmailingli...@googlemail.com>
Cc: r-help@r-project.org
Subject: Re: [R] Turning a Variable into String
Message-ID:<4d0ea4d0.10...@gmail.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
On 19/12/2010 7:21 PM, Paolo Rossi wrote:
I would like to know how to turn a variable into a string. I have tried
as.symbol and as.name but it doesnt work for what I'd like to do
Essentially, I'd like to feed the function below with two variables. This
works fine in the bit working out number of elements in each variable.
In the print(sprintf("OK with %s and %s\n", var1, var2)) line I would
like
var1 and var2 to be magically substituted with a string containing the
name
of var1 and name of var2.
The name of var1 is var1, so I assume you mean the expression passed to
your function and bound to var1. In that case, what you want is
deparse(substitute(var1))
Watch out: if the expression is really long, that can be a vector with
more than one element. See ?deparse for ways to deal with that.
Duncan Murdoch
Thanks in advance
Paolo
haveSameLength<- function(var1, var2) {
if (length(var1)==length(var2))
{
print(sprintf("OK with %s and %s\n", var1, var2))
} else {
print("Problems!!")
}
}
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org 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.
------------------------------
Message: 39
Date: Sun, 19 Dec 2010 20:11:58 -0500
From: Duncan Murdoch<murdoch.dun...@gmail.com>
To: Jeff Breiwick<jeff.breiw...@noaa.gov>
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] system/system2 command
Message-ID:<4d0ead5e.5060...@gmail.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
On 17/12/2010 4:36 PM, Jeff Breiwick wrote:
All,
I had a simple function call I used to open up a dos shell running R under
Win XP:
system("cmd.exe", wait=FALSE, invisible=FALSE).
This does not work with R 2.12.1 - I get a window that briefly flashes
open
but then disappears. Does anyone know the method to open a DOS command
window in running R with Win XP? Thank you.
This is a new bug in 2.12.1, which I am about to fix in R-patched. The
problem was that it was passing a null input stream to cmd.exe, which
saw an immediate EOF, and quit. A similar thing happened in Rterm,
where system("cmd") should drop into a command shell in the same window,
but it would immediately exit.
Duncan Murdoch
------------------------------
Message: 40
Date: Sun, 19 Dec 2010 17:47:20 -0800
From: Dennis Murphy<djmu...@gmail.com>
Cc: r-help@r-project.org
Subject: Re: [R] Time Series of Histograms
Message-ID:
<aanlktiknfhmeuahp_7girmpnkb=emksuednie7jgh...@mail.gmail.com>
Content-Type: text/plain
Hi:
You can get a violin plot in lattice rather straightforwardly. It's easiest
if time is an ordered factor, but you can also do it if time is numeric; in
the latter case, the code associated with Figure 10.14 in the Lattice book
provides a template to start with:
http://lmdvr.r-forge.r-project.org/figures/figures.html
To get horizontal violin plots, use time as the y variable and start by
replacing panel.boxplot with panel.violin; see the help page of the latter
if more specific options are required. It also contains an example using a
panel function.
I don't know how you expect to get horizontal histograms without setting the
time variable to be a factor. If you have enough time periods, the result
will not be pretty. If you have a fairly large number of time periods, the
best distributional displays are boxplots, violin plots, beanplots or some
variation of that general concept.
Since neither data nor code were offered, one can only speculate so far as
to what your intentions might be. A reproducible example with data and code
would undoubtedly elicit more useful responses.
HTH,
Dennis
On Sun, Dec 19, 2010 at 4:03 PM, Enrico R. Crema
Dear List,
I have a set of distributions recorded at an equal interval of time and I
would like to plot them as series of horizontal histograms (with the
x-axis
representing time, and y-axis representing the bins) since the
distribution
shifts from unimodal to multimodal in several occasions. What I would
like
to see is something close to a violinplot, but I do not want a kernel
density estimate...
[[elided Yahoo spam]]
Thanks in Advance,
Enrico
______________________________________________
R-help@r-project.org 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.
[[alternative HTML version deleted]]
------------------------------
Message: 41
Date: Mon, 20 Dec 2010 02:04:22 +0000
To: Dennis Murphy<djmu...@gmail.com>
Cc: r-help@r-project.org
Subject: Re: [R] Time Series of Histograms
Content-Type: text/plain
Many Thanks Dennis,
The distributions are simulated ordinal data all bounded in the same upper
and lower limit, and I wanted to plot how the distribution changes through
time. Since the distributions are often multimodal boxplots were not useful
so I made some violinplots... My practical solution which I'm testing right
now is to create a matrix of frequencies and then plot these as a series of
horrizontal barplots (after normalising each distribution) , using the
offset parameter to control the temporal sequence....It actually works fine,
but I was wondering if there were better ways...
Enrico
On 20 Dec 2010, at 01:47, Dennis Murphy wrote:
Hi:
You can get a violin plot in lattice rather straightforwardly. It's
easiest if time is an ordered factor, but you can also do it if time is
numeric; in the latter case, the code associated with Figure 10.14 in the
Lattice book provides a template to start with:
http://lmdvr.r-forge.r-project.org/figures/figures.html
To get horizontal violin plots, use time as the y variable and start by
replacing panel.boxplot with panel.violin; see the help page of the latter
if more specific options are required. It also contains an example using a
panel function.
I don't know how you expect to get horizontal histograms without setting
the time variable to be a factor. If you have enough time periods, the
result will not be pretty. If you have a fairly large number of time
periods, the best distributional displays are boxplots, violin plots,
beanplots or some variation of that general concept.
Since neither data nor code were offered, one can only speculate so far as
to what your intentions might be. A reproducible example with data and code
would undoubtedly elicit more useful responses.
HTH,
Dennis
wrote:
Dear List,
I have a set of distributions recorded at an equal interval of time and I
would like to plot them as series of horizontal histograms (with the x-axis
representing time, and y-axis representing the bins) since the distribution
shifts from unimodal to multimodal in several occasions. What I would like
to see is something close to a violinplot, but I do not want a kernel
density estimate...
[[elided Yahoo spam]]
Thanks in Advance,
Enrico
______________________________________________
R-help@r-project.org 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.
[[alternative HTML version deleted]]
------------------------------
Message: 42
Date: Sun, 19 Dec 2010 21:11:15 -0500
From: Jorge Ivan Velez<jorgeivanve...@gmail.com>
Cc: r-help@r-project.org
Subject: Re: [R] Time Series of Histograms
Message-ID:
<aanlktikp5zr3_amj7ugehnwruovw1ddnja2jxphpd...@mail.gmail.com>
Content-Type: text/plain
Hi Enrico,
Is this close to what you want to do?
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=109
HTH,
Jorge
On Sun, Dec 19, 2010 at 7:03 PM, Enrico R. Crema<> wrote:
Dear List,
I have a set of distributions recorded at an equal interval of time and I
would like to plot them as series of horizontal histograms (with the
x-axis
representing time, and y-axis representing the bins) since the
distribution
shifts from unimodal to multimodal in several occasions. What I would
like
to see is something close to a violinplot, but I do not want a kernel
density estimate...
[[elided Yahoo spam]]
Thanks in Advance,
Enrico
______________________________________________
R-help@r-project.org 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.
[[alternative HTML version deleted]]
------------------------------
Message: 43
Date: Mon, 20 Dec 2010 13:17:59 +1100
From:<bill.venab...@csiro.au>
To:<emilyhux...@gmail.com>,<r-help@r-project.org>
Subject: Re: [R] monthly median in a daily dataset
Message-ID:
<1bdae2969943d540934ee8b4ef68f95fb27a44f...@exnsw-mbx03.nexus.csiro.au>
Content-Type: text/plain; charset="us-ascii"
I find this function useful for digging out months from Date objects
Month<- function(date, ...)
factor(month.abb[as.POSIXlt(date)$mon + 1], levels = month.abb)
For this little data set below this is what it gives
with(data, tapply(value, Month(date), median, na.rm = TRUE))
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
NA NA NA NA NA NA NA NA NA NA 0.035 0.030
Here is another useful little one:
Year<- function(date, ...)
as.POSIXlt(date)$year + 1900
So if you wanted the median by year and month you could do
with(data, tapply(value, list(Year(date), Month(date)), median, na.rm =
TRUE))
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2008 NA NA NA NA NA NA NA NA NA NA 0.035 0.03
(The result is a matrix, which in this case has only one row, of course.)
See how you go.
Bill Venables.
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of HUXTERE
Sent: Monday, 20 December 2010 8:32 AM
To: r-help@r-project.org
Subject: [R] monthly median in a daily dataset
Hello,
I have a multi-year dataset (see below) with date, a data value and a flag
for the data value. I want to find the monthly median for each month in this
dataset and then plot it. If anyone has suggestions they would be greatly
apperciated. It should be noted that there are some dates with no values and
they should be removed.
Thanks
Emily
print ( str(data$flow$daily) )
'data.frame': 16071 obs. of 3 variables:
$ date :Class 'Date' num [1:16071] -1826 -1825 -1824 -1823 -1822 ...
$ value: num NA NA NA NA NA NA NA NA NA NA ...
$ flag : chr "" "" "" "" ...
NULL
520 2008-11-01 0.034
1041 2008-11-02 0.034
1562 2008-11-03 0.034
2083 2008-11-04 0.038
2604 2008-11-05 0.036
3125 2008-11-06 0.035
3646 2008-11-07 0.036
4167 2008-11-08 0.039
4688 2008-11-09 0.039
5209 2008-11-10 0.039
5730 2008-11-11 0.038
6251 2008-11-12 0.039
6772 2008-11-13 0.039
7293 2008-11-14 0.038
7814 2008-11-15 0.037
8335 2008-11-16 0.037
8855 2008-11-17 0.037
9375 2008-11-18 0.037
9895 2008-11-19 0.034 B
10415 2008-11-20 0.034 B
10935 2008-11-21 0.033 B
11455 2008-11-22 0.034 B
11975 2008-11-23 0.034 B
12495 2008-11-24 0.034 B
13016 2008-11-25 0.034 B
13537 2008-11-26 0.033 B
14058 2008-11-27 0.033 B
14579 2008-11-28 0.033 B
15068 2008-11-29 0.034 B
15546 2008-11-30 0.035 B
521 2008-12-01 0.035 B
1042 2008-12-02 0.034 B
1563 2008-12-03 0.033 B
2084 2008-12-04 0.031 B
2605 2008-12-05 0.031 B
3126 2008-12-06 0.031 B
3647 2008-12-07 0.032 B
4168 2008-12-08 0.032 B
4689 2008-12-09 0.032 B
5210 2008-12-10 0.033 B
5731 2008-12-11 0.033 B
6252 2008-12-12 0.032 B
6773 2008-12-13 0.031 B
7294 2008-12-14 0.030 B
7815 2008-12-15 0.030 B
8336 2008-12-16 0.029 B
8856 2008-12-17 0.028 B
9376 2008-12-18 0.028 B
9896 2008-12-19 0.028 B
10416 2008-12-20 0.027 B
10936 2008-12-21 0.027 B
11456 2008-12-22 0.028 B
11976 2008-12-23 0.028 B
12496 2008-12-24 0.029 B
13017 2008-12-25 0.029 B
13538 2008-12-26 0.029 B
14059 2008-12-27 0.030 B
14580 2008-12-28 0.030 B
15069 2008-12-29 0.030 B
15547 2008-12-30 0.031 B
15851 2008-12-31 0.031 B
