[R] monthly median in a daily dataset

Uwe Ligges ligges at statistik.tu-dortmund.de
Thu Dec 23 11:59:16 CET 2010



On 21.12.2010 08:15, SNV Krishna wrote:
> Hi Dennis,
>
> I am looking for similar function and this post is useful. But a strange
> thing is happening when I try which I couldn't figure out (details below).
> Could you or anyone help me understand why this is so?
>
>> df = data.frame(date = seq(as.Date("2010-1-1"), by = "days", length =
> 250))
>> df$value = cumsum(rnorm(1:250))
>
> When I use the statement (as given in ?aggregate help file) the following
> error is displayed
>> aggregate(df$value, by = months(df$date), FUN = median)
> Error in aggregate.data.frame(as.data.frame(x), ...) :
>    'by' must be a list

The error message is quite helpful, you need a list of all the elements 
you'd have after the "~" in a formula, in this case only the date:

aggregate(df$value, by = list(date = months(df$date)), FUN = median)



> But it works when I use as was suggested
>> aggregate(value~months(date), data = df, FUN = median)
>    months(date)      value
> 1        April 15.5721440
> 2       August -0.1261205
> 3     February -1.0230631
> 4      January -0.9277885
> 5         July -2.1890907
> 6         June  1.3045260
> 7        March 11.4126371
> 8          May  2.1625091
>
> The second question, is it possible to have the median across the months and
> years. Say I have daily data for last five years the above function will
> give me the median of Jan of all the five years, while I want Jan-2010,
> Jan-2009 and so... Wish my question is clear.

Just use Year-Month as the grouping criterion as follows:

aggregate(x=df$value, by = list(date = format(df$date, "%Y-%m")), FUN = 
median)

Uwe Ligges


> 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<djmuser at gmail.com>
> To: HUXTERE<emilyhuxter at gmail.com>
> Cc: r-help at r-project.org
> Subject: Re: [R] monthly median in a daily dataset
> Message-ID:
> 	<AANLkTimXTJhbsE1mq4o121fEKXTf8d1pSYEEgzKKzdYu at 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<emilyhuxter at 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 at 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_crema at yahoo.it>
> To: r-help at 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<statmailinglists at googlemail.com>
> To: r-help at r-project.org
> Subject: [R] Turning a Variable into String
> Message-ID:
> 	<AANLkTi=fa+982ZNie+z-iDwURvq8Ee5zJoQ7OPXLhMK6 at 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<spector at stat.berkeley.edu>
> To: Paolo Rossi<statmailinglists at googlemail.com>
> Cc: r-help at r-project.org
> Subject: Re: [R] Turning a Variable into String
> Message-ID:
> 	<alpine.DEB.2.00.1012191627170.26060 at 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
>   					 spector at 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 at 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.duncan at gmail.com>
> To: Paolo Rossi<statmailinglists at googlemail.com>
> Cc: r-help at r-project.org
> Subject: Re: [R] Turning a Variable into String
> Message-ID:<4D0EA4D0.10507 at 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 at 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.duncan at gmail.com>
> To: Jeff Breiwick<jeff.breiwick at noaa.gov>
> Cc: r-help at stat.math.ethz.ch
> Subject: Re: [R] system/system2 command
> Message-ID:<4D0EAD5E.5060006 at 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<djmuser at gmail.com>
>
> Cc: r-help at r-project.org
> Subject: Re: [R] Time Series of Histograms
> Message-ID:
> 	<AANLkTiknfHmEuaHp_7GirmpNKB=eMKSUeDNiE7jgh6Q_ at 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 at 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<djmuser at gmail.com>
> Cc: r-help at 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 at 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<jorgeivanvelez at gmail.com>
>
> Cc: r-help at r-project.org
> Subject: Re: [R] Time Series of Histograms
> Message-ID:
> 	<AANLkTikp5Zr3_AMJ7uGeHnWRuovW1ddnja2JXPhpD6Uu at 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 at 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.Venables at csiro.au>
> To:<emilyhuxter at gmail.com>,<r-help at r-project.org>
> Subject: Re: [R] monthly median in a daily dataset
> Message-ID:
> 	
> <1BDAE2969943D540934EE8B4EF68F95FB27A44FE07 at 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-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
> Behalf Of HUXTERE
> Sent: Monday, 20 December 2010 8:32 AM
> To: r-help at 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



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