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how to plot annual values directly

6 messages · lily li, PIKAL Petr, Ulrik Stervbo +1 more

Original
lily li
# 
Hi R users,

I have a dataframe, with daily precipitation data, is it possible to plot
annual mean or annual sum values directly? Thanks for your help.

df
year   month   day   precip       time
2010     1          1        0.5     2010-01-01
2010     1          2        0.8     2010-01-02
2010     1          3        1.0     2010-01-03
2010     1          4        0.9     2010-01-04
...

fig1 = ggplot()+ geom_path(data=df, aes(x=time, y= precip)
show(fig1)
PIKAL Petr
# 
Hi

What do you mean to plot annual mean/sum directly? You can compute it by aggregate function and add it to your plot, but I am not sure if it is enough direct.

Cheers
Petr
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lily li
# 
I meant that my dataframe has daily data, but how to plot annual mean/sum
directly? Thanks.
On Wed, Aug 3, 2016 at 4:16 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Ulrik Stervbo
    

    

      
      #
    

  


  

      
You could use dplyr:

library(plyr)
ddply(df, .variables = "year", summarise, mean.precip = mean(precip))

Hope this helps
Ulrik

        
On Wed, 3 Aug 2016 at 17:29 lily li <chocold12 at gmail.com> wrote:

        

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      PIKAL Petr
    

    

      
      #
    

  


  

      
Hi

Other option is aggregate or ave, depending on how do you want to plot mean/sum values.

You should think about posting an example of what do you have and what do you want to achieve (at least partly). Without that we are just guessing your real intention.

Cheers
Petr

From: Ulrik Stervbo [mailto:ulrik.stervbo at gmail.com]
Sent: Wednesday, August 3, 2016 9:03 PM
To: lily li <chocold12 at gmail.com>; PIKAL Petr <petr.pikal at precheza.cz>
Cc: R mailing list <r-help at r-project.org>
Subject: Re: [R] how to plot annual values directly

You could use dplyr:

library(plyr)
ddply(df, .variables = "year", summarise, mean.precip = mean(precip))

Hope this helps
Ulrik

        
On Wed, 3 Aug 2016 at 17:29 lily li <chocold12 at gmail.com<mailto:chocold12 at gmail.com>> wrote:

        
I meant that my dataframe has daily data, but how to plot annual mean/sum
directly? Thanks.

        
On Wed, Aug 3, 2016 at 4:16 AM, PIKAL Petr <petr.pikal at precheza.cz<mailto:petr.pikal at precheza.cz>> wrote:

        

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        
________________________________
Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m.
Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu.
Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat.
Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu.

V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?:
- vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu.
- a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou.
- trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech.
- odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?.

This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients.
If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system.
If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner.
The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email.

In case that this e-mail forms part of business dealings:
- the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning.
- if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation.
- the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects.
- the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.

  
  


  


      

    

    

      
      

        


  

        

      Duncan Mackay
    

    

      
      #
    

  


  

      
Hi

By coincidence I will very soon have to  do something similar. So I thought
this would be a practice run

Data is 10 years of daily rainfall data
str(arm)
'data.frame':   3640 obs. of  7 variables:
 $ date : Date, format: "1990-01-01" "1990-01-02" "1990-01-03" "1990-01-04"
...
 $ year : int  1990 1990 1990 1990 1990 1990 1990 1990 1990 1990 ...
 $ month: int  1 1 1 1 1 1 1 1 1 1 ...
 $ day  : int  1 2 3 4 5 6 7 8 9 10 ...
 $ rain : num  0 19 0 0 0 0 0 8.2 5.6 1 ...
 $ doy  : num  1 2 3 4 5 6 7 8 9 10 ...
 $ ym   : Date, format: "1990-01-01" "1990-01-01" "1990-01-01" "1990-01-01"
...

head(arm)
            date year month day rain doy         ym
48578 1990-01-01 1990     1   1    0   1 1990-01-01
48579 1990-01-02 1990     1   2   19   2 1990-01-01
48580 1990-01-03 1990     1   3    0   3 1990-01-01
48581 1990-01-04 1990     1   4    0   4 1990-01-01
48582 1990-01-05 1990     1   5    0   5 1990-01-01
48583 1990-01-06 1990     1   6    0   6 1990-01-01

The ym is calculated using ?as.yearmon from the zoo package. 
The zoo package may be useful because of its aggregating functions

Need mean daily rain - use whatever aggregation factor/function that you
need

arm.A <- aggregate(rain ~ year, arm, function(x) mean(x[x>0]) )

library(lattice)

xyplot(rain ~ date, arm,
       groups = factor(arm$year),
       ylim  = c(0,60),
       avg   = arm.A$rain, # 
       panel = panel.superpose,
       panel.groups = function(x, y,  type, col,group.number,avg, ...) {
                  
                 # plot daily values
                  panel.xyplot(x, y, col = "grey80", type = "h") # delete if
necessary
                 # plot average
                  panel.xyplot(x, avg[group.number],  type = "l", col =
"black")

       }
)

Regards

Duncan

Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mackay at northnet.com.au

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of lily li
Sent: Wednesday, 3 August 2016 04:10
To: R mailing list
Subject: [R] how to plot annual values directly

Hi R users,

I have a dataframe, with daily precipitation data, is it possible to plot
annual mean or annual sum values directly? Thanks for your help.

df
year   month   day   precip       time
2010     1          1        0.5     2010-01-01
2010     1          2        0.8     2010-01-02
2010     1          3        1.0     2010-01-03
2010     1          4        0.9     2010-01-04
...

fig1 = ggplot()+ geom_path(data=df, aes(x=time, y= precip)
show(fig1)


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