Function that create day of the year column.
Dear Federic
You need to do something like
function(p, weights = NULL, data = NULL, subset = NULL, na.action =
na.fail) { # but you will have Day, Month, Year, data
# instead of the first three
if(is.null(data)) data <- sys.frame(sys.parent())
mf <- match.call()
mf$data <- NULL
mf$subset <- NULL
mf$na.action <- NULL
mf[[1]] <- as.name("data.frame")
mf <- eval(mf, data)
if(!is.null(subset)) mf <- mf[subset,]
mf <- na.action(mf)
p <- as.numeric(mf$p) # here you will have Day <- mf$Day and so on
weights <- mf$weights
After this code (stolen from somewhere, it is definitely not original to
me) you will find that the variables you need (in my case p and weights)
have been found from data. So if your variables happen to be called
jour, mois and an you will then be able to write Day = jour, Month =
mois, Year = an, data = mesdonnees and have them picked up from your
data frame mesdonnees.
You do not need to attach data in this system and in general it is a bad
idea.
On 05/11/2014 08:34, Frederic Ntirenganya wrote:
Dear PIKAL,
I am using a data frame not a single value. I wanted to add a new column
"DOY" on a dataset. I made it but I want to make it General. i.e. You will
see in the dataset I am using columns names (Year,month and day.) What if I
have this information in other columns names? In this function I will be
pushed to change my columns names which I don't think it is a good idea.
How can I make it in such away that I can use any column's name which
contains the informations I want to use?
In the previous comment this is what I mean saying a single date "single
date column(ex:2009-08-02)"
The following is the function and head-tail of the data after running the
function.
library(lubridate)
DOY=function(data){
attach(data)
#=================================================================
#This function create day of teh year from a single date
column(ex:2009-08-02) or/and
#from the date in 3 columns (Year, month, Day).
#================================================================
if (is.data.frame(data)) { # check if the data is a dataframe
dt1=yday(as.Date(paste(Year, Month, Day,sep="-")))
# check if the year is leap or not.
datelp= dt1>59 & !leap_year(year(as.Date(paste(Year,Month, Day,sep="-"))))
datelp[is.na(datelp)]=FALSE
dt1[datelp]=dt1[datelp]+1 # add one for non leap_year
data2=cbind(data, dt1) # combining columns of data
n<-ncol(data)
colnames(data2)[n+1] = "DOY"
}
data2
}
head(DOY(Kitale))
Year Month Day Rain DOY
1 1979 1 1 0 1
2 1979 1 2 0 2
3 1979 1 3 0 3
4 1979 1 4 0 4
5 1979 1 5 0 5
6 1979 1 6 0 6
tail(DOY(Kitale))
Year Month Day Rain DOY
11315 2009 12 26 40.7 361
11316 2009 12 27 13.6 362
11317 2009 12 28 42.2 363
11318 2009 12 29 53.6 364
11319 2009 12 30 19.8 365
11320 2009 12 31 0.5 366
Thanks for the help.
Regards,
Frederic.
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fredo at aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
On Tue, Nov 4, 2014 at 6:55 PM, PIKAL Petr <petr.pikal at precheza.cz> wrote:
Hi
You want a function with input of either single value or a data frame?
This is very strange. One option is to make a method for function but it is
quite beyond my capabilities and it is used by "professional" developers.
methods(c) or methods(plot)
So you need to set method for your function DOY something like
DOY.single_value and DOY.data.frame
Anyway
Function yday takes some value in time/date format regardless it is single
value or a vector, so the problem is how do you want your function to work
and more importantly how general you want it. I simple case you need to
check if data is single value or data frame.
DOY=function(data){
if (is.data.frame(data)) {
do stuff for data frame} else {
do stuff for single value}
return something
}
Other comments see in line
data$Rain=as.numeric(as.character(data$Rain))
dt=yday(data$Date) # single date of the data
If data is data frame dt shall be vector uf numbers not a single date
datelp= dt>59 & !leap_year(data$Date)# tell us that the date
datelp is (maybe) a logical vector with the same length as dt
occurs
during a leap year dt[datelp]=dt[datelp]+1 # add one for non leap_year
Why you do this?
dt<-1:365 # e.g. length is 365
datelp<-dt>59 # logical vector
dt[datelp]<-dt[datelp]+1 # change all numbers greater than 59
dt
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18
[19] 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
36
[37] 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53
54
[55] 55 56 57 58 59 61 62 63 64 65 66 67 68 69 70 71 72
73
^^^^^^
now you have day 60 missing
[73] 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
91
[91] 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108
109
[109] 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126
127
[127] 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144
145
[145] 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162
163
[163] 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
181
[181] 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198
199
[199] 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216
217
[217] 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234
235
[235] 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252
253
[253] 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270
271
[271] 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288
289
[289] 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306
307
[307] 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324
325
[325] 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342
343
[343] 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360
361
[361] 362 363 364 365 366
But length of dt stays the same
length(dt)
[1] 365
Cheers
Petr
cbind(data, dt) # combining columns of data conames(data)="DOY" # name of new column. ??I have a problem on
how
I can precise the column in gerenal. }
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
project.org] On Behalf Of Frederic Ntirenganya
Sent: Tuesday, November 04, 2014 1:31 PM
To: daniel
Cc: r-help at r-project.org
Subject: Re: [R] Function that create day of the year column.
Hi Daniel,
How can I add an if conditiopn or for loop to implement this in my
function?
ex: option1 : single date
option2: date in 3 columns (year,moth,day)
yday(Sys.Date())
yday(as.Date(paste(y,m,d,sep="-")))
Regards,
Frederic.
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fredo at aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
On Tue, Nov 4, 2014 at 3:20 PM, daniel <daniel319 at gmail.com> wrote:
Frederic,
Check the lubridate library.
install.packages("lubridate")
library(lubridate)
yday(Sys.Date())
d <- 4
m <- 11
y <- 2014
yday(as.Date(paste(y,m,d,sep="-")))
Daniel Merino
2014-11-04 7:01 GMT-03:00 Frederic Ntirenganya <ntfredo at gmail.com>:
Dear All, I would like to make a function that create Day of the year column
in
a dataset.
State of the problem: write a function that create a day column
either from a single date column (string/or factors) or from date in
3 columns (year, month, day).
I made the following function for a single date. I would like to add
a condition for date in 3 columns (year, month, day). My data is
daily climate data.
#write a function that create a day column either from a single date
column (string/or factors) #or from date in 3 columns (year, month,
day).
DOY=function(data){
#=================================================================
#This function create day of teh year from a single date
column(ex:2009-08-02) or/and
#from the date in 3 columns (Year, month, Day).
#================================================================
data$Rain=as.numeric(as.character(data$Rain))
dt=yday(data$Date) # single date of the data
datelp= dt>59 & !leap_year(data$Date)# tell us that the date
occurs
during a leap year dt[datelp]=dt[datelp]+1 # add one for non leap_year cbind(data, dt) # combining columns of data conames(data)="DOY" # name of new column. ??I have a problem on
how
I can precise the column in gerenal.
}
ex: year month day Date Rain Tmin Tmax
1971 1 1 1971-01-01 0 8.2 15
1971 1 2 1971-01-02 0 4.2 11
. . . . . . .
. . . . . . .
. . . . . . .
Any ideal on how I can make this function is welcome. thanks!
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fredo at aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
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-- Daniel
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