R help - Jun 2009 - Getting a column of values from a list - think I'm doing it the hard way

Example code it shown below.

I think I am doing this the hard way.  I'm just trying to get the full year
value from an array of dates.  An example array is shown below.  Right now,
I'm using a "for" loop to pull the year out of a list where the
dates were split up into their individual components.

This seems to work, but just wondering if there is an easier way.  

Thanks for any insights. 

#*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
HouseDates <- c("02/27/90", "02/27/91",
"01/14/92", "02/28/93", "02/01/94",
"02/01/95", "02/01/96")

# ?as.Date
HouseDatesFormatted<-as.Date(HouseDates, "%m/%d/%y")

HouseDatesFormatted

HouseDatesList<-strsplit(as.character(HouseDatesFormatted), "-",
fixed=TRUE)

HouseYear_array<-NULL
length_array<-length(HouseDatesList)
for(ii in 1:length_array)
{   
        HouseYear<-HouseDatesList[[ii]][1]
        
        HouseYear_array<-c(HouseYear_array, HouseYear)
}

as.character(HouseYear_array)

# Desired:
# [1] "1990" "1991" "1992" "1993"
"1994" "1995" "1996"

2009/6/4 Jason Rupert <jasonkrupert at yahoo.com>:
>
> Example code it shown below.
>
> I think I am doing this the hard way. ?I'm just trying to get the full
year value from an array of dates. ?An example array is shown below. ?Right now,
I'm using a "for" loop to pull the year out of a list where the
dates were split up into their individual components.
>
> This seems to work, but just wondering if there is an easier way.
>
> Thanks for any insights.
>
> #*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
> HouseDates <- c("02/27/90", "02/27/91",
"01/14/92", "02/28/93", "02/01/94",
"02/01/95", "02/01/96")
>
> # ?as.Date
> HouseDatesFormatted<-as.Date(HouseDates, "%m/%d/%y")
>
> HouseDatesFormatted
>
> HouseDatesList<-strsplit(as.character(HouseDatesFormatted),
"-", fixed=TRUE)
> sapply(HouseDatesList,function(x) x[[1]])
[1] "1990" "1991" "1992" "1993"
"1994" "1995" "1996"

> HouseYear_array<-NULL
> length_array<-length(HouseDatesList)
> for(ii in 1:length_array)
> {
> ? ? ? ?HouseYear<-HouseDatesList[[ii]][1]
>
> ? ? ? ?HouseYear_array<-c(HouseYear_array, HouseYear)
> }
>
> as.character(HouseYear_array)
>
> # Desired:
> # [1] "1990" "1991" "1992" "1993"
"1994" "1995" "1996"
>
> ______________________________________________
> 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.
>


-- 
HUANG Ronggui, Wincent
PhD Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html

On Jun 3, 2009, at 9:45 PM, Jason Rupert wrote:
>
> Example code it shown below.
>
> I think I am doing this the hard way.  I'm just trying to get the  
> full year value from an array of dates.  An example array is shown  
> below.  Right now, I'm using a "for" loop to pull the year
out of a
> list where the dates were split up into their individual components.
>
> This seems to work, but just wondering if there is an easier way.
>
> Thanks for any insights.
>
> #*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
> HouseDates <- c("02/27/90", "02/27/91",
"01/14/92", "02/28/93",
> "02/01/94", "02/01/95", "02/01/96")
>
> # ?as.Date
> HouseDatesFormatted<-as.Date(HouseDates, "%m/%d/%y")
>
> HouseDatesFormatted
>
> HouseDatesList<-strsplit(as.character(HouseDatesFormatted),
"-",
> fixed=TRUE)
>
> HouseYear_array<-NULL
> length_array<-length(HouseDatesList)
> for(ii in 1:length_array)
> {
>        HouseYear<-HouseDatesList[[ii]][1]
>
>        HouseYear_array<-c(HouseYear_array, HouseYear)
> }
>
> as.character(HouseYear_array)
>
> # Desired:
> # [1] "1990" "1991" "1992" "1993"
"1994" "1995" "1996"


Yep, definitely an easier way:

HouseDates <- c("02/27/90", "02/27/91",
"01/14/92", "02/28/93",
"02/01/94", "02/01/95", "02/01/96")

# You can convert the entire *vector* to a Date and then format the  
result to Year only.
# See ?format.Date

 > format(as.Date(HouseDates, "%m/%d/%y"), "%Y")
[1] "1990" "1991" "1992" "1993"
"1994" "1995" "1996"


One of the key things to bear in mind about R, is that in general  
(there are exceptions), avoiding explicit 'for' loops is a paradigm. R  
is by design, a vectorized language. That means that many operations  
are designed to take a vector, matrix or array and operate on it in a  
"whole object" fashion. Essentially, iterate an operation over the  
entire object with a single function call.

If you are coming from a different programming language, this is  
perhaps one of the more challenging perspectives to achieve. But once  
you make that leap, you will look back and be amazed at how you ever  
managed to get things before (and how many lines of code it took...)

If you have not, read through 'An Introduction to R', provided with  
your R distribution or on the R web site. It's a good place to start  
and there are other contributed materials and books listed on the R  
web site that can supplement that as you progress.

HTH,

Marc Schwartz

Dear Jason,

Have a look at years() from the chron package.

library(chron) 
HouseDates <- c("02/27/90", "02/27/91",
"01/14/92", "02/28/93",
"02/01/94", "02/01/95", "02/01/96")
HouseDatesFormatted<-as.Date(HouseDates, "%m/%d/%y")
years(HouseDates)

HTH,

Thierry
------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
Thierry.Onkelinx at inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey

-----Oorspronkelijk bericht-----
Van: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
Namens Jason Rupert
Verzonden: donderdag 4 juni 2009 4:45
Aan: R-help at r-project.org
Onderwerp: [R] Getting a column of values from a list - think I'm doing
it thehard way


Example code it shown below.

I think I am doing this the hard way.  I'm just trying to get the full
year value from an array of dates.  An example array is shown below.
Right now, I'm using a "for" loop to pull the year out of a list
where
the dates were split up into their individual components.  

This seems to work, but just wondering if there is an easier way.  

Thanks for any insights. 

#*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
HouseDates <- c("02/27/90", "02/27/91",
"01/14/92", "02/28/93",
"02/01/94", "02/01/95", "02/01/96")

# ?as.Date
HouseDatesFormatted<-as.Date(HouseDates, "%m/%d/%y")

HouseDatesFormatted

HouseDatesList<-strsplit(as.character(HouseDatesFormatted), "-",
fixed=TRUE)

HouseYear_array<-NULL
length_array<-length(HouseDatesList)
for(ii in 1:length_array)
{   
        HouseYear<-HouseDatesList[[ii]][1]
        
        HouseYear_array<-c(HouseYear_array, HouseYear) }

as.character(HouseYear_array)

# Desired:
# [1] "1990" "1991" "1992" "1993"
"1994" "1995" "1996"

______________________________________________
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.

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