Hi,
I got a problem. Using loop, I'm trying to feed data frames, returned from
the function, into the list but not all data frames are getting captured.
I have vector with some string values which I want to pass to the function.
*groups* <- c('cocacola', 'youtube','facebook)
for(i in 1:length(gr
Hi Ragia,
If you have read in your data frame with read.table or similar and not
specified stringsAsFactors=FALSE, the two columns will already be
factors. However, unless they both contain the same number of unique
values, the numbers associated with those levels won't be the same.
Probably the ea
Hi carol,
You could use the "I" function, which will just return what you pass to it.
Jim
On Thu, Mar 10, 2016 at 12:28 AM, carol white via R-help
wrote:
> What should be FUN in aggregate as no function like mean, sum etc will be
> applied
> Carol
>
> On Wednesday, March 9, 2016 1:59 PM, S
How much do you care about dealing with misformatted date strings, like
"111-Oct"
or "12-Mai"? Flagging those may be more important than milliseconds of CPU
time.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Mar 9, 2016 at 5:24 PM, Dalthorp, Daniel wrote:
> How about the following, whi
How about the following, which is both faster and simpler than either the
as.Date(...) or sub(...) solutions discussed earlier:
substring(x,first=nchar(x)-2)
require(microbenchmark)
microbenchmark(format(as.Date(paste0(x,"-2016"),format='%d-%b-%Y'),'%b'))
# 59.3 microseconds on my computer
microb
How about slower? That is objective.
I use dates all the time so I am quite familiar with what they are good for.
However, I prefer to avoid inventing information such as which year the date
should have included unless I have to based on knowledge of the data source. It
is not good to mislead
Good point about 29-Feb...fixed in the following:
format(as.Date(paste0(x,"-2016"),format='%d-%b-%Y'),'%b')
# Also: The date functions can be used to easily calculate passage of time
and offer good flexibility for formatting output.
-Dan
P.S. "harder to understand" is in the eye of the beholder
Still not recommended. That takes more steps, is harder to understand, and
will break when given "29-Feb" as input.
--
Sent from my phone. Please excuse my brevity.
On March 9, 2016 4:15:31 PM PST, "Dalthorp, Daniel" wrote:
>Or:
>
>x <- c( "3-Oct", "10-Nov" )
>format(as.Date(paste0(x,rep("-19
Or:
x <- c( "3-Oct", "10-Nov" )
format(as.Date(paste0(x,rep("-1970",length(x))),format='%d-%b-%Y'),'%b')
# the 'paste0' appends a year to the text vector
# the 'as.Date' interprets the strings as dates with format 10-Jun-2016
(e.g.)
# the 'format' returns a string with date in format '%b' (which
?as.Date
On Wed, Mar 9, 2016 at 10:14 AM, KMNanus wrote:
> I have a series of dates in format 3-Oct, 10-Oct, 20-Oct, etc.
>
> I want to create a variable of just the month. If I convert the date to a
> character string, substr is ineffective because some of the dates have 5
> characters (3-Oct
Your dates are incomplete (no year) so I suggest staying away from the date
functions for this. Read ?regex and ?sub.
x <- c( "3-Oct", "10-Nov" )
m <- sub( "^\\d+-([A-Za-z]{3})$", "\\1", x )
--
Sent from my phone. Please excuse my brevity.
On March 9, 2016 10:14:25 AM PST, KMNanus wrote:
>I h
> On Mar 8, 2016, at 1:48 PM, Sarah Goslee wrote:
>
> Thanks for the complete reproducible example.
>
> Here's one way to approach the problem; there are many others.
>
> recodeDat <- function(x, invals, outvals) {
> x <- as.character(x)
> invals <- as.character(invals)
> outvals <- as.charact
I have a series of dates in format 3-Oct, 10-Oct, 20-Oct, etc.
I want to create a variable of just the month. If I convert the date to a
character string, substr is ineffective because some of the dates have 5
characters (3-Oct) and some have 6 (10-Oct).
Is there a date function that accompli
> On 09 Mar 2016, at 18:52 , John Hillier wrote:
>
> Dear All,
>
>
> I am attempting to describe a distribution of height data. It appears
> roughly linear on a log-log plot, so Pareto seems sensible. However, the
> data are only reliable in a limited range (e.g. 2000 to 4800 m). So, I wou
The variable DISPLAY is what is causing problems. Run the command 'unset
DISPLAY' before running R .
Stephen
On 09/03/16 01:06 PM, Uwe Ligges wrote:
> I do not get this: If it works it is OK to use it. If it does not work,
> you can't
>
> Best,
> Uwe Kigges
>
>
>
>
>
> On 09.03.2016 17
I do not get this: If it works it is OK to use it. If it does not work,
you can't
Best,
Uwe Kigges
On 09.03.2016 17:44, Santosh wrote:
Thanks for your response. Since the test failed due to X11 connectivity
reasons, is it okay to use it in applications where X11 server connectivity
is
Dear All,
I am attempting to describe a distribution of height data. It appears roughly
linear on a log-log plot, so Pareto seems sensible. However, the data are only
reliable in a limited range (e.g. 2000 to 4800 m). So, I would like to fit a
Pareto distribution to the reliable (i.e. trunca
On Tue, 8 Mar 2016, stefania innocenti wrote:
Hello,
I am trying to use the ivregress function to estimate a second stage
model which looks like the following:
LogGDP=GEO+RULE+OPENNESS
I would like to instrument Rule (RULE) with Settler mortality (SETTLER)
and Openness (OPENNESS) with logFr
So in both my solution and your second option, lm prints that it evaluated the
regression in a function context (using dfr) which the user of the function
might prefer to be unaware of (they know what X.des is). Your first solution
avoids that but hardcodes access to the global variable so if t
Thanks for your response. Since the test failed due to X11 connectivity
reasons, is it okay to use it in applications where X11 server connectivity
is not required?
Thanks and much appreciated,
Santosh
On Tue, Mar 8, 2016 at 10:41 PM, Uwe Ligges wrote:
>
>
> On 09.03.2016 02:19, Santosh wrote:
>
> > What I need is this:
> > [[1]]
> > [1] 1 2 3
> > [[1]]
> > [2] 1 2 3
> > [[1]]
> > [2] 1 2 3
Try
rep(list(1:3), 3)
S Ellison
***
This email and any attachments are confidential. Any use...{{dropped:8}}
Hi,
On Wed, Mar 9, 2016 at 10:22 AM, Jan Kacaba wrote:
> Hello I would like to assign a vector to list sequence. I'm trying my code
> bellow, but the output is not what inteded.
>
> # my code
> mls=vector(mode="list") # my list
> cseq=c(1:3) # my vector
> mls[cseq]=cseq
>
> I get following:
> [
Hello I would like to assign a vector to list sequence. I'm trying my code
bellow, but the output is not what inteded.
# my code
mls=vector(mode="list") # my list
cseq=c(1:3) # my vector
mls[cseq]=cseq
I get following:
[[1]]
[1] 1
[[1]]
[2] 2
[[1]]
[2] 3
What I need is this:
[[1]]
[1] 1 2 3
[[1]
Just to help next time some comments in-line
On 09/03/2016 14:31, Venky wrote:
Hi
I want to find the information value for my overall data.
I have data like this.my data set name is data
Best not to call your data.frame data as it may conflict with the
function data()
Id smoked
A search on rseek.org for information value turns up several
contributed packages that may do what you want.
Sarah
On Wed, Mar 9, 2016 at 9:31 AM, Venky wrote:
> Hi
>
> I want to find the information value for my overall data.
>
> I have data like this.my data set name is data
>
> Id smoke
On Wed, Mar 9, 2016 at 8:28 AM, carol white wrote:
> What should be FUN in aggregate as no function like mean, sum etc will be
> applied
I have no idea, since you haven't told us what you want the results to
look like.
> Carol
>
>
> On Wednesday, March 9, 2016 1:59 PM, Sarah Goslee
> wrote:
>
Hi
I want to find the information value for my overall data.
I have data like this.my data set name is data
Id smokedrink nothing
1 yesno no
2 no yesyes
3 yesyesno
. . . .
Etc
Hi
Without posting some data by dput, you hardly get any sensible answer.
You can use any function which gives you appropriate result, even your own.
Or even silly one like
aggregate(iris[,1:3], list(iris$Species), paste, collapse="")
Cheers
Petr
> -Original Message-
> From: R-help [m
> How is it possible to group rows of a matrix or a data frame by the same
> values
> of the first column?
If you mean _group_ as in SQL GROUP BY, use aggregate() with a count or summary
statistic.
If you mean _sort_, just to get similar values close together, use order()
For example, to sort
I found the right usage.
Thanks
On Wednesday, March 9, 2016 2:28 PM, carol white wrote:
What should be FUN in aggregate as no function like mean, sum etc will be
applied
Carol
On Wednesday, March 9, 2016 1:59 PM, Sarah Goslee
wrote:
Possibly aggregate(), but you posted in HT
What should be FUN in aggregate as no function like mean, sum etc will be
applied
Carol
On Wednesday, March 9, 2016 1:59 PM, Sarah Goslee
wrote:
Possibly aggregate(), but you posted in HTML so your data were mangled.
Please use dput(), post in plain text, and try to explain more clear
Possibly aggregate(), but you posted in HTML so your data were mangled.
Please use dput(), post in plain text, and try to explain more clearly
what you want the result to look like.
Sarah
On Wed, Mar 9, 2016 at 7:09 AM, carol white via R-help
wrote:
> How is it possible to group rows of a matri
How is it possible to group rows of a matrix or a data frame by the same values
of the first column?
1 14331 453452 653 3762 45
1 1433,453452 45, 653 376
Thanks
Carol
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
> In a nutshell, formulas carry the environment in which they are defined along
> with the variable names, and your dfr was defined in the test.FN environment,
> but the formulas were defined in the global environment. I got this to work by
> defining the formula character strings in the global env
In a nutshell, formulas carry the environment in which they are
defined along with the variable names, and your dfr was defined in the
test.FN environment, but the formulas were defined in the global
environment. I got this to work by defining the formula character strings
in the global environ
Hi
As you did not provide the data it is only a guess. The values are actually
factors and in that case they already have unique numeric code.
you can try
as.numeric(yourdata)
If the values are character you can transfer them to factor and use as.numeric.
as.numeric(as.factor(yourdata))
Chee
Dear group
kindly if i had data frame of 2 columns that has repeated URLS ..and want to
replace those urls with ID's for each one so the url will have
unique ID, how can I do this
thanks in advance
e.g otf data
pcworld.com open.itworld.com
pcworld.com storage.itworld.com
salon.com
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