Thanks, Ista. I thought there might be a "tidy" way to do this, but I
hadn't use stringr.
-- Mike
On Tue, Apr 25, 2017 at 8:47 PM, Ista Zahn wrote:
> stringr::str_count (and stringi::stri_count that it wraps) interpret
> the pattern argument as a regular expression by default.
>
> Best,
> Ista
stringr::str_count (and stringi::stri_count that it wraps) interpret
the pattern argument as a regular expression by default.
Best,
Ista
On Tue, Apr 25, 2017 at 11:40 PM, Michael Hannon
wrote:
> I like Boris's "Hadley" solution. For the record, I've appended a
> version that uses regular expres
I like Boris's "Hadley" solution. For the record, I've appended a
version that uses regular expressions, the only benefit of which is
that it could be generalized to find more-complicated patterns.
-- Mike
counts <- sapply(text1, function(next_string) {
loc_example <- length(gregexpr("Exampl
I should add: there's a str_count() function in the stringr package.
library(stringr)
str_count(text1, "Example")
# [1] 5 5 5 5
I guess that would be the neater solution.
B.
> On Apr 25, 2017, at 8:23 PM, Boris Steipe wrote:
>
> How about:
>
> unlist(lapply(strsplit(text1, "Example"), func
How about:
unlist(lapply(strsplit(text1, "Example"), function(x) { length(x) - 1 } ))
Splitting your string on the five "Examples" in each gives six elements.
length(x) - 1 is the number of
matches. You can use any regex instead of "example" if you need to tweak what
you are looking for.
B.
Hi all,
I am looking for a streamlined way of counting the number of enumerated
items are each element of a character vector. For example:
text1<-c("This is an example.
List 1
1) Example 1
2) Example 2
10) Example 10
List 2
1) Example 1
2) Example 2
These have been examples.","This is another ex
On 25 April 2017 at 18:05, Duncan Murdoch wrote:
| On 25/04/2017 5:41 PM, Dirk Eddelbuettel wrote:
| >
| > On 25 April 2017 at 16:04, Jeff Reichman wrote:
| > | R Users
| > |
| > | Having problems converting the following DTG into an R recognized
date/time
| > | field
| > |
| > | 01-01-2016T14:02
On 25/04/2017 5:41 PM, Dirk Eddelbuettel wrote:
On 25 April 2017 at 16:04, Jeff Reichman wrote:
| R Users
|
| Having problems converting the following DTG into an R recognized date/time
| field
|
| 01-01-2016T14:02:23.325
|
| Would I separate it into a date field and time filed then put it back
On 25 April 2017 at 16:04, Jeff Reichman wrote:
| R Users
|
| Having problems converting the following DTG into an R recognized date/time
| field
|
| 01-01-2016T14:02:23.325
|
| Would I separate it into a date field and time filed then put it back
| together???
The anytime package (on CRAN) do
> z <- as.POSIXct("01-01-2016T14:02:23.325", format="%d-%m-%YT%H:%M:%OS")
> dput(z)
structure(1451685743.325, class = c("POSIXct", "POSIXt"), tzone = "")
> z
[1] "2016-01-01 14:02:23 PST"
> format(z, "%H:%M:%OS3 on %b %d, %Y")
[1] "14:02:23.325 on Jan 01, 2016"
(Don't separate the date and time pa
On 25/04/2017 5:04 PM, Jeff Reichman wrote:
R Users
Having problems converting the following DTG into an R recognized date/time
field
01-01-2016T14:02:23.325
Would I separate it into a date field and time filed then put it back
together???
This appears to work (though I'm not sure whe
R Users
Having problems converting the following DTG into an R recognized date/time
field
01-01-2016T14:02:23.325
Would I separate it into a date field and time filed then put it back
together???
Jeff
[[alternative HTML version deleted]]
__
How about:
d_sample_1 <- floor(d_sample/100) * 100
for (i in 1:nrow(d_sample_1)) {
d_sample_1[i, duplicated(unlist(d_sample_1[i, ]))] <- NA
}
B.
> On Apr 25, 2017, at 1:10 PM, Bert Gunter wrote:
>
> If I understand you correctly, one way is:
>
>> z <- rep(LETTERS[1:3],4)
>> z
> [1] "A
If I understand you correctly, one way is:
> z <- rep(LETTERS[1:3],4)
> z
[1] "A" "B" "C" "A" "B" "C" "A" "B" "C" "A" "B" "C"
> z[!duplicated(z)]
[1] "A" "B" "C"
?duplicated
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it
Dear Altruist,
Thank you so much.
This does it.
I shall keep connected.
Regards,
Miah
From: Anthoni, Peter (IMK)
Sent: Tuesday, April 25, 2017 8:38 AM
To: Saifuddin, Miah Mohammad
Cc: Thomas Mailund; r-help@r-project.org; PIKAL Petr
Subject: Re: [R] aski
Hello!
I welcome the new *R* 3.4.0. I installed it on my Windows 7 Home
Premium [Service Pack 1, updated to latest, running on an HP AMD(Phenom)
II 955 X4 Processor, 3.20 GHz, 16 GB RAM, 64-bit, with lots of free
storage on disk and a solid state disk for virtual cache]. It was
installed atop the
If anyone are interested, I found a solution for lazy lists. A simplified
version of their construction and access looks like this:
nil <- function() NULL
cons <- function(car, cdr) {
force(car)
force(cdr)
function() list(car = car, cdr = cdr)
}
is_nil <- function(lst) is.null(lst())
car <
Hi All,
in my current project I am working with multiple-response questions
(MRSets):
-- Coding --
100 Main Code 1
110 Sub Code 1.1
120 Sub Code 1.2
130 Sub Code 1.3
200 Main Code 2
210 Sub Code 2.1
220 Sub Code 2.2
230 Sub Code 2.3
300 Main Code 3
310 Sub Code 3.1
320 Sub Code 3.2
The coding
Hi,
the cut function might be helpful.
vec=1: 163863
fcut=cut(vec,seq(1, 163863+1,by= 6069),include.lowest = T,right=F)
aggregate(vec,by=list(fcut),min)
aggregate(vec,by=list(fcut),max)
cheers
Peter
On 25. Apr 2017, at 14:33, PIKAL Petr
mailto:petr.pi...@precheza.cz>> wrote:
Hi
-Origin
On 2017-04-25 10:34, Martin Maechler wrote:
Göran Broström
on Tue, 25 Apr 2017 10:22:48 +0200 writes:
> I installed R-3.4.0 and got problems with the survival package, for
instance
>
>> library(survival)
>> mort <- data.frame(exit = 1:4, event = rep(1, 4), x
Hi
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Thomas
> Mailund
> Sent: Tuesday, April 25, 2017 11:20 AM
> To: r-help@r-project.org; Saifuddin, Miah Mohammad
>
> Subject: Re: [R] asking for help
>
> If you write something like
>
> indices <- rep(
Right, normally this is how it works for me when I install R from
source. In this case, on this computer, I use the debian/ubuntu
packaging, and then it is necessary to 'rebuild' packages, obviously.
Thanks, Göran
On 2017-04-25 12:20, Viechtbauer Wolfgang (SP) wrote:
Sort of an obvious approa
Dear Mustafa,
Please keep the mailing list in cc.
Since you claim to have written the code, you can share the code so we can
review it. That makes more sense than having us to write code for you...
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute fo
Sort of an obvious approach, but after every upgrade (regardless if it is
major/minor), I just delete my entire personal library and reinstall everything
from scratch. For this, I have a script that includes just a bunch of
install.packages() calls. Like:
install.packages(c("lme4", "glmmML", "M
> Ashim Kapoor
> on Tue, 25 Apr 2017 14:02:18 +0530 writes:
> Dear all,
> I am not able to understand the interplay of absolute vs relative and
> tolerance in the use of all.equal
> If I want to find out if absolute differences between 2 numbers/vectors
are
> big
If you write something like
indices <- rep(1:(163863/6069), each = 6069)
you can get the i’th block of rows with
table[indices == i,]
It looks like a little more work than a loop would be since you have to run
through all rows for each block, but the implicit loop in this approach is
likely t
Thanks Martin,
that helped!
Göran
On 2017-04-25 10:34, Martin Maechler wrote:
Göran Broström
on Tue, 25 Apr 2017 10:22:48 +0200 writes:
> I installed R-3.4.0 and got problems with the survival package, for
instance
>
>> library(survival)
>> mort <- data.fra
> Göran Broström
> on Tue, 25 Apr 2017 10:22:48 +0200 writes:
> I installed R-3.4.0 and got problems with the survival package, for
instance
>
>> library(survival)
>> mort <- data.frame(exit = 1:4, event = rep(1, 4), x = c(0, 1, 0, 1))
>> fit <- coxph
Dear all,
I am not able to understand the interplay of absolute vs relative and
tolerance in the use of all.equal
If I want to find out if absolute differences between 2 numbers/vectors are
bigger than a given tolerance I would do:
all.equal(1,1.1,scale=1,tol= .1)
If I want to find out if relat
I installed R-3.4.0 and got problems with the survival package, for instance
> library(survival)
> mort <- data.frame(exit = 1:4, event = rep(1, 4), x = c(0, 1, 0, 1))
> fit <- coxph(Surv(exit, event) ~ x, data = mort)
Error in fitter(X, Y, strats, offset, init, control, weights = we
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