Write a for-loop that tests your condition and carries the necessary parameters
forward. break() when your condition is met.
B.
> On Oct 29, 2017, at 10:24 PM, li li wrote:
>
> Dear all,
> The function f() below is a function of m1 and m2, both of which are
> matrices with 3 rows. The func
Matti -
Since you are asking about looping through a column, not looping across
columns, it is simply the following:
# Note: data.frame() turns strings into factors by default.
myDF <- data.frame(type = c("a", "j", "a", "a", "j"),
weight = c(12.3, 6.8, 10.5, NA, "5.5"))
myDF
ep coming along and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
> On Mon, Nov 6, 2017 at 3:28 PM, Boris Steipe wrote:
> Matti -
>
> Since you are asking about looping through a column, not looping acros
Hi -
This list has a "no homework" policy, but we can certainly help once you show
what effort you have put into this yourself. Read the posting guide, especially
regarding the instructions re. reproducible examples, and follow them exactly.
Cheers,
B.
> On Nov 9, 2017, at 10:05 AM, Hye Joo
|> x <- sample(0:2, 10, replace = TRUE)
|> x
[1] 1 0 2 1 0 2 2 0 2 1
|> tabulate(x)
[1] 3 4
|> table(x)
x
0 1 2
3 3 4
B.
> On Nov 10, 2017, at 4:32 AM, Allaisone 1 wrote:
>
>
>
> Thank you for your effort Bert..,
>
>
> I knew what is the problem now, the values (1,2,3) were only an e
If you consider the definition of a DTM, and that findAssoc() computes
associations between words as correlations across documents(!), you will
realize that you can't what you want from a single document. Indeed, what kind
of an "association" would you even be looking for?
B.
> On Nov 15, 20
t; Election speech) in one single TEXT document, and i want to find the
> association of the top 3 most frequently occurring words with the other
> words in the speech, what method do I adopt ?
>
> On Wed, Nov 15, 2017 at 7:08 PM, Boris Steipe
> wrote:
>
> > If you co
Large packages sometimes mask each other's functions and that creates a
headache, especially for teaching code, since function signatures may depend on
which order packages were loaded in. One of my students proposed using the idiom
<- ::
... in a preamble, when we use just a small subset of
Combine columns 1 and 2 into a column with a single ID like "33.55", "44.66"
and use split() on these IDs to break up your dataset. Iterate over the list of
data frames split() returns.
B.
> On Nov 17, 2017, at 12:59 PM, Allaisone 1 wrote:
>
>
> Hi all ..,
>
>
> I have a large dataset of
gt;
> when I run this code as before :-
>
> maf <- apply(SeparatedGroupsofmealsCombs, 2, function(x)maf(tabulate(x+1)))
>
> an error message says : dim(X) must have a positive length . I'm not sure
> which length
> I need to specify.. any suggestions to correc
gives this error :-
> Error in FUN(left, right) : non-numeric argument to binary operator
>
> I have been trying since yesterday but but until now I'm not able to identify
> the correct syntax.
>
>
>
>
> From: David Winsemius
> Sent: 18 November 2017 20:06:56
&
gt; AllMAFs[[i]] <- apply( SeparatedGroupsofmealsCombs[[i]], 2,
> function(x)maf( tabulate( x+1) ))
>
> refers to a variable "i" that has never been defined.
>
> Duncan Murdoch
>
>>
>>
>> The lapply function :
>> results<-lapply(Separ
I have noticed that when I iterate permutations of short vectors with the same
seed, the cycle lengths are much shorter than I would expect by chance. For
example:
X <- 1:10
Xorig <- X
start <- 112358
N <- 10
for (i in 1:N) {
seed <- start + i
for (j in 1:1000) { # Maximum cycle length to c
You can either use positive lookahead/lookbehind - but support for that is a
bit flaky. Or write a proper regex, and use
backreferences to keep what you need.
R > x <- "abc 1,1 ,1 1, x,y 2,3 "
R > gsub("(\\d),(\\d)", "\\1.\\2", x, perl = TRUE)
[1] "abc 1.1 ,1 1, x,y 2.3 "
B.
> On Feb 12, 20
I haven't been able to reproduce this because you posted in HTML and no data
arrived. Use the dput() function if you want to post data.
But: a frequent and trivial reason for disappearing category labels is that the
plot window is too small to print them all. Try increasing the plotting window,
If you are talking about the "Supporting Information" - that contains only one
small piece of matlab code that looks pretty trivial to translate if necessary.
The rest are R scripts.
What then is the problem you need to solve?
B.
> On Feb 21, 2018, at 9:37 AM, JoyMae Gabion
> wrote:
>
> D
good with R software.
> Since I will be using R in my thesis, it will be helpful if there are R codes
> available for that computation.
> Can I get your help for this, Sir?
>
>
> Joy Mae
>
> On Thu, Feb 22, 2018 at 4:08 AM, Boris Steipe
> wrote:
> If you are
R > rowMeans(roop)
[1] 1.67 5.33 3.00
R > mean(as.numeric(roop[1,]))
[1] 1.67
:-)
> On Mar 20, 2018, at 10:18 PM, Sorkin, John wrote:
>
> I am trying to get the mean of a row of a data frame. My code follows:
>
>
> roop <- data.frame(x=c(1,2,3),y=c(4,5,2),z=c(0,9,4))
> roo
Should not the result be NULL if you have removed the NA with na.rm=TRUE ?
B.
> On Mar 21, 2018, at 11:44 AM, Stefano Sofia
> wrote:
>
> Dear list users,
> let me ask you this trivial question. I worked on that for a long time, by
> now.
> Suppose to have a data frame with NAs and to sum so
you cannot compute on NULL so no, that doesn't
> work.
>
> See the note under the "Value" section of ?sum as to why zero is returned
> when all inputs are removed.
> --
> Sent from my phone. Please excuse my brevity.
>
> On March 21, 2018 9:03:29 AM PDT, Bor
ust
> cause trouble.
>
> -pd
>
>> On 21 Mar 2018, at 18:05 , Boris Steipe wrote:
>>
>> Surely the result of summation of non-existent values is not defined, is it
>> not? And since the NA values have been _removed_, there's nothing left to
>> sum ov
If one is equal to the reverse of another, keep only one of the pair.
B.
> On Mar 29, 2018, at 9:48 PM, Ranjan Maitra wrote:
>
> Dear friends,
>
> I would like to get all possible arrangements of n objects listed 1:n on a
> circle.
>
> Now this is easy to do in R. Keep the last spot fixed
Just for completeness: there are several ways to open files for editing in
RStudio -
* Configure your computer to be able to double-click on a file and open it
in RStudio: this is OS dependent. Google for "change file associations"
to get advice.
* Drag and drop a file icon onto the RStu
Interesting problem.
I would discretize the x-values and interleave them. Lines from one dataset
still overlap, so you see high- density and low-density regions, but lines from
the other dataset are drawn into the interval. Like so:
interleave <- function(x, MIN, MAX, N, nChannel = 2, channel)
Q[j-2] gives you Q[0] in your first inner loop iteration.
R arrays start at one.
B.
> On 2018-06-13, at 07:21, Maija Sirkjärvi wrote:
>
> Amat[J-1+j-2,j-1]= 1/(Q[j] - Q[j-1]) + 1/(Q[j-1] - Q[j-2])
__
R-help@r-project.org mailing list -- To UNSUBSC
o do it?
>
> Thanks again!
>
> Maija
>
> 2018-06-13 15:52 GMT+03:00 Boris Steipe :
> Q[j-2] gives you Q[0] in your first inner loop iteration.
> R arrays start at one.
>
> B.
>
>
> > On 2018-06-13
Wasn't there also the requirement that the numbers be drawn from a uniform
distribution? These sequences are not. I wonder whether this can for all
practical purposes be simplified to consider only the sequence with maximum
entropy.
B.
> On 2018-07-11, at 06:23, Rolf Turner wrote:
>
> On
This looks opaque and hard to maintain.
It seems to me that a better strategy is to subset your vector with modulo
expressions, use a normal sort on each of the subsets, and add the result to
each other. 0 and 1 need to be special-cased.
myPrimes <- c(2, 3, 5)
mySource <- sample(0:10)
# specia
Yes.
> On Mar 31, 2017, at 11:53 PM, Art U wrote:
>
> Hi,
> I'm trying to create number of vectors, part of them are binary and part
> are continuous. Is there a way in R to generate them with specific
> correlation between each pair?
> Thank you in advance.
> Ariel
> --
> *I like to pretend
Your code is syntactically correct but goes against all R style guides I know.
I've changed that - but obviously you don't have to.
x1 <- c(a1 = 0, b3 = 2, e2 = -2)
x2 <- c(c = 3, d = 4, f = 5)
N <- c("a1", "b3", "d", "e2", "c", "f")
x3 <- c(x1, x2) # concatenate
x3 <- x3[N
the attachment.
>
>
> Wolny od wirusów. www.avast.com
>
> On Sat, Apr 1, 2017 at 2:38 PM, Boris Steipe wrote:
> The modulo operator returns remainder after division. The goal is to select a
> number if the remainder is zero. Casting a number to logical returns FALSE
I discourage the use of print() for debugging.
Put a browser() statement into your loop and when execution takes you to the
debugger interface, examine your variables and expressions one by one.
B.
> On Apr 4, 2017, at 10:09 AM, DANIEL PRECIADO wrote:
>
> To your first comment: Yes, the fu
The "whatever information" would be the usual minimal reproducible example.
Since you think your code works at the shell level, make up an example with a
system call to "echo".
Then state what you expect to happen and what happens instead.
B.
> On Apr 4, 2017, at 12:44 AM, stephen sefick wro
Is the function linear between the discontinuities?
Can you give an example of how the function is specified?
B.
> On Apr 9, 2017, at 8:38 PM, li li wrote:
>
> Hi Burt,
>Yes, the function is monotone increasing and points of discontinuity
> are all known.
> They are all numbers between 0
Are you sure this is trivial? I have the impression the combination of an
ill-posed problem and digital representation of numbers might just create the
illusion that is so.
B.
> On Apr 10, 2017, at 12:34 AM, Bert Gunter wrote:
>
> Then it's trivial. Check values at the discontinuities and
e I suggested fail?
>
> (Always happy to be corrected).
>
> -- Bert
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip
t; "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Mon, Apr 10, 2017 at 7:56 AM, Boris Steipe
> wrote:
>> Well - the _proce
F is f). The infimum of a set is not
> necessarily a member of the set.
>
> -pd
>
>> On 10 Apr 2017, at 16:56 , Boris Steipe wrote:
>>
>> Well - the _procedure_ will give a result.
>>
>> But think of f(x) = {-1; x <= 1/3 and 1; x > 1/3
>>
Here's my crossword-puzzle for the day:
# A sample monotonous discontinuous function with a single root
# (vectorized)
F <- function(x) {
return((as.numeric(x >= 0.112233445566778899) * 2) - 1)
}
discRoot <- function(xL, xR, F, k = 10) {
# Return the interval containing a single root of
Fie, knave, dost thou bid us do thine homework!
gsub("[^A-Za-z]", "", )
> On Apr 10, 2017, at 7:11 PM, Olu Ola via R-help wrote:
>
> Hello,I have a column of my data of the Following format
> 10. Arnold125. Jessica1. Romeo117. Juliet
> I need to eliminate the numbers, period, and the space be
As with R, do with RStudio: Read The Beautiful Manual, and peruse The Google.
For example, searching Google with the two (admittedly hard to guess)
cryptograms:
"RStudio Rprofile"
will present more than a dozen most enlightening links to fulfil your desire.
Perhaps the following link works be
Moreover, setting the seed once, then evaluating two functions means you are
sampling from the same distributions, but you do in fact have different values.
Outliers in the rarefied tails of the distribution may lie quite considerably
off the expected diagonal. Try
set.seed(123)
qqplot(rbinom(n
That's not how qqline() works. The line is drawn with respect to a
_reference_distribution_ which is the normal distribution by default. For the
binomial distribution, you need to specify the distribution argument. There is
an example in the help page that shows you how this is done for qchisq()
Ramnik,
a final mail is actually really important: this is to document in the archives,
for the benefit of those who found the thread at a later time, that the
responses indeed solved the problem.
Other than that, the single most important advice is to
- provide a minimal working example of the
ed how often
> out of 100 times did the data fall outside the CI.This I expect to be
> binomial with n=100,p=.05. I repeated this a 100 times and obtained
> count1_vector.
>
> Best Regards,
> Ashim.
>
>
> On Mon, Apr 17, 2017 at 7:51 PM, Boris Steipe
> wrote:
&
The concept of equality for numbers that are represented on a computer is
frequently misapplied. Consider:
a <- seq(from=0, to=1, by=0.02)
print(a[36])
[1] 0.7
a[36] == 0.7
[1] FALSE
print(a[36], digits=22)
[1] 0.7000666134
a[36] == 0.7001
[1] TRUE
All clear?
B.
Here you go:
https://www.google.ca/search?q=r+partial+least+squares
https://www.google.ca/search?q=r+ridge+regression
> On Apr 18, 2017, at 3:45 PM, SAIRA SALEEM wrote:
>
> i required r codes to calculate partial least squares and ridge regression
> _
Can you provide a small reproducible example and explain what exactly is going
wrong?
Just a handful of data points will do.
B.
> On Apr 18, 2017, at 2:16 PM, Santiago Bueno wrote:
>
> Thanks Boris, the following is an extract of my data. I have developed
> biomass models using codes like:
What about the advice I sent you this morning?
B.
> On Apr 19, 2017, at 1:49 PM, Boris Steipe wrote:
>
> No attachments please. Use the dput() function. And keep the correspondence
> on the list, there are many colleagues who are far more knowledgeable than I
> am.
>
This can be easily done in base R.
The solution below is pedestrian, transparent and explicit. Thus it's easy to
debug and validate(!) each step.
Prepare:
(1) Save your Excel spreadsheet as a text file with tab-separated values.
(2) Initialize a 128 * 128 matrix to hold your results. It should ha
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
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.
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(l
;> 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) {
>&
Your mental model of what R is and does appears to be misaligned with
actuality. Quoting from Wikipedia:
R is an open source programming language and software environment
for statistical computing and graphics [...]. The R language is
widely used among statisticians and data miners for d
r a period (.). If
> the delimiter is a period and there are hard returns (example 2), then I
> expect that will be easy enough to differentiate sentences ending with a
> number from enumerated items. However, I imagine it would be much more
> difficult to differentiate the two for
Given that, you can be certain that your file contains more than one or two
numbers, thus your way of reading it must be wrong.
B.
> On Apr 29, 2017, at 11:34 AM, Ragia . wrote:
>
>
> files size is 7682 kb
>
> sorry for multiple emails
>
> Ragia
>
>
>
>
> ___
There's a lot that doesn't make sense here. I think what you need to do is
produce a small, reproducible example, post that with dput() and state your
question more clearly - including what you have tried and what didn't work.
You'll probably be amazed how quickly you will get good advice if
_y
Did you try using the table() function, possibly in combination with sort() or
rank()?
Consider:
myNouns <- c("proper", "nouns", "domain", "ontology", "dictionary",
"dictionary", "corpus", "patent", "files", "proper", "nouns",
"word", "frequency", "file", "preprocess",
polish head", "polish system", "polym pad", "polyurethan
> pad", "porous", "process paramet", "process path", "process time",
> "recoveri", "rotat speed", "rough", "scatter", &
myDf1 <- data.frame(drugs = c("Ibuprofen", "Simvastatin", "Losartan"),
indications = c("pain", "hyperlipidemia", "hypertension"),
stringsAsFactors = FALSE)
myDf2 <- data.frame(drugs = c("Simvastatin", "Losartan", "Ibuprofen",
"Metformin"),
Pedestrian code, so you can analyze this easily. However entirely untested
since I have no ambitionto recreate your input data as a data frame. This code
assumes:
- your data _is_ a data frame
- the desired column is called food.price, not "food price" (cf. ?make.names )
# define a function th
t;
> On Tuesday, May 9, 2017 4:20 PM, Boris Steipe
> wrote:
>
>
> Pedestrian code, so you can analyze this easily. However entirely untested
> since I have no ambitionto recreate your input data as a data frame. This
> code assumes:
> - your data _is_ a data frame
&
ad
> on another site.
>
> Cheers,
> Bert
>
>
>
> On May 9, 2017 3:36 PM, "Boris Steipe" wrote:
> Great.
>
> I am CC'ing the list - this is important so that others who may come across
> this thread in the archives know that this question has bee
If I understand you correctly what you are really asking is how to embed quotes
in a string so that it can be parse()'d as an expression. The answer would be:
escape the quotes.
R > myOptions <- "Hello"
R > eval(parse(text = paste( "print(", myOptions, ")" )))
Error in print(Hello) : object 'H
Does:
rainbow(3)[1]
rainbow(3)[2]
rainbow(3)[3]
... solve your issue?
B.
> On Jun 8, 2017, at 8:20 AM, WRAY NICHOLAS wrote:
>
> Hi R folk I have a distance time graph for a locomotive and at various times
> different events occur on board the loco. I want to put a vertical line on
Run it through a loop. I assume the cell contents is NA (Not Available). Test
for it with is.na(). Whenever that returns TRUE, replace the NA value with the
value from the previous row.
Cheers,
B.
> On Jun 24, 2017, at 1:49 PM, Christophe Elek
> wrote:
>
> Hello Total newbie here... I hope
I don't think OP asked an unreasonable question at all.
Civility!
> On Jun 27, 2017, at 2:00 PM, Suzen, Mehmet wrote:
>
> Why don't you implement and uplad the package to CRAN?
>
> On 27 Jun 2017 17:45, "Chris Buddenhagen" wrote:
>
> Does anyone know of some code, and examples that implem
In principle what you need to do is the following:
- break down the time you wish to simulate into intervals.
- for each interval, and each failure mode, determine the probability of an
event.
Determining the probability is the fun part, where you make your domain
knowledge explicit and i
t packages exist because others just
> like them contributed something count as being uncivil? Terse, perhaps, since
> it bypassed the obvious suggestion to use a search engine, but not rude.
> --
> Sent from my phone. Please excuse my brevity.
>
> On June 28, 2017 5:08:16
I'd do it this way ... let me know if you need explanations.
minSize <- 15
maxSize <- 100
minSample <- 0.1
maxSample <- 0.8
# setup dataframe with totals, and cases as fractions
myStudies <- data.frame(study = 1:Nstudies,
cases = runif(Nstudies,
Have a look at the functions available in the igraph package.
B.
> On Jul 13, 2017, at 11:08 PM, SEB140004 Student
> wrote:
>
> Greeting.
>
> Dear Mr/Mrs/Miss,
>
> I want to create a network by using R but I only have a table that contain
> OTU ID and the abundance value of two sample
You need a stemming algorithm. See here:
https://cran.r-project.org/web/views/NaturalLanguageProcessing.html
Myself, I've had good experience with Rstem.
B.
> On Jul 31, 2017, at 4:47 PM, Riaan Van Der Walt
> wrote:
>
> I am new to R.
> Busy with Text Analysis.
>
> Need a script to fin
gt; Tel / Phone / Mogala : 27+72+2172429
> Email / Epos / Emeile: riaan.vanderw...@nwu.ac.za
> Url: http://www.nwu.ac.za/
>
> >>> Boris Steipe 31 Jul 2017 23:37 >>>
> You need a stemming algorithm. See here:
> https://cran.r-project.org/web/views/NaturalLanguag
There are many ways to get the output you want, so your question is ill
defined. But it is easy to see where your code goes wrong. And it should be
easy for you to fix it.
If you subset a vector with the '[' operator, this is like putting a vector of
indices "into" the square brackets. So, to d
> a <- c("<0.1", NA, 0.3, 5, "Nil")
> a
[1] "<0.1" NA "0.3" "5""Nil"
> b <- as.numeric(a)
Warning message:
NAs introduced by coercion
> b
[1] NA NA 0.3 5.0 NA
> b[! is.na(b)]
[1] 0.3 5.0
B.
> On Sep 22, 2017, at 11:48 AM, Shane Carey wrote:
>
> Hi,
>
> How do I extract just n
ns
>
> a <- c("<0.1", NA, 0.3, 5, "Nil")
> b <- c("<0.1", 1, 0.3, 5, "Nil")
>
> And I just want to remove the rows from the dataframe where there were NAs in
> the b column, what is the syntax for doing that?
>
> Thanks in adva
; to remove 0's rather than NAs , what would it be?
>
> Thanks
>
> On Mon, Sep 25, 2017 at 12:41 PM, Boris Steipe
> wrote:
> myDF <- data.frame(a = c("<0.1", NA, 0.3, 5, "Nil"),
>b = c("<0.1", 1, 0.3, 5, "
Is this a fixed width format?
If so, read.fwf() in base, or read_fwf() in the readr package will solve the
problem. You may need to trim trailing spaces though.
B.
> On Oct 5, 2017, at 10:12 AM, jean-philippe
> wrote:
>
> dear R-users,
>
>
> I am facing a quite regular and basic problem
Since you have an authoritative description of the format, by all means use
that - not a guess based on a visual inspection of where data appears in a
sample row.
B.
> On Oct 5, 2017, at 11:02 AM, jean-philippe
> wrote:
>
> dear Jim,
>
> Thanks for your reply and your proposition.
>
>
Just for the record - and posterity: this is the Wrong way to go about defining
a fixed width format and the strategy has a significant probability of
corrupting data in ways that are hard to spot and hard to debug. If you _have_
the specification, then _use_ the specification.
Consider what yo
This article may be helpful, at least to get you started:
https://www.r-bloggers.com/ordinal-data/
Cheers,
Boris
> On Oct 5, 2017, at 3:35 PM, Bert Gunter wrote:
>
> I would consider this is a question for a statistics forum such as
> stats.stackexchange.com, not R-help, which is about R pr
It's generally a very good idea to examine the structure of data after you have
read it in. str(data2) would have shown you that read.csv() turned your strings
into factors, and that's why the == operator no longer does what you think it
does.
use ...
data_2 <- read.csv("excel_data.csv", strin
This is not a regular expression but simply a conjunction (sequence of '&') of
logical expressions. Moreover it's not wrong. Consider:
xyz <- data.frame(municipio = c('Limeira'), mesincident = c('marco'),
trechoklmetros = c(3.00, -4.00, 30))
xyz
municipio mesincident trechoklmetros
1 Limeir
I can reproduce your problem. It affects both sine-waves if the graph is
plotted as the second plot, so it seems that plotting the graph affects all
subsequent plots. It affects all plots in subsequent plots to the same window.
The window needs to be closed to correct this. I would file a bug re
will reveice all strings to filter in a dataset by
> subset below:
>
>
> a loop each line:
>
> subset(dataset_read, desired_fomart[i])
>
> My question: taking into consideration this cenario, in your oppinion your
> reply is my suitable for my problem, whereas i wil
You haven't provided nearly enough information for us to be able to help.
Please see here for some hints on how to ask questions productively:
http://adv-r.had.co.nz/Reproducibility.html
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
... and please read the po
> 1 & xyz$b == 4)
xyz[eval(expp), ] # evaluate an expression in place
a b
2 2 4
... but really, it's just a guess at what you might be trying to do.
B.
On Apr 18, 2015, at 1:30 PM, Boris Steipe wrote:
> Sorry - it's not entirely clear to me
That would be the "each" argument to rep()...
n <- 5
rep(1:n, each=4)
[1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5
Cheers,
B.
On Apr 19, 2015, at 9:44 AM, John Sorkin wrote:
> Windows 7 64-bit
> R 3.1.3
> RStudio 0.98.1103
>
>
> I am trying to generate a list of length 4n which consists of
Cross Validated at http://stats.stackexchange.com/
B.
On Apr 19, 2015, at 10:37 AM, CHIRIBOGA Xavier
wrote:
> Dear members,
>
>
>
> Since this is not a Forum for Stats questions...does anyone can recomend me a
> good forum to post questions about statistics?
>
>
>
> Thank you for info
That does not sound like a clustering problem at all since you already know the
desired characteristics and are not trying to discover structure in your data.
Simply define a score as a suitably weighted sum of individual features, order
your passengers by that score, and pick the top few, or an
e the passengers from one level to another
> level. But I don't know what method exactly need to follow .
>
> -Original Message-
> From: Boris Steipe [mailto:boris.ste...@utoronto.ca]
> Sent: Wednesday, April 22, 2015 9:08 PM
> To: Lalitha Kristipati
> Cc: R-
Have a look at the help page for the function ts()
(type ?ts at the R prompt). Other than that, you haven't provided nearly
enough information for us to diagnose your issue.
Please see here for some hints on how to ask questions productively:
http://adv-r.had.co.nz/Reproducibility.html
http://
If you google for "invalid 'description' argument" it should be pretty clear
what's going on - most likely your DirPath_Matrix is not a single item.
B.
On Apr 28, 2015, at 7:02 PM, catherine peters
wrote:
> Hi thank you in advance for your assistance.
>
> I am using the following section o
A: Look at the function seq() e.g. seq(0.01, 1, by=0.005)
B: Don't post in HTML. Seriously. Do not.
C: You need to invest more time getting the basics down. Google for R
introduction.
D: Ow. pi is a predefined constant. Your re-defintion loses accuracy. (See C:)
Please see here for some hints
For crying out loud Dawood - can you please stop posting screenshots? Do you
not know how to copy your output and paste it into the body of a (not HTML
formatted!) email? Let us know if you need help with that.
B.
On May 5, 2015, at 11:05 AM, dawood ahmad wrote:
> Dear Daniel,
>
> Thank yo
Wrong.
B.
On May 5, 2015, at 2:54 PM, lejeczek wrote:
> hi eveybody
>
> I'm trying something simple (Biocunductor packages), so simple I believe it's
> example from docs but I get segfault.
> I don't suppose incorrect scripting can cause segfault, right?
>
> regards
>
> _
Here's a discussion on stack overflow with more hints ...
http://stackoverflow.com/questions/19226816/how-can-i-view-the-source-code-for-a-function
... and a link to an R-News article by Uwe Ligges on the topic
http://cran.r-project.org/doc/Rnews/Rnews_2006-4.pdf (p. 43 f)
Cheers,
Boris
On Ma
How about %in% ?
# preparing something that looks like I think your data looks like:
ap <- c("aajkss", "dfghjk", "sdfghk", "xxcvvn")
af <- matrix(1:10, nrow=2)
colnames(af) <- c("aajkss", "b", "c", "dfghjk", "e")
# doing what I think you need done:
ap[ap %in% colnames(af)]
Cheers,
B.
(PS. a r
I think the "record number" i.e. the indices of the elements were asked for.
That would be:
which(5 <= allrecords & allrecords <= 9)
Cheers,
B.
On May 10, 2015, at 7:33 AM, Rui Barradas wrote:
> Hello,
>
> You should learn about indexing in R. Read the pdf R-intro.pdf that comes
>
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