, to have an idea of the data I would rather further limit
the number of matrix lines printed with
head(object)
tail(object)
Hope this helps,
Rui Barradas
On Mon, Jul 8, 2019 at 10:16 AM Kevin Thorpe
wrote:
On Jul 8, 2019, at 10:06 AM, Spencer Brackett <
spbracket...@saintjosephhs.
of 173 variables:
There are 173 variables eac h with the same number of observations, 20530.
When you read in data files with any form of read.table (read.csv is the
same function with different arguments default values set), you will
always get a data.frame.
Hope this helps,
Rui Barra
untry'.
See in base R
?aggregate
?tapply
?ave
?by
And there's also contributed packages, such as dplyr and data.table.
If you can repost your data, setting your e-mail client to plain text,
we will be able to say more.
Hope this helps,
Rui Barradas
Às 20:33 de 20/07/19, Faradj Kolie
}
A[nrow(A), 1:remainder] <- x[(len*passes + 1):n]
A
}
print0(rnorm(23), 10)
print0(rnorm(23), 10, fill = 0)
Hope this helps,
Rui Barradas
Às 21:34 de 20/07/19, Steven escreveu:
Dear All:
Below is what I meant. Procedure print0 allows me to print a vector of
length 53 in four rows of 10 plu
en*m + 1):n], rep(fill, len*(m + 1) - n)))
}
A
}
Hope this helps,
Rui Barradas
Às 07:47 de 22/07/19, Rui Barradas escreveu:
Hello,
Maybe something like the following is what you want.
I have added an extra argument 'fill' to allow to choose what to print
in the end. It's defa
m <- n %/% len
remainder <- n %% len
A <- matrix(x[seq_len(len*m)], ncol = len)
if(remainder > 0){
A <- rbind(A, c(x[(len*m + 1):n], rep(fill, len*(m + 1) - n)))
}
print(A, na.print = "")
}
print0c(rnorm(23), 10)
print0c(2^(1:23), 10)
Hope this helps,
Rui Barr
ps, reserve memory
beforehand.
wc <- vector("list", length = length(mc_list))
tmp <- vector("list", length = length(tmp_list))
are much better than your
wc <- list()
tmp <- list()
Maybe I will find ways to save time with the really slow instructions.
Hope this
Hello,
Instead of read.table use
data.table::fread
It's an order of magnitude faster and all you have to do is to change
the function, all arguments are the same (in this case).
Hope this helps,
Rui Barradas
Às 20:18 de 24/07/19, Rui Barradas escreveu:
Hello,
This is far f
Hope this helps,
Rui Barradas
Às 18:41 de 27/07/19, Agustín Alonso Rodriguez escreveu:
Estimados miembros de la Lista:
Me dirijo a vosotros para que, si pod�is, ilumin�is mi confusi�n.
Estoy trabajando con los modelos neural netwok del libro Forecasting,
Principles and Practic
learn Monte Carlo simulation and go through the many
examples that can be found on-line. Then come back with *code* that
doesn't work, or that doesn't do what you expect it to do, or a similar
problem.
Hope this helps,
Rui Barradas
Thank you
[[alternative HTML versi
ion_dodge()) +
theme_minimal() +
scale_fill_manual(values = c("blue", "red")) +
scale_x_continuous(breaks = 1:12) +
scale_y_discrete(limits = c(0, 100))
Hope this helps,
Rui Barradas
Às 10:50 de 30/07/19, Knut Krueger via R-help escreveu:
Hello to all,
how can I can set both y a
4 4 -1.3 -1.13 -1.78 -0.05 -0.23 -0.23
5 5 -1.6 -1.3 -1.34 -0.23 -0.45 -0.45
", header = TRUE)
Hope this helps,
Rui Barradas
Às 03:25 de 06/08/19, reichm...@sbcglobal.net escreveu:
R Help Forum
I have output from hierarchal clustering and want to plo
pch = 20,
col = rgb(0, 0, 0, 0.5),
jitter = 0.001)
Hope this helps,
Rui Barradas
Às 18:49 de 08/08/19, Ana Marija escreveu:
Hello,
I made plot in attach using:
boxplot(flcn_M~subject,data=dx,col =
c("royalblue1","palevioletred1"),xlab="subjec
_j[[k]]$abs.error
}
}
}
Hope this helps,
Rui Barradas
Às 21:14 de 10/08/19, ravi via R-help escreveu:
Bert,Thanks a lot for your help. I have a few follow-up questions (shown in
the comment lines). Numerical values and error for a (i) vector and (ii) array.
a <- seq(from=0,to=1,
= factor(fill) and the graph is exactly the same.
Which leads me to propose an easier way of creating the vector 'fill',
in the first place.
fill <- as.integer((5 < x) & (x < 7.5))
Will give the same result, coerce to factor if you like.
Hope this helps,
Rui Barradas
Às
Hello,
This seems to be the site you want, not CRAN.
http://gecon.r-forge.r-project.org/
And please no HTML, post in plain text.
Hope this helps,
Rui Barradas
Às 02:01 de 11/08/19, tanyi_cm2000--- via R-help escreveu:
Hi,my name is; William, a graduate student in the Department of
esides, if
this goes into a function all temporary variables will be gone and the
memory released, in which case there will be no problem.
(The with equivalent is i <- with(temp, order(patid, time)), btw.)
Hope this helps,
Rui Barradas
Às 03:20 de 13/08/19, Sorkin, John escreveu:
I want to
t;- c(1, rep(0, n - 1))
f <- rep(f, length.out = length(x))
f <- cumsum(f)
}
split(x, f)
}
breakVec(LETTERS)
breakVec(LETTERS, 4)
breakVec(LETTERS, 1)
breakVec(LETTERS, -1)
Hope this helps,
Rui Barradas
Às 19:01 de 18/08/19, Christofer Bogaso escreveu:
Hi,
Let say I have a ve
Hello,
Bert's modular arithmetic way simplifies my code a lot.
breakVec <- function(x, n = 5){
if(n < 1) stop(paste("Illegal value n:", n))
f <- if(n == 1) "" else (seq_along(x) - 1) %/% n
split(x, f)
}
Hope this helps,
Rui Barradas
Às 20:08 de 18/08/
)
s2E <- s2a$Estimate
Is s1as1E a typo? The way you define and use s2E suggests that it should be
s1E <- s1a$Estimate
Hope this helps,
Rui Barradas
Time <- c(1,2)
SBP <- c(s1E,s2E)
plot(Time,SBP)
title("Predicted Sys SBP, age 5.99 yr (SS 524 obs with BP 190mmHg removed)"
fill = col), show.legend = FALSE) +
scale_fill_manual(values = levels(col)) +
facet_wrap(~ Country, ncol = 3) +
theme(axis.text.x = element_text(angle = 50, hjust = 1))
Hope this helps,
Rui Barradas
Às 04:21 de 25/08/19, Eric Berger escreveu:
This seems to work
ggplot(t1) +
geom_col(ae
(x = as.factor(DATE),
y = abundance, colour = site)) +
geom_line(aes(group = year)) +
theme_bw()
Ab + facet_grid(year ~ .) +
theme(axis.text.x = element_text(angle = 90,
hjust = 1,
size = 4))
Hope this helps
he sense of
function identical(), meaning they are *exactly* the same. Why shouldn't
they?
Hope this helps,
Rui Barradas
Às 17:06 de 05/09/19, Boo G. escreveu:
Hello,
I am trying to perform a Kolmogorov–Smirnov test to assess the difference
between a distribution and samples drawn prop
th(d_2))
#kolmogorov-smirnov distance
x <- f_d_1(d_2)
y <- f_d_2(d_2)
ht <- ks.test(x, y)
d_stat[i, 2] <- ht$statistic
d_stat[i, 3] <- ht$p.value
}
Hope this helps,
Rui Barradas
Às 19:29 de 05/09/19, Rui Barradas escreveu:
Hello,
I don't have the algorithms
ser to a uniform with
increasing sizes. Try
hist(d_stat[801:1000, 3])
[1]https://www.jstatsoft.org/article/view/v008i18
Hope this helps,
Rui Barradas
2) Also in this case the p-values don’t make much sense, according to my
previous interpretation.
Again, I could be wrong in my inter
ered a chi-squared GOF test?
Hope this helps,
Rui Barradas
Às 20:51 de 05/09/19, Boo G. escreveu:
Hello and thanks for your patience.
As far as I understand, the paper of Marsiglia and colleagues refers to CDF
samples (i.e. from a hypothetical distribution — e.g. a Poisson), while I have
an ECDF s
Hello,
There is no attached file, R-help is doesn't allow many file types.
Try reposting with the extension .txt or post the output of
dput(head(dataset, 20))
directly in your e-mail.
Hope this helps,
Rui Barradas
Às 10:58 de 07/09/19, farshad goudarzian escreveu:
Hello.
Here I att
14, 20, 8 ", stringsAsFactors = FALSE)
dat2 <- read.csv(text="
ID, weight
A, 0.25
B, 0.42
C, 0.65 ", stringsAsFactors = FALSE)
Simpler, no?
Hope this helps,
Rui Barradas
Às 20:23 de 07/09/19, Val escreveu:
Hi all
Correction for my previous posting.
dat2 should be read
Hello,
Bert:
Quoting Jeff, "Val has been posting to this list for almost a decade". I
didn't know about the "decade" part (didn't look it up) but that's why I
answered the question.
Val:
It's not that hard to run the code you post *before* posting i
emed
more appropriate (?).
s <- strsplit(as.character(daT1), "[[:blank:]]+")
x <- lapply(s, function(.s) .s[!grepl("[[:alpha:]]+", .s)])
daT3 <- as.data.frame(do.call(cbind, x))
daT3
Hope this helps,
Rui Barradas
Às 12:54 de 09/09/19, Marna Wagley escreveu:
Hi R User,
- nms
daT3
Hope this helps,
Rui Barradas
Às 16:28 de 09/09/19, Marna Wagley escreveu:
Thank you Rui. it helped a lot.
Sincerely,
MW
On Mon, Sep 9, 2019 at 6:47 AM Rui Barradas <mailto:ruipbarra...@sapo.pt>> wrote:
Hello,
You could try a character class instead of one
17 -2.25359807972754
1998-05-21 -2.55799006865995
1999-08-22 -2.25114162617707
1999-08-25 -1.47905397376409
1999-09-05 -0.641589808755325
1999-09-09 -0.648954682695949
1999-09-13 -0.726364489272492
1999-09-16 -1.28445236942011
")
Hope this helps,
Rui Barradas
Às 20:14 de 09/09/19, Ogbos Okike
Hello,
Please include data, say
dput(head(data, 20)) # post the output of this
But, is the problem as simple as
rowMeans(data[2:3], na.rm = TRUE)
?
Hope this helps,
Rui Barradas
Às 15:53 de 12/09/19, Subhamitra Patra escreveu:
Dear R-users,
I have daily data from 03-01-1994 to 29-12
lt;- merge(aggA, aggB)
head(MonthReturns)
Final clean up.
rm(date, month, aggA, aggB)
Hope this helps,
Rui Barradas
-- Bert
On Thu, Sep 12, 2019 at 8:41 AM Rui Barradas <mailto:ruipbarra...@sapo.pt>> wrote:
Hello,
Please include data, say
dput(head(data, 20)) # post
Not tested, since there is no data.)
Hope this helps,
Rui Barradas
Às 17:27 de 19/09/19, Phillip Heinrich escreveu:
Attached is every at bat for the Arizona Diamondback’s first three games of
2018 – BBdata1.rda. I added the Date and DHCode variables by parsing the first
variable labeled Gam
#7 2018-03-30 2
#8 2018-03-30 2
#9 2018-04-01 3
#10 2018-04-01 3
#11 2018-04-01 3
#12 2018-04-01 3
Hope this helps,
Rui Barradas
Às 22:09 de 19/09/19, Rui Barradas escreveu:
Hello,
There was no attachment, R-Help allows only a limited number of file
t
Hello,
Maybe I am not understanding but isn't this what you have asked in your
previous question and my 2nd post (adapted) does?
If not, where does it fail?
Hope this helps,
Rui Barradas
Às 18:46 de 20/09/19, Phillip Heinrich escreveu:
With the data snippet below I’m trying to incr
Hello,
Something like this?
ctab <- function(data) {
gmodels::CrossTable(as.matrix(data), prop.chisq = FALSE, prop.c =
FALSE, prop.t = FALSE, format = "SPSS")
}
mtcars %>% select(cyl, gear) %>% ctab()
Hope this helps,
Rui Barradas
Às 16:30 de 20/09/19, Zachary Lim
Hello,
Is this what you are looking for?
ci95 <- apply(my.data, 2, quantile, probs = c(0.025, 0.975))
Hope this helps,
Rui Barradas
Às 20:42 de 23/09/19, varin sacha via R-help escreveu:
Dear R-Experts,
Here is my reproducible R code to get the Mean squared error of GAM and MARS
afte
bootResults <- boot(data=data, statistic=mse, R=1000)
boot.ci(bootResults, type = boot.ci.type)
Hope this helps,
Rui Barradas
Às 13:43 de 25/09/19, varin sacha via R-help escreveu:
Dear R-experts,
Below the reproducible example. I have tried to write a function that returns the
statistic of
ew environment and load the data
there might be sound advice. Check if your data set exists in the
.GlobalEnv first, then load() it.
ls(pattern = "^ari", envir = .GlobalEnv)
Hope this helps,
Rui Barradas
Às 18:06 de 26/09/19, Duncan Murdoch escreveu:
On 26/09/2019 12:55 p.m., Phill
23 1
Hope this helps,
Rui Barradas
Às 18:05 de 27/09/19, David J. Birke escreveu:
Dear R community,
I just stumbled upon the following behavior in R version 3.6.0:
set.seed(42)
y <- rep(0, 30)
x <- rbinom(30, 1, prob = 0.91)
# The following will not show any
pply(phdf[-6], 1, function(x){
1*all(x == 0) + # rule 1
2*(x[1] == 1 & x[2] == 1) + # rule 2
3*(x[1] == 0 & x[2] == 1 & x[3] == 2) # rule 3
})
phdf$code <- ifelse(i == 0, phdf$code, new_values[i])
Hope this helps,
Rui Barradas
Às 04:24 de 02
Hello,
You have to use is.na to get the NA values.
t1 <- data.frame(sex_chromosome_aneuploidy_f22019_0_0 = c(NA, "Yes"),
other = 1:2)
i <- t1$sex_chromosome_aneuploidy_f22019_0_0 == "Yes" &
!is.na(t1$sex_chromosome_aneuploidy_f22019_0_0)
i
t
Hello,
Then it's easier, is.na alone will do it.
j <- is.na(t1$sex_chromosome_aneuploidy_f22019_0_0)
t1[j, ]
Hope this helps,
Rui Barradas
Às 20:29 de 03/10/19, Ana Marija escreveu:
Hi Rui,
sorry for confusion, I would only need to extract from my t1 dataframe
rows which ha
a(.)
!i means (.) != "Yes"
Hope this helps,
Rui Barradas
Às 21:21 de 03/10/19, Rui Barradas escreveu:
Hello,
Then it's easier, is.na alone will do it.
j <- is.na(t1$sex_chromosome_aneuploidy_f22019_0_0)
t1[j, ]
Hope this helps,
Rui Barradas
Às 20:29 de 03/10/19, Ana Marija
estions on R code, not to
search (R related?) stuff for you.
3. At an R prompt run
install.packages('sos')
sos::findFn('pvcm file')
It yields nothing of relevance.
Hope this helps,
Rui Barradas
Às 12:00 de 04/10/19, Jui-Kun Chiang via R-help escreveu:
Dear Sir,
Hello,
Try the following
cols <- sapply(tot, function(x) any(grepl("^E94", x)))
To have the column numbers,
which(cols)
Hope this helps,
Rui Barradas
Às 19:50 de 05/10/19, Ana Marija escreveu:
Hello,
I have a data frame tot which has many columns and many rows.
I am tryin
, x))
e10 <- rowSums(e10) > 0
e11 <- sapply(tot, function(x) grepl("^E11", x))
e11 <- rowSums(e11) > 0
tot$newcol <- -9
tot$newcol[e10] <- 1
tot$newcol[e11] <- 2
On both cases the 2 lines sapply/rowSums can be made one with
rowSums(sapply(...)) > 0
Hope this he
tmp), text = unlist(strsplit(tmp, ",")))
})
n <- max(sapply(tmp, '[[', 'n'))
tmp <- lapply(tmp, function(x) as.numeric(x$text))
tmp[[1]] <- rep(tmp[[1]], each = n)
res <- do.call(cbind.data.frame, tmp)
names(res) <- lns[i_title]
res
If you have hundreds of fil
t;- cbind.data.frame(first, res)
names(res) <- sub("\\[.*\\]$", "", lns[i_title])
res
}
df1 <- readCSVFile("strange.csv")
If this function doesn't do it, please try to make an effort on your
own, R-Help is not a code writing service, it's a mail li
Hello,
Here are 3 ways.
The first are almost the same, they use base graphics.
x <- 1:6
y <- c(73,53,42,67,41,50)
barplot(setNames(y, x))
Or
names(y) <- x
barplot(y)
And 3:
library(ggplot2)
ggplot(data.frame(x, y), aes(x, y)) +
geom_col()
Hope this helps,
Rui Barradas
Às
UE)[nvec]
x[i[order(i)]]
}else{
sort(x, decreasing = TRUE)[nvec]
}
}
barplot(TopN(CONTBR_RESULT, 15))
barplot(TopN(CONTBR_RESULT, 15, TRUE))
Hope this helps,
Rui Barradas
Jim
On Wed, Oct 16, 2019 at 11:21 AM wrote:
r-help forum
I have a database that I have performed a "
na.strings = "8888")
})
Hope this helps,
Rui Barradas
Às 02:38 de 18/10/19, ani jaya escreveu:
Dear R-Help,
I have a list of data frame that I import from excel file using read.xlsx
command.
sheets <- openxlsx::getSheetNames("rainfall.xlsx")
test <- lapply(sheets,functio
a2") <- "second_a"
str(X)
ok <- c(1, 3, 4)
X <- X[ok, ]
str(X)
Hope this helps,
Rui Barradas
Às 03:13 de 20/10/19, Rolf Turner escreveu:
On 20/10/19 3:00 PM, Richard M. Heiberger wrote:
Look at
methods(as.data.frame)
Define your specialized columns to have a newly defi
df1$DATE %/% 100
aggregate(Spread ~ PERMNO + MONTH, df1, mean)
Hope this helps,
Rui Barradas
Às 17:17 de 20/10/19, jim holtman escreveu:
Does this do what you want:
library(tidyverse)
input <- read_delim("PERMNO DATE Spread
+ 111 19940103 0.025464308
+ 111 19940104 0.064424296
+
{
attributes(x[[nm]]) <- Sat[[nm]]
x[[nm]]}, x = x, Sat = SaveAt)
names(lX) <- names(x)
x <- as.data.frame(lX)
x
}
The (frequent) error comes from tests where a X was created in the
globalenv and found by the method.
Rui Barradas
Às 22:55 de 20/10/19, Rolf Turner escre
t the right place to ask, R-Help is about R
*code* not statistics. For questions about statistics see Cross
Validated [1]
[1] https://stats.stackexchange.com
Hope this helps,
Rui Barradas
Às 10:21 de 21/10/19, javed khan escreveu:
I have two samples A and B in excel and I want to use the vargh
Hello,
Just do col = rev(heat.colors(256)) ?
Hope this helps,
Rui Barradas
Às 12:55 de 21/10/19, Eleftheria Dalmaris escreveu:
Hello all,
I have made the attached heat map without a problem.
On this heat map, the yellows are my high values and the oranges are my low
values. Is there a way
asign <- data.table::fread(asign_file)
# extract the relevant rows and merge
res <- l4[i1, ]
res2 <- asign[i2, setdiff(names(asign), names(l4))]
merge(res, res2, by.x = c("X1", "X2"), by.y = c("chr", "pos"))
Hope this helps,
Rui Barradas
Hello,
That's a floating-point issue.
See FAQ 7.31.
See also [1], [2] and the links therein.
[1]
https://stackoverflow.com/questions/9508518/why-are-these-numbers-not-equal
[2] https://stackoverflow.com/questions/588004/is-floating-point-math-broken
Hope this helps,
Rui Barradas
Às
Hello,
Is this as simple as
if ( (!is.na(dts[1,j-1])) & (!is.na(dts[3,j-1])) ) {
(change the logical operator from '|' to '&')?
The result vseries1 still has some NA's at the end of some of its series.
Hope this helps,
Rui Barradas
Às 02:05 de 30/10/19, p
Hello,
Another option that prevents
length(x) == 1
is to use NROW(x) instead of length(x).
Hope this helps,
Rui Barradas
Às 16:58 de 30/10/19, William Dunlap escreveu:
Your EXTEND() function appears to expect that its 'x' argument will be a
numeric vector, but you pass it a
the same for all vector elements, if not
y <- c('11/7/2016', '14-07-16', '2016/7/11')
lubridate::dmy(y)
#[1] "2016-07-11" "2016-07-14" NA
#Warning message:
# 1 failed to parse.
Hope this helps,
Rui Barradas
Às 02:25 de 03/11/19, reichm...@sbcglo
Hello,
If performance is important, and with 73M rows it probably is, take a
look at this StackOverflow post.
[1] https://stackoverflow.com/a/36058634/8245406
Hope this helps,
Rui Barradas
Às 21:33 de 08/11/19, Martin Morgan escreveu:
With this example
df = data.frame(a = c(1, 1, 2, 2
(dirname(tmpfile))
system2('java', '-version', '>', tmpfile)
readLines(tmpfile)
unlink(tmpfile) # final cleanup
Hope this helps,
Rui Barradas
Às 16:51 de 09/11/19, Dennis Fisher escreveu:
R 3.6.3
OSX and Windows
Colleagues
I want to identify if Java is in
Hello,
This error can be a 32 bits vs 64 bits thing. R and java must have the
same architecture. If you have 64b Java installed, don't run 32b R. And
vice-versa.
Hope this helps,
Rui Barradas
Às 18:53 de 09/11/19, Dennis Fisher escreveu:
Dirk
I am now more confused (and I appreciate
n1, function(DF){
format(DF[["Tanggal"]][1], "%Y-%m")
})
pon1 <- pon1[order(inx_ym)]
# get the minimum and maximum of every "RR"
min.RR <- sapply(pon1, function(DF) min(DF[["RR"]], na.rm = TRUE))
max.RR <- sapply(pon1, function(DF) max(DF[["RR&qu
Hello,
Try which.max?
Hope this helps,
Rui Barradas
Às 14:10 de 13/11/19, ani jaya escreveu:
Thank you very much Mr. Rui, but for delete the duplicated row I use:
...
library(tidyverse)
alldata<-data.frame(Reduce(rbind, pon1))
c<-(which(duplicated(alldata$Tanggal))) #duplicate
a
bels = scales::percent_format())
Hope this helps,
Rui Barradas
Às 23:56 de 15/11/19, Josh B escreveu:
Hello,
I am trying to include the count labels on stacked bar plots which
represent percentages. I want to show x-amount of individuals make up the
graphed percentages. However, when I include
ncol = ncols)+
ylab("Success fraction") +
theme_bw()
print(ggplot(X) +
geom_point(aes(y = Sfrac , x = x)) +
facet_grid_paginate(facets = Species ~ LifeStage:degC,
page = page,
nrow = nrows, ncol = ncols)+
ylab("Succ
w()
)
}
dev.off()
print(plotObj[[1]])
print(plotObj[[2]])
Hope this helps,
Rui Barradas
Às 13:07 de 17/11/19, Rui Barradas escreveu:
Hello,
I don't know why but use an explicit nrow = nrows with 'nrows' defined as
nrows <- length(levels(X$Species))
It seems to work (rem
Hello,
See the docs: get_amp returns a ggmap object and ggmap returns a ggplot
object. So ggsave() should do what you want.
Hope this helps,
Rui Barradas
Às 09:48 de 17/11/19, Kristi Glover escreveu:
Hi R users,
I am struggling to save the map with georeferenced (TIFF) that was imported
Hello,
It's dificult to tell without data. Can you post the output of
dput(head(tagSummary, 20)) # or 30
?
(If it's private data, something mimicking its structure.)
Rui Barradas
Às 02:06 de 18/11/19, Josh B escreveu:
Hello Rui,
I worked through your suggestion and appear to be
Hello,
There is
system.time {base}
CPU Time Used
Description
Return CPU (and other) times that expr used.
There are also packages microbenchmark or bench.
Hope this helps,
Rui Barradas
Às 15:44 de 20/11/19, Jeff Reichman escreveu:
R- Help
Is there a command or way to obtain the
;n",
xlab="", ylab="")
axis(side=4, at=c(-5,-1,0), labels=c("98%","100%","102%"))
mtext("blue line", side=4, line=2.5, at=0)
par(op)
Hope this helps,
Rui Barradas
Às 02:53 de 27/11/19, Ogbos Okike escreveu:
Dear Contribut
n't work",values=c(16,3),
labels=c("clyde","irving")) +
facet_grid(cols=vars(grp))
[1]
https://stackoverflow.com/questions/2185252/reshaping-data-frame-from-wide-to-long-format
Hope this helps,
Rui Barradas
Às 00:04 de 01/12/19, Rolf Turner
:n){
print(g + facet_wrap_paginate(~Trt, ncol = 4, nrow = 3, page = i))
}
print(g)
Hope this helps,
Rui Barradas
[1] https://stackoverflow.com/a/58373858/8245406
Às 11:46 de 01/12/19, Rolf Turner escreveu:
I am trying to produce a ggplot2 graphic in which there is a single
conditio
e","irving")) +
scale_shape_manual("Doesn't work",values=c(16,3),
labels=c("clyde","irving")) +
facet_grid(cols=vars(grp))
Hope this helps,
Rui Barradas
Às 01:07 de 02/12/19, Rolf Turner escreveu:
On 2/12/19 3:03 am, Rui Barradas w
more only with data.
Hope this helps,
Rui Barradas
Às 09:18 de 03/12/19, Francesca escreveu:
Dear Contributors,
I would like to ask help on how to create a plot that is the overlapping
of two other plots.
It is a geom_bar structure, where I want to count the occurrences of two
vari
0, 0, 0, 0, 0, 10, 10, 10, 30, 10, 0, 0, -10, 40, 0, 0,
10, 10, 40, 30, -10, 0, 0, 10, 20, 0, 0, 10, 40, 0, 0, -10,
-20)), row.names = c(NA, -236L), class = "data.frame")
Hope this helps,
Rui Barradas
Às 14:10 de 03/12/19, Francesca escreveu:
Hi
here it is;. THANKS!
dput(
a package, and in this case you
would even have more examples, if needed.
Hope this helps,
Rui Barradas
Às 19:07 de 04/12/19, Nestor Toledo escreveu:
Hello everyone, even I'm not fluent in coding, R has become a
fundamental part of my daily work as a researcher and I'm very much
m_density(aes(alpha = 0.2)) +
scale_alpha_continuous(guide = "none") +
facet_wrap( ~ participation)
Hope this helps,
Rui Barradas
Às 10:19 de 05/12/19, Francesca escreveu:
This is a consolation, because I cannot get it in ggplot either!
Thanks for the code!
F.
%>%
set_width(0.99) %>%
set_col_width(1:5, 0.99) %>%
set_number_format(2)
hx
quick_docx(hx, file = "test.docx")
There are ways of doing the same without pipes, those functions don't
have the prefix 'set_'. But I believe that what's important is function
?qui
")
hx
```
Thank you,
Ashim
On Thu, Dec 12, 2019 at 12:57 PM Ashim Kapoor <mailto:ashimkap...@gmail.com>> wrote:
On Wed, Dec 11, 2019 at 9:11 PM Rui Barradas mailto:ruipbarra...@sapo.pt>> wrote:
Hello,
This works for me:
library(dplyr)
//www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hello,
Are you looking for something like this?
DFA <- data.frame(id = 1:2)
DFA[["ar"]] <- array(1:8, c(2,2,2))
DFA$ar[1, , ]
#> [,1] [,2]
#> [1,]1 5
#> [2,
w up on top in the legend
# mutate(name = factor(name, levels = c("K", "J"))) %>%
ggplot(aes(Year, value, color = name)) +
geom_smooth(
formula = y ~ x,
method = lm,
se = FALSE
) +
geom_point() +
scale_color_manual(values =
emove or not NA's by adding a
na.rm argument:
my_fun <- function(wd1, ws1, na.rm = FALSE) {
[...]
mean_u <- mean(u_component, na.rm = na.rm)
mean_v <- mean(v_component, na.rm = na.rm)
[...]
}
aggregate(wd ~ day + month, data=df, FUN = my_fun, ws1 = ws, na.
p but like this it is not possible to do so.
Rui Barradas
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and pr
FALSE
)
Other ways to subset the data are
# dplyr
data = w %>% filter(year(date) %in% 2010:2015)
# base R
data = subset(w, year(date) %in% 2010:2015)
Hope this helps,
Rui Barradas
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and
[1] "Jan 2021" "Mar 2021" "Apr 2021"
Hope this helps,
Rui Barradas
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http:/
ich(data2$Layer == "Level 12")
data2$LU[i] <- "Park"
# equivalent one-liner
data2$LU[which(data2$Layer == "Level 12")] <- "Park"
Hope this helps,
Rui Barradas
__
R-help@r-project.org mailing list -
quot;)] <- "Agri";
...
...
...
.
Is there any other way to expand the code in order to consider all of the
levels simultaneously? Like the below code:
data2$LU[which(data2$Layer == c("Level 1","Level 2", "Level 3", ...))] <-
c("Park", "Agri&qu
Às 13:18 de 11/06/2023, Rui Barradas escreveu:
Às 22:54 de 11/06/2023, javad bayat escreveu:
Dear Rui;
Many thanks for your email. I used one of your codes,
"data2$LU[which(data2$Layer == "Level 12")] <- "Park"", and it works
correctly for me.
Actually I
ll be
used
data3 = data2[!grepl(c("_esmdes","_Des Section","0"), data2$Layer),]
Warning message:
In grepl(c("_esmdes", "_Des Section", "0"), data2$Layer) :
argument 'pattern' has length > 1 and only the first eleme
do? I expected the function to remove 3 rows of the
dataframe.
I do not know the reason.
On Mon, Jun 12, 2023 at 5:16 PM Rui Barradas wrote:
Às 23:13 de 12/06/2023, javad bayat escreveu:
Dear Rui;
Many thanks for the email. I tried your codes and found that the length
of
the "Values&qu
tor(Annee)))+
geom_bar(position="dodge",stat="identity")+
ylab("Simulation Progress (%)") +
facet_wrap(~Type,nrow=3)
myplot + theme(panel.grid.major = element_blank(),
legend.title=element_blank(), panel.grid.minor = element_blank()
=Conc)) +
geom_point(size=6) +
geom_line(aes(group = Conc)) +
scale_colour_manual(values = c("darkslategray3", "darkslategray4",
"deepskyblue4")) +
ggtitle("Working example") +
xlab(expression(bold("Time (h)"))) +
ylab(expression(
e commented, minimal, self-contained, reproducible code.
Hello,
You don't need so many calls to paste, one is enough.
And you don't need the for loop at all, paste and paste0 are vectorized.
myvalues <- matrix(nrow=2,ncol=4)
cnames <
ng <- c(string, name)
print(string)
}
#> [1] "xxx1"
#> [1] "xxx1" "xxx2"
# Now creation of xxx1 and xxx2 does work
string
#> [1] "xxx1" "xxx2"
# 2. create a vector of the appropriate length beforehand, my preferred
string <- chara
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