Try this
pdf("~/graph.pdf")
par(mar=c(8, 4, 4, 2))
barplot(d2, legend= c("SYCL", "CUDA"), beside=
TRUE,las=2,cex.axis=0.7,cex.names=0.7,ylim=c(0,80), col=c("#9e9ac8",
"#6a51a3"))
dev.off()
See ?par to see the details for adjusting margins and other plot features.
David
On Tue, Jun 13, 2023 at 5
Bert,
I stand corrected. What I said may have once been true but apparently the
implementation seems to have changed at some level.
I did not factor that in.
Nevertheless, whether you use an index as a key or as an offset into an
attached vector of labels, it seems to work the same and I th
Thank you so much David, here is correction:
d1=suppressWarnings(read.csv("/Users/anamaria/Downloads/B1.csv",
stringsAsFactors=FALSE, header=TRUE))
d1$X <- NULL
d2=as.matrix(sapply(d1, as.numeric))
pdf("~/graph.pdf")
b<-barplot(d2, legend= c("SYCL", "CUDA"), beside=
TRUE,las=2,cex.axis=0.7,cex.nam
Below.
On Tue, Jun 13, 2023 at 2:18 PM wrote:
>
>
> Javad,
>
> There may be nothing wrong with the methods people are showing you and if
it satisfied you, great.
>
> But I note you have lots of data in over a quarter million rows. If much
of the text data is redundant, and you want to simplify s
Your first data column appears to contain character data (e.g. SYCL) which
cannot be converted to numeric. You also appear to have 0's in the numeric
columns which will cause problems since log(0) is -Inf. Barplots are useful
for categorical data, but not continuous, numeric data which are better
h
Javad,
There may be nothing wrong with the methods people are showing you and if it
satisfied you, great.
But I note you have lots of data in over a quarter million rows. If much of the
text data is redundant, and you want to simplify some operations such as
changing some of the values to
Hello,
I have a data frame like this:
d11=suppressWarnings(read.csv("/Users/anamaria/Downloads/B1.csv",
stringsAsFactors=FALSE, header=TRUE))
> d11
X Domain.decomp. DD.com..load Neighbor.search Launch.PP.GPU.ops.
Comm..coord.
1 SYCL 2. 10 3.7
It is safer to use !grepl(...) instead of -grep(...) here. If there are no
matches, the latter will give you a zero-row data.frame while the former
gives you the entire data.frame.
E.g.,
> d <- data.frame(a=c("one","two","three"), b=c(10,20,30))
> d[-grep("Q", d$a),]
[1] a b
<0 rows> (or 0-lengt
Dear Kevin,
actually you're mixing markdown and LaTeX syntax, which is the reason
why you see LaTeX code in the PDF. You have to choose...
1) Either you wish to produce an RMarkdown document and your sections,
subsections... should read:
# Abstract
In this document, ...
## Boundaries of the Ra
Hi, all,
I'm trying to compose an Rmarkdown document and render it as a PDF file.
My first block of R code seems to work okay, but the second on seems to
be interpreted as LaTeX code, and not executed as R code. In the output,
the three back-ticks that mark the R code block are interpreted as a
Às 17:18 de 13/06/2023, javad bayat escreveu:
Dear Rui;
Hi. I used your codes, but it seems it didn't work for me.
pat <- c("_esmdes|_Des Section|0")
dim(data2)
[1] 281549 9
grep(pat, data2$Layer)
dim(data2)
[1] 281549 9
What does grep function do? I expected the funct
Dear all;
I used these codes and I get what I wanted.
Sincerely
pat = c("Level 12","Level 22","0")
data3 = data2[-which(data2$Layer == pat),]
dim(data2)
[1] 281549 9
dim(data3)
[1] 244075 9
On Tue, Jun 13, 2023 at 11:36 AM Eric Berger wrote:
> Hi Javed,
> grep returns the positions of
Hi Javed,
grep returns the positions of the matches. See an example below.
> v <- c("abc", "bcd", "def")
> v
[1] "abc" "bcd" "def"
> grep("cd",v)
[1] 2
> w <- v[-grep("cd",v)]
> w
[1] "abc" "def"
>
On Tue, Jun 13, 2023 at 8:50 AM javad bayat wrote:
>
> Dear Rui;
> Hi. I used your codes, but it
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