similar to: about saving format

In the lower right panel of R-studio interface, there is the "Export" button. I saved as PDF from there directly, rather than using functions On Sat, Aug 5, 2017 at 6:18 PM, Ismail SEZEN <sezenismail at gmail.com> wrote: > > > On 6 Aug 2017, at 03:01, lily li <chocold12 at gmail.com> wrote: > > > > I am using the plot() function, but have a problem.

If the OP is using RStudio and not using R (i.e. pdf()) directly, it sounds like this query should be directed to RStudio support, not here. Cheers, 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 ) On Sat, Aug 5, 2017 at 5:54 PM, Ismail

> On 6 Aug 2017, at 03:47, lily li <chocold12 at gmail.com> wrote: > > In the lower right panel of R-studio interface, there is the "Export" button. I saved as PDF from there directly, rather than using functions > > On Sat, Aug 5, 2017 at 6:18 PM, Ismail SEZEN <sezenismail at gmail.com> wrote: > > > On 6 Aug 2017, at 03:01, lily li <chocold12 at

On 05/08/2017 9:10 PM, Bert Gunter wrote: > If the OP is using RStudio and not using R (i.e. pdf()) directly, it > sounds like this query should be directed to RStudio support, not > here. Two things: First, to Lily: "?" is the "per mil" symbol, not percent; it's not an ASCII symbol, which is why there are issues displaying it. If you really mean percent, use

> On 6 Aug 2017, at 03:01, lily li <chocold12 at gmail.com> wrote: > > I am using the plot() function, but have a problem. When saving as pdf > format, the ? sign in the x-axis label becomes (...) sign. I prefer to save > in pdf, as this format has a higher resolution than jpeg or other picture > formats. Could anyone tell me how to do then? Thanks. Please, share minimal

... Or if you just want to stick with basic regex's without extra packages: > x <- "\"cm_ffm\":\"563.77\"" > sub("[^[:digit:]]*([[:digit:]]*.?[[:digit:]]*).*","\\1",x) [1] "563.77" Cheers, Bert On Wed, Aug 2, 2017 at 5:16 PM, Ismail SEZEN <sezenismail at gmail.com> wrote: > >> On 3 Aug 2017, at 02:59,

> On 27 Oct 2017, at 10:43, Engin YILMAZ <ispanyolcom at gmail.com> wrote: > > Dear R Staff > > My working file is in the annex. "g1.csv" > I have only 2 columns. Rice and coke. > I try to execute following(below) function, but do not work. > Because "Coke" value has NA values. > I try to add "na.rm=True" to the function but do not

... and Marc's solution is **much** better than mine. -- 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 ) On Wed, Aug 2, 2017 at 5:59 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > ... Or if you just want to stick with

Dear R Staff My working file is in the annex. "g1.csv" I have only 2 columns. Rice and coke. I try to execute following(below) function, but do not work. Because "Coke" value has NA values. I try to add "na.rm=True" to the function but do not work How can I solve this problem with this function or another algorithm? (Note: I have normally 450 columns) Sincerely

> On 3 Aug 2017, at 02:59, Christofer Bogaso <bogaso.christofer at gmail.com> wrote: > > Hi again, > > I am struggling to extract the number part from below string : > > "\"cm_ffm\":\"563.77\"" > > Basically, I need to extract 563.77 from above. The underlying number > can be a whole number, and there could be comma separator as

> On 27 Nov 2017, at 13:59, Engin YILMAZ <ispanyolcom at gmail.com> wrote: > > Dear Berger and Jim > > Can you see my eviews example in the annex? (scattersample.jpg) > > Sincerely > Engin Please, use an image hosting service (i.e. https://imgbb.com/) to share images in the list and share the link in the email.

> On 15 Sep 2017, at 12:38, yadav neog <yadavneog at gmail.com> wrote: > > hello to all. I am working on macroeconomic data series of India, which in > a yearly basis. I am unable to convert my data frame into time series. Do you really need to convert your data to time series/xts/zoo? I don?t know you try what kind of an analysis but perhaps you don?t have to. > kindly

> On 27 Nov 2017, at 11:56, Engin YILMAZ <ispanyolcom at gmail.com> wrote: > > Dear > > I try to realize one scatter matrix which draws *one single variable to all > variables* with *regression line* . You can see my eviews version in the > annex . > > How can I draw this graph with R studio? A tiny note; You do calculations in R not RSudio. RStudio is a tool

Hi again, I am struggling to extract the number part from below string : "\"cm_ffm\":\"563.77\"" Basically, I need to extract 563.77 from above. The underlying number can be a whole number, and there could be comma separator as well. So far I tried below : > library(stringr) > str_extract("\"cm_ffm\":\"563.77\"",

Hi all, I'm trying to add single points with known coordinates to a level plot, but could not find the proper answer. I got to know that layer() function is good for this, but I don't know which package is related to this function. The source is here: https://stackoverflow.com/questions/28597149/add-xy-points-to-raster-map-generated-by-levelplot but my question is a little different as I

Hi Bert, I think you are correct that I can use levelplot, but I have a question about converting data. For example, the statement: levelplot(Z~X*Y), Z is row-wise from the lower left corner to the upper right corner. My dataset just have gridded Z data as a txt file (or can be called matrix?), how to convert them to the vector in order for levelplot to use? Thanks. On Mon, Jan 15, 2018 at 6:04

Hi Jim, For true color, I meant that the points in the figure do not correspond to the values from the dataframe. Also, why to use rainbow(9) here? And the legend is straight in the middle, is it possible to reformat it to the very bottom? Thanks again. On Sun, Jul 16, 2017 at 2:50 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi lily, > As I have no idea of what the "true

Hi all, I ran the code: > s <- stack(replicate(2, raster(matrix(runif(100), 10)))) > xy <- data.frame(coordinates(sampleRandom(s, 10, sp=TRUE)), + z1=runif(10), z2=runif(10)) > levelplot(s, margin=FALSE, at=seq(0, 1, 0.05)) + + layer(sp.points(xy, pch=ifelse(pts$z1 < 0.5, 2, 3), cex=2, col=1), columns=1) + + layer(sp.points(xy, pch=ifelse(pts$z2 < 0.5, 2,

First, see the example at https://isezen.github.io/PCA/ > On 15 Sep 2017, at 13:43, Shylashree U.R <shylashivashree at gmail.com> wrote: > > Dear Sir/Madam, > > I am trying to do PCA analysis with "iris" dataset and trying to interpret > the result. Dataset contains 150 obs of 5 variables > > Sepal.Length Sepal.Width Petal.Length Petal.Width

For more than 10 records, how to reformat the colors? Also, how to show the first legend only, but at the bottom, while the second legend in your code is not necessary? In all, the same A values have the same color, but different symbols in DF==1 and DF==2. Thanks for your help. On Sun, Jul 16, 2017 at 9:28 AM, lily li <chocold12 at gmail.com> wrote: > Hi Jim, > > For true color,