Hi Luigi, I may be missing the point, but: matrix((z1+z2+z3)/3,ncol=10) gives you the mean rating for each item, and depending upon what distribution you choose, the confidence intervals could be calculated in much the same way. Jim On Sun, Oct 24, 2021 at 7:16 AM Luigi Marongiu <marongiu.luigi at gmail.com> wrote:> > Hello, > I have a series of classifications of the same data. I saved this > classification in a single dataframe (but it could be a list). X and Y > are the variable and Z is the classification by three raters. `I` is > the individual identifier of each entry: > ``` > z1 = c(0,0,0,0,0,1,0,0,0,2, > 0,1,1,1,0,0,0,1,0,2, > 0,1,1,2,0,0,0,1,0,2, > 1,1,1,2,1,0,0,1,1,2, > 1,0,0,2,1,1,0,1,2,0) > z2 = c(0,0,0,0,0,1,0,0,1,1, > 0,1,1,2,0,0,0,1,1,2, > 0,0,0,1,0,0,0,1,0,0, > 1,2,1,2,1,0,0,1,1,2, > 1,0,1,2,1,1,0,1,2,0) > z3 = c(0,0,0,2,0,0,0,0,0,2, > 0,1,0,2,0,0,0,1,0,2, > 0,1,1,2,0,0,0,1,0,2, > 1,1,1,2,1,0,0,2,1,2, > 2,0,1,1,1,1,0,1,1,0) > df = data.frame(X=rep(1:5,3), Y=rep(1:5,3), Z=factor(c(z1,z2,z3)), I =1:150) > ``` > Is there a way to obtain a kind of heath map for each point? Let's say > for the point (x=1,y-1), what was the most common (average) > classification? Is it possible to get the 95% CI of that mean? > Would Two-Dimensional Kernel Density Estimation be the right path? > Thank you > > ______________________________________________ > R-help at 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 provide commented, minimal, self-contained, reproducible code.
Luigi Marongiu
2021-Oct-24 05:39 UTC
[R] generate average frame from different data frames
Thank you. Sorry for the fuzziness of the question but I find it difficult to give a proper definition of the problem. I have given a graphical rendering on this post https://www.researchgate.net/post/How_to_find_95_CI_of_a_matrix_of_classification_data As you can see in the figure, there are dots where the same value is represented all the time, and others where the values fluctuate. I would like to generate the "mean" merge of the figures. (Perhaps also with lines saying: this value comes out 9/10 of times, this 5/10 of times...). The problem is that the Z values are factors, not numbers. On Sun, Oct 24, 2021 at 12:08 AM Jim Lemon <drjimlemon at gmail.com> wrote:> > Hi Luigi, > I may be missing the point, but: > > matrix((z1+z2+z3)/3,ncol=10) > > gives you the mean rating for each item, and depending upon what > distribution you choose, the confidence intervals could be calculated > in much the same way. > > Jim > > On Sun, Oct 24, 2021 at 7:16 AM Luigi Marongiu <marongiu.luigi at gmail.com> wrote: > > > > Hello, > > I have a series of classifications of the same data. I saved this > > classification in a single dataframe (but it could be a list). X and Y > > are the variable and Z is the classification by three raters. `I` is > > the individual identifier of each entry: > > ``` > > z1 = c(0,0,0,0,0,1,0,0,0,2, > > 0,1,1,1,0,0,0,1,0,2, > > 0,1,1,2,0,0,0,1,0,2, > > 1,1,1,2,1,0,0,1,1,2, > > 1,0,0,2,1,1,0,1,2,0) > > z2 = c(0,0,0,0,0,1,0,0,1,1, > > 0,1,1,2,0,0,0,1,1,2, > > 0,0,0,1,0,0,0,1,0,0, > > 1,2,1,2,1,0,0,1,1,2, > > 1,0,1,2,1,1,0,1,2,0) > > z3 = c(0,0,0,2,0,0,0,0,0,2, > > 0,1,0,2,0,0,0,1,0,2, > > 0,1,1,2,0,0,0,1,0,2, > > 1,1,1,2,1,0,0,2,1,2, > > 2,0,1,1,1,1,0,1,1,0) > > df = data.frame(X=rep(1:5,3), Y=rep(1:5,3), Z=factor(c(z1,z2,z3)), I =1:150) > > ``` > > Is there a way to obtain a kind of heath map for each point? Let's say > > for the point (x=1,y-1), what was the most common (average) > > classification? Is it possible to get the 95% CI of that mean? > > Would Two-Dimensional Kernel Density Estimation be the right path? > > Thank you > > > > ______________________________________________ > > R-help at 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 provide commented, minimal, self-contained, reproducible code.-- Best regards, Luigi