similar to: ggplot2: Plot multiple lines using stacked data.

Displaying 20 results from an estimated 1000 matches similar to: "ggplot2: Plot multiple lines using stacked data."

2008 Sep 05
1
Confidence Intervals on Hazard Plots
Hello, Is it possible to create confidence intervals for hazard rates? ?I'm creating two muhaz objects:? haz1 <- muhaz(NumDaysCustomer[cRV=="true"],status[cRV=="true"]) haz2 <- muhaz(NumDaysCustomer[cRV=="false"],status[cRV=="false"]) and plotting them. ? There are many, many more observations in the cohort cRV=="false" than
2012 Apr 24
2
Combined grouped and stacked bargraph
Dear R list, I am having some trouble drawing a bar-graph with two groups, both of which are stacked. A sample of my data (IUCN and national conservation status for marine and terrestrial species) look like this: Status IUCN Terrestrial IUCN Marine National CS Terrestrial National CS Marine NE 168 69 16 69 LC 41 6 148 6 NT 5 4 7 4 VU 4 2 7 2 EN 4 2 4 2 CR 1
2009 Jan 20
2
Stacked barplot with two stacked bars besides each other
Hi, I have a particular barplot I would like to generate, but I am having trouble getting it to work. What I would like is in effect two barplots with stacked bars merged into one. For example, I have two samples (yoda1,yoda2) on which I measure whether two variables (var1,var2) are present or absent for a number of measurements on that sample. > var1 <- data.frame(yoda1=c(3,7),
2009 May 28
2
ggplot2 legend
Hi: I need some help with the legend. I got 14 samples(Muestreo) and I am trying to plot a smooth line for each sample. I am able to accomplish that but the problem is that the legend only displays every other sample. How can I force the legend to show all of my Muestreos? Thanks in advance. fish_ByMuestreo <- structure(list(data = structure(list(SampleDate = structure(c(3L, 3L, 3L, 3L,
2011 Jan 19
1
Using subset to filter data table
I am having difficulty understanding how I would constrain a data set by filtering out 'records' based on certain criteria. Using SQL I could query using 'select * from my.data where LithClass in ('sand', 'clay')' or some such. Using subset, there seem to be ghosts left behind (that is, all of the LithClass *.Labels* remain after subset) > dput(tcc)
2012 Jan 20
1
Stacked barchart in ggplot (or other library)
Hey, I want to create a stacked barchart in R for the following dataset (http://pastebin.com/pyHUNgr2): # usage capacity diff 1 4 10 6 2 2 20 18 3 5 10 5 The stacked barchart should, in one plot show each line of the dataset as a stacked bar using data from 'usage' and 'diff' to create the stacked bar. I can't find a good example of how to do this on the ggplot2 site.
2009 Oct 09
1
Placing text in a ggplot
I am attempting to graph 12 months of temperatures, delineate the months with a vline and place the names of the months at the top of the graph. So far I have gotten everything to work except the names, despite getting a similar graph to work yesterday the day before yesterday with Baptise A's help. Can anyone suggest what I am doing wrong. Data set is below code. Thanks. Code
2012 Nov 08
2
Stacked Barchart as relative share
Hi @ all, I'm looking for a solution to plot a dataframe as a stacked bar chart like on this picture: http://4.bp.blogspot.com/_79SognVSu7A/S6OtzgksPSI/AAAAAAAABrw/-IuFNewdZFE/s400/Stacked%2BBar%2BChart.png My dataframe example2.csv <http://r.789695.n4.nabble.com/file/n4648854/example2.csv> My questions: Is there a standard function to convert the values to 100 percent relate to
2023 May 02
1
Error in percentage stacked barplot
Dear all,? I am trying to plot the following table in stacked barplot in percentages and also horizontal. Component? ? ? Sample 1? ? ? Sample 2? ? ? Sample 3CaO????????????????? ? 45????????????? ? 52????????????? ? 48SiO2????????????????? ?25????????????? ? 22????????????? ? 18Al2O3?????????????????15????????????? ? 11????????????? ? 14TiO2? ? ? ? ? ? ? ? ? ? ?6? ? ? ? ? ? ? ? ? 5? ? ? ? ? ? ? ?
2008 Dec 02
1
ggplot2 facet_wrap problem
Hadley, I don't know if I am doing something wrong or if it is ggplot please see the two graphs at the bottom of the page (code). melt.nut <- (structure(list(RiverMile = c(119L, 119L, 119L, 119L, 119L, 119L, 119L, 119L, 119L, 148L, 148L, 148L, 148L, 148L, 148L, 148L, 179L, 179L, 179L, 179L, 179L, 179L, 179L, 185L, 185L, 185L, 185L, 185L, 185L, 185L, 190L, 190L, 190L, 190L, 190L, 190L,
2006 Jul 06
3
Stacked Area graph?
Hi, How do I make stacked area graph? For stacked bar graph gruff has seperate class "StackedBar" but I could not find anything as "StackedArea"? Is there any other way? Please let know. Thanks -- Posted via http://www.ruby-forum.com/.
2008 Feb 29
1
Scruffy Stacked Bar Chart
Hi- Trying to do a stacked bar chart, but nothing is rendering when I add more then 1 bar: This works: graph = Scruffy::Graph.new(:title => "My Chart") graph.add :stacked do |stack| stack.add(:bar, ''Stack1'', [10,10]) end This doesn''t: graph = Scruffy::Graph.new(:title => "My Chart") graph.add :stacked
2017 Apr 26
0
error ssl stacked error routines
Hi I have configured one line in postfix main.cf (after configure each line I check /var/log/mail.err): smtpd_tls_dh1024_param_file = /etc/postfix/dh2048.pem After setup above line I have error in above log file (these 4 lines looped): Apr 25 14:08:09 serwer-1 dovecot: imap-login: Error: SSL: Stacked error: error:140760FC:SSL routines:SSL23_GET_CLIENT_HELLO:unknown protocol Apr 25 14:08:09
2011 Jun 09
1
lattice plot query
Dear R Group I have the following data for which I am trying to create subject wise lattice plot for a given attribute and product . though the lattice plot is generated, for some reasons that i dont understand in each plot the subject panels take a random order, I would rather want all the plots to display the subject order in the same order as how i have ordered this particular factor level.
2002 Jan 08
1
dimensions of stacked barplot
I'm plotting stacked barplots using: > barplot(tmp, beside=F) This gives me several stacked bars. Each stack is divided into several segments. I'd like to add a dotted line from the edge of one segment to the same edge on the next stacked bar. This might help to see which stack shrinks or expands from one stacked bar to the next. I'm thinking that I could overlay dotted lines
2003 Dec 16
2
Fw: [S] plot stacked bar chart in R
posting the question in r-help@ to get more feedback :-) thanks, Yun-Fang ----- Original Message ----- From: "Yun-Fang Juan" <yunfang at yahoo-inc.com> To: <s-news at wubios.wustl.edu> Sent: Tuesday, December 16, 2003 2:04 PM Subject: [S] plot stacked bar chart in R > Hi, > I am trying to plot a stacked bar chart in R but am not able to find the > documentation.
2012 Apr 27
1
Hyperspec package: need to change spectra names in a stacked plot
Hi all, I need to insert the name of spectra in a stacked plot obtained with hyperspec. I use this command plot(spectra [c(-1:-4, -6:-8, -10:-12, -14:-16)], stacked = T) but, in this way R draw nameless spectra on the Y axis. How can I solve the problem? Thank you for any suggestion. Best regards, Roberto -- View this message in context:
2011 Jun 13
1
Heatmap in R and/or ggplot2
I have a dataframe df with columns x, y, and height. I want to create a heatmap-like plot that creates a grid of x by y, and then color codes the grid depending on the value of height. Is there a ggplot2 object to do this? I'm able to easily do this in Excel with pivot tables and conditional formatting so I'm including an image that is close to the output I want. I want to be able to
2008 Jun 04
1
dotchart
I am trying to plot the following data using dotchart intersect.data<-structure(list(X = structure(c(1L, 3L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 2L, 4L, 5L, 6L, 7L), .Label = c("1-100", "1001-1100", "101-200", "1101-1200", "1201-1300", "1301-1400", "1401-1500", "201-300", "301-400", "401-500",
2012 Aug 06
5
sapply() and by()
Hello everyone, I have a dataset with 5 colums (4 colums with thresholds of weather stations and one with month - data of 5 years). Now I would like to calculate the average for each month. I tried this unsuccessfully: lf.med <- sapply(LF[,1:4],mean,LF[,5]) Error in mean.default(X[[1L]], ...) : 'trim' must be numeric and have length 1 With lf.med <- by(LF[,1:4],LF[,5],mean)