similar to: R newbie attempting to plot data

Displaying 20 results from an estimated 11000 matches similar to: "R newbie attempting to plot data"

2010 Jan 14
2
Newbie mistakes saving images to files
I am encountering problems using the png() function to save around 20 charts to separate files. My script is conceptually structured as follows: ### <script> png("Image %03d.png") # the following are calls to user-defined functions I wrote that call plot, barchart, etc. with special arguments. myPlot("a") myPlot("b") myPlot("c")
2010 Jun 18
2
Drawing paths through a grid
I would like to draw a set of points that are equally spaced in a 2-D grid. Then I would like to draw lines that illustrate different directed paths through subsets of points. Imagine that the points correspond to booths in a conference center, and I want to show the various paths people took to visit the booths (using color to highlight different types of paths). An example path might be: [(1,1),
2010 Jan 14
3
Barchart bar lengths not proportionate
When I use barchart (with default formatting options), I get bars whose lengths/heights are not proportional to their value. For example: http://drop.io/wbagm6s/asset/capture-png Many of the values in this chart are 1; however, because the blue bars extend to the left of the "0" tick mark, those bars appear to represent higher numeric values. Is there a way to make the length of the
2006 Mar 25
1
Newbie question about SQL and data sources
Hi, I just downloaded R, and am wondering about data sources. Where do people typically get their data for analysis? It seems to me most people would have their data somehow automatically gathered and stored in an SQL database (e.g. MySQL), but this seems not to be the case. Does everyone just use the plain-text tab-separated values format? If so, how are these tables typically created in the
2010 Jan 15
1
Possible to write text inside a bar of a barplot?
How can I write text inside a bar of a barplot? I tried using text(), but I am only able to specify the numeric y-coordinate. The different columns of my barplot correspond to factors and not numbers, so I don't know how to access the horizontal positions of the bars. I tried fiddling with different values of the "adj" argument, but this appears unreliable. Background: I have
2007 Sep 18
1
Where is the interpreter (in Instant Rails)?
Hello, I just installed Instant Rails on my Windows XP box. I would like to test out the Ruby interpreter on the command line. However, when I run C:\InstantRails\ruby\bin\ruby.exe from the Windows command prompt, it hangs. Any ideas on how I can get a Ruby command line? Thanks, Rex --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to
2010 Jan 13
1
Recommended visualization for hierarchical data
Let's say I have data in the following schema that describes the number of purchases a company has received from each County in the US: State | County | Purchases --------------------------------------- NJ | Mercer | 550 CA | Orange | 23 .... I would like to visualize what states contribute the most to the overall total, and furthermore within those states, what Counties contribute the most.
2009 Apr 07
2
newbie query: simple crosstabs
I've been playing around with various table tools, trying to construct a fairly simple cross-tab. It shouldn't be hard, but for some reason it turning out to be (for me). If I want to see how many men and how many women agree with a agree/disagree question (coded 1,0), I can do this: >attach(mydata) >mytable <- table(male, q1.bin) # gender and a binary response variable
2015 Oct 22
2
C_LogLin (stats/loglin)
Hi everyone, I have a question regarding a C function of the "stats" package in R. I tried to understand the ?loglin? basic function of the ?stats? package implemented in R. The implemented function itself runs without any problem (perhaps see sample). When I ran it line by line it stopped at the lines 23-24 of the loglin-function; (the following line): z <- .Call(C_LogLin,
2013 Jul 15
1
pmatch inconsistency
The pmatch help (see also section 4.3.2 in the R Language Definition) claims that pmatch with duplicates.ok=FALSE provides the same functionality as R's argument matching algorithm, modulo how empty strings are matched. Here's an undocumented inconsistency between pmatch and R's argument matching algorithm: > sessionInfo() R version 3.0.1 (2013-05-16) Platform:
2009 Oct 16
1
Frequencies, proportions & cumulative proportions
Dear R-Helpers, I've looked high and low for a function that provides frequencies, proportions and cumulative proportions side-by-side. Below is the table I need. Is there a function that already does it? Thanks, Bob > # Generate some test scores > myValues <- c(70:95) > Score <- ( sample( myValues, size=1000, replace=TRUE) ) > head(Score) [1] 77 71 81 88 83 93 > >
2012 Feb 26
6
loop for a large database
Yes, I am a newbie. I have a data.frame (MyTable) of 1445846 rows and 15 columns with character data. And a character vector (MyVector) of 473491 elements. I want simply to get a data.frame with the count of how many times each element of MyVector appears in MyTable. I've tried a loop with : for (i in 1 : length (myvector)) sum (MyTable== i) but it crashes my computer. I've also
2023 Dec 16
2
Partial matching performance in data frame rownames using [
On Wed, 13 Dec 2023 09:04:18 +0100 Hilmar Berger via R-devel <r-devel at r-project.org> wrote: > Still, I feel that default partial matching cripples the functionality > of data.frame for larger tables. Changing the default now would require a long deprecation cycle to give everyone who uses `[.data.frame` and relies on partial matching (whether they know it or not) enough time to
2023 Dec 19
1
Partial matching performance in data frame rownames using [
Hi Hilmar and Ivan, I have used your code examples to write a blog post about this topic, which has figures that show the asymptotic time complexity of the various approaches, https://tdhock.github.io/blog/2023/df-partial-match/ The asymptotic complexity of partial matching appears to be quadratic O(N^2) whereas the other approaches are asymptotically faster: linear O(N) or log-linear O(N log N).
2008 Jun 23
3
One-to-one matching?
Hi folks, Can anyone suggest an efficient way to do "matching without replacement", or "one-to-one matching"? pmatch() doesn't quite provide what I need... For example, lookupTable <- c("a","b","c","d","e","f") matchSample <- c("a","a","b","d") ##Normal match()
2013 Mar 14
1
error: object of type 'closure' is not subsettable
Hi all, when i run this script: >read.table("Angelika.txt",header=T,sep="\t") >mytable=read.table("Angelika.txt",header=T,sep="\t") >for ( dye in c("A","B","C","F","G","K","L","M")) + { + for (cond in 1:8) + { + measurement =
1999 May 25
1
Muliple partial matches with pmatch
I am using R0.64.1 under Windows 95. The help for `pmatch' states that: If `duplicates.ok' is false multiple matches will result in the value of `nomatch' being returned, and if it is true, the index of the first matching value will be returned. The help for pmatch gives the following examples: pmatch("m", c("mean",
2006 Mar 06
4
Contingency table and zeros
Hello, Let's assume I have a vector of integers : > myvector <- c(1, 2, 3, 2, 1, 3, 5) My purpose is to obtain the cumulative distribution of these numerical data, i.e. something like : value nb_occur. <=1 2 <=2 4 <=3 6 <=4 6 <=5 7 For this, I create a table with ; > mytable <- table(myvector) 1 2 3 5 2 2 2 1 However, table() returns an array
2012 Mar 06
1
frequency count by row
I feel this is a very easy thing but I've never done it before and it is getting frustrating. I have a big data.frame (1445846 rows, 15 col) that looks like this: V1 V2 V3 V4 V5 1 home sister brother chair 0 2 cat dog animal 0 0 3 girl boy 0
2010 Jul 26
4
using string variable as order() function argument
Hello, In my script I would like to use a loop, which sorts the dataframe according to different columns, pointed by the string variable. id col1 col2 col3 1 10 0 4 8 2 11 1 2 2 3 12 0 8 3 4 13 0 5 5 Usually the order() function can be used like this: sorted = mytable**[order(column3) , ] which results in properly sorted table: **