similar to: Problems with large data frames?

Displaying 20 results from an estimated 10000 matches similar to: "Problems with large data frames?"

2009 Jul 09
9
Population pyramids
Hi, I hope somebody can help me with this issue: I am doing population pyramids using the barplot command, so in the left side I have male age structure and in the right side the female age structure. To plot the male age structure I put the data in negative numbers. Now, I want to change the sign in the bar plot in such way that I have no-sign numbers, both in left and right side of the graph. I
2008 Feb 27
4
Error in cor.default(x1, x2) : missing observations in cov/cor
Hello, I'm trying to do cor(x1,x2) and I get the following error: Error in cor.default(x1, x2) : missing observations in cov/cor A few things: 1. I've used cor() many times and have never encountered this error. 2. length(x1) = length(x2) 3. is.numeric(x1) = is.numeric(x2) = TRUE 4. which(is.na(x1)) = which(is.na(x2)) = integer(0) {the same goes for is.nan()} 5. I also try
2009 Jul 24
4
CI wiskers
I have a matrix containing means and CIs (lower and upper in two columns, so three columns for every data point) for several points. I have to build a graph of these means accompained by the CIs (as wiskers). No problems with making the graph of means, but I don't know how to introduce CIs. Can anybody advise? -- View this message in context:
2008 Mar 31
2
Finding a mean value of a variable holding a dummy variable fixed
I have time-series data on approval ratings of British Prime Ministers. The prime ministers dating from MacMillan onward till today are coded as dummy variables and the approval ratings are entered for each month. I want to know the mean value of the approval rating of each Prime Minister in the dataset and the approval rating during his/her first month and last month as PM. What R code should
2009 Jun 16
1
Running stats on individual data.frames from the split() function list
Hello, and thanks in advance. I have a data.frame from which I want to count observations that occur on each day and determine the mean and std.error of said counts. For instance: x<-split(my.df, my.df$julian.days) Although I'm still in my R learning infancy I am under the impression that x is a list of data.frames subsetting my.df by group (i.e.,julian.day) where day 1:366 are
2009 Sep 20
3
statistics
The myoglobin sequence, with reference number NM_005368 in Gen bank, has the following frequencies of DNA nucleotides: A C G T 237 278 309 242 Do these data provide sufficient evidence, at the 1% level of significance, that the DNA nucleotides have an unequal distribution, that is the DNA nucleotides are not evenly utilised? Clearly state your hypothesis, test statistic and conclusion.
2008 Feb 26
3
OLS standard errors
Hi, the standard errors of the coefficients in two regressions that I computed by hand and using lm() differ by about 1%. Can somebody help me to identify the source of this difference? The coefficient estimates are the same, but the standard errors differ. ####Simulate data happiness=0 income=0 gender=(rep(c(0,1,1,0),25)) for(i in 1:100){ happiness[i]=1000+i+rnorm(1,0,40)
2010 Apr 08
2
Problem using elements in a vector
Hi So my particular problem is this: I have a row vector of length 5200 elements - specifically created by x<-rbinom(5200,1,0.5) y<-matrix(x,nrow=1,ncol=5200) y now, each element is either a 0 or a 1 - e.g. it could be (0,1,1,1,1,0,0,0,1,1,1) e.t.c. when the element is a 1, i need to multiply a number (say 1000) by 1.005, and if it is 1 again, multiply it _again_ by 1.005. so for
2009 Nov 11
3
how to use # in a rd doc in url address
I am writing a rd doc, and need to use "#" in a url adress. This would make: \url{http://www.xxxx.org/myfolder/#myanchor} Of course, I suppose this will not work because # is a special character starting a comment line in the rd dialect. I did not found a similar example in "Writing R exentions". I am not sure bout using \dQuote{a quotation}), and use \sQuote and \dQuote
2008 Feb 23
1
clarification about glm
Hello, I have a question about glm: if i have a binary covariate (unit=1,0) the reference group would be 0? (prediction for unit=1) example: dat1<-data.frame(y,unit,x1,x2) log_u <- glm(y~.,family=binomial,data=dat1) summary(log_u) Estimate Std. Error z value Pr(>|z|) (Intercept) -0.54247 0.24658 -2.200 0.0278 * unit1 -0.13052 0.22861 -0.571 0.5680 aps
2009 Sep 04
2
Nested Fixed Effects - basic questions
Hi R people, I have a very basic question to ask - I'm sorry if it's been asked before, but I searched the archives and could not find an answer. All the examples I found were much more complicated/nuanced versions of the problem - my question is much more simple. I have data with multiple, nested fixed effects (as I understand it, fixed effects are specified by the experimental design
2007 Nov 16
1
Nonparametric manova
Hi, I have seen a discussion in the R-help asking whether nonparametric manova has been implemented in R yet. This discussion is form March 2006 and there seemed to be no package or function implemented at the time. Has this changed? Is there a package that provides nonparametric manova as in McArdle and Anderson (2001) and Anderson (2001) now? Thanks, Daniel ------------------------- cuncta
2008 Nov 20
1
sub / gsub - extracting between identical symbols
Hi, I am trying to extract some numbers from a text string. The problem is that the delimiting symbols are identical so that I do not know how to tell "sub" between which of them to extract. The string looks like this 12/01/03/08 The extracted variables should look like: x1=12 x2=01 x3=03 x4=08 If anybody could help or point me to useful help, I would be greatful. Cheers, Daniel
2009 Aug 05
4
for loop
I am trying to get the function "Models" to work each time there is an instance of k. This code will stop after the first model is complete. I need it to come back and pass the next value of c into the "Initial.State" function. any ideas? col<-c(23:28) #Setup for(k in col){ Initial.State(Response=zample[,c(k,29)], Explanatory=zample[,variable_columns],
2008 Nov 24
3
count the cumulative for each subject
I have a data set like the following: subject visit x1 1 1 0.5 1 2 1.2 1 3 0.7 2 1 0.4 2 2 0.6 2 3 1.0 ..... where x1 is the interval between the two visits. Now I want to calculate the cumulative intervals since the beinging, for example subject visit x1 cum 1 1 0.5 0.5 1 2 1.2 0.5+1.2 1 3 0.7 0.5+1.2+0.7 2 1 0.4 0.4 2 2 0.6 0.4+0.6 2 3 1.0 0.4+0.6+1.0 ..... is there an easy to generate the
2009 Jul 14
5
plotting confidence intervals
Hi R People: If I have a fitted values from a model, how do I plot the (1-alpha)100% confidence intervals along with the fitted values, please? Also, if the intervals are "shaded" gray, that would be nice too, please? I check confint, but that doesn't seem to do what I want. Thanks in advance, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and
2007 Dec 31
3
Survival analysis with no events in one treatment group
I'm trying to fit a Cox proportional hazards model to some hospital admission data. About 25% of the patients have had at least one admission, and of these, 40% have had two admissions within the 12 month period of the study. Each patients has had one of 4 treatments, and one of the treatment groups has had no admissions for the period. I used:
2008 Dec 11
1
Extract rows from data frame based on row names from another data frame
Hi all, Is there a function to extract row names from a data frame based on row names from another data frame? I can write a loop function to do this, but this may be inefficient in terms of processing. thanks for any information, Wade
2009 Jul 22
1
How do I delete a row from a data frame when varA == "TRUE"
I want to delete a row from a data frame XYZ where A == "TRUE". Let's assume there are five variables in this data frame XYZ. My deletion is based on the TRUE vs. FALSE value under variable A. How do I do this? Thanks! My current code: for (i in 1:length(XYZ$A)) if (XYZ$A == "FALSE") XYZ<-XYZ else XYZ<- XYZ[-i,] Error message is: Error in if (XYZ$A ==
2008 Jun 13
1
parsing - input buffer overflow
Hi, I am trying to parse a large amount of text using gregexpr(). Unfortunately, I get an "input buffer overflow" message when I attempt that with too large an amount of text. The error messages occurs before the parsing. The problem is that I cannot assign the text to a variable (an object) if the text is too large. This problem has been mentioned before, which I found using the