similar to: calculate sampel size?

On 14/04/16 19:28, Marna Wagley wrote: > Hi R user, > Can we calculate sample size when only mean and SE are given? > Let say one example, I have mean with SE is 0.54+-0.0517 (mean+-SE). Is > there any way to find the samples (sample size n) in that condition in R? > > i think this question is not related to R, I hope you won't mind. You're correct, your question is

Hi Jim, The data set is correct. I took two readings from the "SITE A" within a short time interval, therefore I want to take the first value if there are repeated within a same group of "timeGroup". Therefore I wanted following FinalData1 B1 B2 id_X "A" "B" id_Y "A" "B" thanks, On Tue, May 1, 2018 at 4:05 PM, Jim

Hi Jim, Thank you very much for your suggestions. I used it but it gave me three sites. But actually I do have only two sites "Id_X" and "Id_y" . In fact "A" is repeated two times for "Id_X". If it is repeated, I would like to take the first one among many repeated values. dat<-structure(list(ID = structure(c(1L, 1L, 1L, 2L, 2L), .Label =

Hi Marna, This is a condition that the function cannot handle. It would be possible to reformat the result based on the time intervals, but the stretch_df function doesn't try to interpret the values, just stretches them out to a wide format. Jim On Wed, May 2, 2018 at 9:16 AM, Marna Wagley <marna.wagley at gmail.com> wrote: > Hi Jim, > The data set is correct. I took two

Hi R users, I need to run a analysis using a data for each folder. I do have several folders. Each folder contains several files but these files name are similar to the files that is saved into another folders. I need to repeat the analysis for every folder and would like to save the output in that particular folder. After completing the analysis in one folder, I want to move another folder.

Hi Marna, I think this is due to having three rows for id_X and only two for id_Y. The function creates a data frame with enough columns to hold the greatest number of values for each ID variable. Notice that the SITE_n columns contain three values for id_X (A, A, B) and two for id_Y (A, B, NA) as there was no third occasion of measurement for the latter. Even though there are only two _values_

Hi R user, I was trying to convert a long matrix to wide? I have an example and would like to get a table (FinalData1): FinalData1 B1 B2 id_X "A" "B" id_Y "A" "B" but I got the following table using the following code. FinalData1 B1 B2 id_X "A" "A" id_Y "A" "B" the code and the

Hi Marna, Assuming that you are descending into different subdirectories from the same directory: directories<-c("dir1","dir2","dir3") for(directory in directories) { setwd(directory) # do whatever you want to do # then return to the directory above setwd("..") } This will allow you to start and finish in the same directory. Jim On Fri, Apr 13,

Hi Marna, Try this: library(prettyR) stretch_df(dat,idvar="ID",to.stretch=c("EventDate","SITE")) Jim On Wed, May 2, 2018 at 8:24 AM, Marna Wagley <marna.wagley at gmail.com> wrote: > Hi R user, > I was trying to convert a long matrix to wide? I have an example and would > like to get a table (FinalData1): > > > FinalData1 > B1

Hi R user, Would you mind to help me on how I can change a value in a specific column and row in a big table? but the column of the table is a factor (not numeric). Here is an example. I want to change dat[4:5,3]<-"20" but it generated NA> do you have any suggestions for me? dat<-structure(list(Sites = structure(1:5, .Label = c("Site1", "Site2",

Hi R users, I need to create more than 20 figures (one for each group) in one page. I have a common legend for 20 figures using the facet_wrap. However the range of the values among the groups are very wide. For example one group has the value of 0 to 3, but the values of some of the groups has ranged from 0 to 20 so that when I used a single common legend for all 20 figures, I could not display

Building on Petr's suggestion, you could modify his code to get all 10 samples at once in a compact format: > Samples <- lapply(lll, function(x) replicate(10, sample(x, rep=TRUE))) # Samples is a list containing 3 matrices, one for each group # Each column gives the index (row) numbers for a particular sample > str(Samples) List of 3 $ A: int [1:3, 1:10] 2 1 3 1 1 3 2 3 1 2 ... $

Hi R User, I am new in R and trying to create tables with selecting rows randomly (but with replacement) for each group but each group should have same number as original. Is it possible to create it using the following example data set? Your help is highly appreciated. dat1<-structure(list(RegionA = structure(c(1L, 1L, 2L, 3L, 3L, 4L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L),

Hi maybe there is another more elegant solution but something like this > idx <- 1:nrow(dat1) > lll <- split(idx, dat1$group) > dat1[unlist(lapply(lll, sample, rep=TRUE)),] gives you selected rows. You could use for cycle or save those data frames manually Cheers Petr > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Marna

Hi R Users, I do have very big data sets and wanted to run some of the analyses many times with randomization (1000 times). I have done the analysis using an example data but it need to be done with randomized data (1000 times). I am doing manually for 10000 times but taking so much time, I wonder whether it is possible to perform the analysis with creating a loop for many replicated datasets?

The simplest would be to convert precip to character and then back to a factor if you really want it to be a factor. This will also remove the levels that no longer exist. str(dat) # 'data.frame': 5 obs. of 3 variables: # $ Sites : Factor w/ 5 levels "Site1","Site2",..: 1 2 3 4 5 # $ temp : num 14 15 12 12.5 17 # $ precip: Factor w/ 5 levels

I really like the ease of use with the boxplot command in R. I would rather have a boxplot that shows the average value and the standard deviation then the median value and the quartiles. Is there a way to do this? Chad Junkermeier, Graduate Student Dept. of Physics West Virginia University PO Box 6315 210 Hodges Hall Morgantown WV 26506-6315 phone: (304) 293-3442 ext. 1430 fax: (304)

Howdy, I just gem unpacked merb and have configured the sample_app to use an sqlite3 database. I see that the Rakefile has a task named "schema", but it points to /dist/schema/schema1.rb", which doesn''t exist. I copied /dist/schema/schema1.rb to /dist/schema/schema.rb then ran "rake schema". What''s the recommended way to create the database tables?

> My understanding is that this represents bivariate normal > approximation of the data which uses the kernel density function to > test for inclusion within a level set. (please correct me) You can fit a bivariate normal distribution by computing five parameters. Two means, two standard deviations (or two variances) and one correlation (or covariance) coefficient. The bivariate normal

I recommend that you consult with a local statistical expert. Much of what you say (outliers?!?) seems to make little sense, and your statistical knowledge seems minimal. Perhaps more to the point, none of your questions can be properly answered without subject matter context, which this list is not designed to provide. That's why I believe you need local expertise. Bert Gunter "The