similar to: Get the: "standard deviation" and "mean value" of 3D-measurments

Hello, I've got a matrix (mail end) with the colnames x, y, z. In this matrix are different measurements. x and y are risign coordinates. With the following line I got the median value of z for all "x" AND "y" witch are the same (not every measurment in my list hast the same number of "x" and "y" values. Sometimes lines are missing. >MEDIAN <-

Hello, I have got a question. I've got a matrix (mail end) with the colnames x, y, z. In this matrix are different measurements. x and y are risign coordinates. My question. Always, if the "x" AND "y" coordinates are the same, I want to get the mean of their z values. e.q. " x" AND "y" in line1 and line8 are identical: 29 4.5 --> mean of

Hello, I have a problem with readLines. I have a data file with many informations added with a different number of measurments (example at the end). I only want to read the measurments witch start with "START OF HEIGHT DATA" and end with "END OF HEIGHT DATA". The difficulty is: -I want to read the file with "readLines", because the measurments have letters and

Hello, I have a data set where there are multiple "cycles" per "patient," and I want to exclude from my data set instances where a variable was not measured every cycle. The difficulty is that the patients have different cycles; some have cycles 1,2, and 3, others only have 1 and 3 (and everything in between). Therefore, I'm having difficulty in in distinguishing between

Dear all, is there an implemented function to compute a weighted standard deviation (-like weighted.mean-) in R ? Thank's a lot in advance Johannes Schnitzler

Hello, I am using urca package to run cointegration. I would like to find the standard error in the (normalized, Johansen) cointegration relationship. How can I do it? As far as I know, The function "cajorls" in the "urca" package provides the normalized cointegrating relationships. Nevertheless, it does not provide the standard deviation of the coefficient for each

Dear R-users, I have a data set where each observation consists of a number of trials (n.trials) that varies between 5 and 7, 6 being most common. Each trial can take either of two outcomes, success or failure. A dummy data set: n.trials <- sample(5:7, 50, replace=T, prob=c(0.2, 0.6, 0.2)) success <- rbinom(50, n.trials, p=0.5) failure <- n.trials - success I know I could test for a

Hi, Could anyone give suggestions how to plot a scatter plot with 1 standard deviation for each point. To make it clearer, here is a simple example: the scatterplot is plot(X, Y), but I want to add 1 standard deviation according to the value of Z for each Y. X Y Z 1 3.5 1.1 . . . . . . . . . Thanks a lot in advance. FD [[alternative HTML

I am trying to do logistic regression with a categorical predictor variable with the glm() function, family=binomial. Using glm() I would like to be able to calculate the confidence intervals of all three possible odds ratios for a factor (the factor has three categories). Three categories imply two columns of 0's and 1's in the design matrix, and two parameter estimates with their

Hi, I have a dataset of 78.903 news articles pertaining to 340 corporate takeovers. Mean 231.3871 [articles per takeover] Std. Dev. 673.6395 I would like to calculate the probability of a certain number of news articles if I had more takeovers available. How likely is it to have X articles if I had Y takeovers? How can I do that? Thank you very much, Michael

Hi all, I've got clusters and would like to match individual records to each cluster based on a sum of squares deviation. For each cluster and individual, I've got 50 variables to use (measured in the same way). Matrix 1 is individuals and is 25000x50. Matrix 2 is the cluster centroids and is 100x50. The same variables are found in each matrix in the same order. I'd like to

Dear list, i'm looking for how to get the mean squared deviation in R. I tried a lot of (complicated) instructions, sometimes the worked, but now i'm looking for an easier one....Hope you can help - greetings.

Hello, I have a vector of values with significant autocorrelation, and I want to calculate an unbiased standard deviation that adjusts for the autocorrelation. The formula linked below purports to provide what I want: https://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Effect_of_autocorrelation_.28serial_correlation.29 However, rather than just implementing this equation in

Hi I could not understand how "active bitrate" and "bitrate standard deviation" is calculated in demo program. Can anyone please explain ? average bitrate: 5.489 kb/s maximum bitrate: 10.400 kb/s *active bitrate: 7.266 kb/sbitrate standard deviation: 1.559 kb/s* -- Thanks & Regards Vittal Prasad B R 9066662597

Hello! I'd like to print certain kinds of diagrams with R. They should look like barplots with besides=TRUE, but also have the variance or standard deviation on every bar so that one could better compare the bars. How could I apply this? thanks for your help Raoul -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hello everyone, I know that to calculate the linear regression between x and y I have to use the function "lm". But how to do if x and y have a standard deviation like in the following example? How to compute the chi square test in this case? Thank you, Fulvio. example: x +- dx 5 2 13 4 17 4 23 6 y +- dy 6.3 0.6 9.2 0.9 14 1 21 4

Dear R-users, I need to standardize residuals using standard deviation. Is the 'stdres' the proper function? Beside other methods (for standardization and normalization), are there some approaches how to standardize using standard deviation? Here is the sample of my residuals: Zres040 Zres0820 Sres040 Sres0820 PCres040 -2.101740 -2.0682900 1.6328500 0.5046730

Hi, I have the two vectors mean and sd of individual columns, but I am unsure how to generate bar charts with the standard deviation, even after looking the help of barplot and barplot.2. [[alternative HTML version deleted]]

Hi list Maybe someone can help with the following problem (thanks in advance): I need to analisy a long list of data time \fluoscence I want to make one plot with standard deviation and the average data... Could you suggest me how to make that.. The data have this distribution: How can read and extrapolate this data? pippo.csv ps <- read.table(pippo.csv) time fluorescence 0; 0 4.013; 62.96

Hi, R seems to use the "1/n-1"-factor calculating the standard-deviation sd(). If i wat to get the "non-sample" standard-deviation i use sqrt(sd(x)^2*((n-1)/n)) Is there a parameter to get the sd()-function using the "1/n" factor directly? Or is there any other function to do so? Thank you in advance :-) Best, Eli