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Is R not recognizing data for some reason?
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csv itself and this doesn't appear to have helped. I have made sure to change the cell formatting to "number" in the. I am quite puzzled because I have gone back and checked the. My other concern, which is more problematic than missing zeroes is that 'x1' has a length of '0' according the the length() function.
Error in taply code#
I have trawled for the error and the function 'tapply' here, on other forums, and actually copied the code from "A beginners guide to R" and replaced it with my own data. However, the missing values are not missing values as such, they are real 0's. I suspect that this is because there are missing values. LoginAsk is here to help you access Error In Tapply quickly and handle each specific case you encounter. $ chemical: Factor w/ 2 levels "aquis","benzocaine": 1 1 1 1 1 1 1 1 1 1. Error In Tapply will sometimes glitch and take you a long time to try different solutions. Originally, I had an error message as follows: > data str(data) But note that the standard errors of the estimates are not identical with the standard errors of the data.I have imported a set of data into R as a data frame from a. Multiple R-squared: 0.9785, Adjusted R-squared: 0.9763į-statistic: 440.9 on 3 and 29 DF, p-value: < 2.2e-16Īs you can see, these estimates are identical to the means of the factor levels. Residual standard error: 3.223 on 29 degrees of freedom
Error in taply how to#
codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 tApply Function in R: How to Use the 'tapply()' Function in R, to Apply a Function to subsets of a Variable or Vector in R. ): summary(lm(mpg ~ 0 + as.factor(cyl), mtcars))
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If you want the lm function to calculate the means of the factor levels, you have to exclude the intercept term ( 0 +. The effect cyl8 is related to the difference between cyl = 8 and cyl = 4. Hence, the estimate and standard error for cyl6 are related to the difference between cyl = 6 and cyl = 4. But the other effects result from a comparison of one factor level with the reference category. The intercept of the linear model corresponds to the mean of the dependent variable in the reference category. By default, the first level, 4, is used as reference category. The factor mtcars$cyl has three levels (4,6, and 8). This is the default for categorical data. If no contrast is specified manually, treatment contrasts are used in R. The lm function does not estimate means and standard errors of the factor levels but of the contrats associated with the factor levels. (when editing my question, should I delete the original text or adding my edition as I did ) My question is why are they different and not the same? The means are exactly the same but the standard errors are different for these 2 methods (as Sven also notices). With(mtcars, tapply(mpg, cyl, sd)/sqrt(summary(mtcars$cyl)) ) We can compare this with an direct calculation of the means and their standard errors: with(mtcars, tapply(mpg, cyl, mean)) What is going on here? It is related to lm() fitting the mean for each group and an error term?Īfter Svens answer (below) I can formulate my question more concise and clearly.įor categorical data we can calculate the means of a variable for different groups is by using lm() without an intercept. The direct calculation gives the same mean but the standard error is different for the 2 approaches, I had expected to get the same standard error. To get the standard errors for the means I calculate the sample standard variation and divide by the number of observations in each group: with(mtcars, tapply(mpg, cyl, sd)/sqrt(summary(mtcars$cyl)) ) To get the means by direct calculation I use this: with(mtcars, tapply(mpg, cyl, mean)) Then the length of the covariates and the count is not the same anyore and this causes problem in a subsequent tapply call in the getcovcounterrlist function. The intercept is the mean for the first group, the 4 cylindered cars. The problem is that these sites are removed only for the count, site.id, time and weights vectors but the covariates are ignored. Here is an example (taken from here Predicting the difference between two groups in R )įirst calculate the mean with lm(): mtcars$cyl |t|) But this standard error differs from what I get from a calculation by hand. This also gives the standard errors for the estimated means. When dealing with data with factors R can be used to calculate the means for each group with the lm() function. BuchananThis video covers the basic ideas of functions using R - topics include:- ggplot2- bar graphs with one indep.