I’ve posted an issue on dplyr github page. I can reproduce the results using the code below. It has to do with whether the csv contains a column of rownames without a header. read_csv and read.csv handle this differently, thus producing differing results with filter. First the case when it works write_csv to read_csv or read.csv; both work fine with filter Now for when it fails … Read more
Have a look at these lines here: What you see is that their lengths differ. How is this supposed to work? Think about what you are trying to do: You either need to evaluate the RMSE on the train data, or on the test data, but you are mixing them. That is, either this would … Read more
You added the as.factor(dataCancer$Classification) in the script, but even if the dataset dataCancer is attached, a command like the one above does not transform the dataset variable Classification into a factor. It only returns a factor on the console. Since you want to fit the model on the training dataset, you either specify or use the as.factor function in the glm line code
I ran into this same error and solved it by running: and then running the plot again. I think the graphics device was messed up earlier somehow by exporting some graphics and it didn’t get reset. This worked for me and it’s simpler than reinstalling ggplot2.
are you doing this: when you mean to do this In general use the ~ then you have data like and the , when you have data like I assume you’re doing something like: because the error suggests you have no variation in the grouping factor x
if all your dates are the same width, you can put the dates in a vector and use substring output
The column went_on_backorder is a factor. Linear regression requires a numeric response variable. To use logistic regression, use glm in base R or a package such as vgam. Here’s a brief example: And you can predict your classes:
It happens because R coerces last_visit[,2] to a dimensionless vector, whereas apply expects the object to have some dimensions. You can prevent the coercion by adding drop=F to your command, i.e.: Another way would be just to use lapply or sapply on the vector:
The call to datasetInput1 and datasetInput2 in the two last lines is the reason of the error. You should instead call datasetInput1() and datasetInput2(). Otherwise R tries to convert the function to char. It should be: The () allows you to get the value of the reactive element rather than interact with the reactive element itself. This … Read more
The error message lm.fit(x,y,offset = offset, singular.ok = singular.ok, …) : 0 (non-NA) cases is generated by the lm(y ~ x) command when variables x or y (or both) have only NAs.Here is an example: In your code I suggest to test (just before your lm commands) if one of your variables has all NAs using: In my example: