harrelfe/Hmisc

Median if labelled

zachariae opened this issue · 3 comments

zachariae commented

Dear all,

during the last days, I faced a problem. I was no longer able to calculate the median from a column when it was labelled using the Hmisc function.

Greetings, Silke

SamGG commented

Hi,
From a basic R point of view, if the "labelled" objects are the values of the columns, then this column is a factor, and IIRC no numerical computation could be done on a factor. Do you have a short and reproducible example?
Best.

zachariae commented

Hi,
thank you for the clarification. Actually the problem with calculating the median occurs not with df, but when the df has other classes as well such as tbl_df...- So you may not be the right address to report the problem?

MG, Silke

test_data <- data.frame(AGE = c(40.9, 38.6, 33.4, 25.0, 29.5, 31.5,
                                31.5, 25.9, 31.4, 27.6, 42.4, 29.8,
                                31.6, 43.4, 31.3, 35.7, 35.7, 29.4, 
                                42.4, 47.7, 40.3, 45.2, 36.1, 38.6, 48.4, 28.1,
                                41.8, 41.0, 32.4, 34.8, 28.3))



str(test_data)

test_median <- test_data %>%
  dplyr::select(AGE) %>%
  median()

test_mean <- test_data %>%
  dplyr::select(AGE) %>%
  mean()
SamGG commented

IMHO the error relates to the use of dplyr, not Hmisc.
The select() returns a data.frame, not a vector, that's why median() is complaining.
You should use summarise to apply the median

test_data %>%
  summarise(median(AGE))

Answers at
https://stackoverflow.com/questions/61158926/median-in-r-needs-numeric-data

Personal point of view: I didn't move to those nice functions, because every time I tried I get such error due to that I am so familiar with the base functions.

Please close the issue unless you feel it's related to the use of a Hmisc function.
Have a nice day.

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