Description Usage Arguments Value
Convenience function for formatting age values: In child language acquisition studies, age values ofte come in formats like 2;1.10 (year;month.day), which causes problems when sorting the data because they will typically be recognized as character strings, and 2;10.5 will precede 2;1.10 because of the 0 in the first character string. This function helps to circumvent this problem by adding trailing zeros to all one-digit month and day values. Its input can be either a vector or a dataframe. Its output is a vector of formatted numbers in the former case or a dataframe with an added column age_formatted in the latter case. Alternatively, if you use a dataframe as input, you can specify factors = TRUE, in which case the original formatting will be kept but the factor levels will be rearranged in such a way that the temporal order is accurately reflected.
1 | format_age(x, col, month_separator = ";", day_separator = ".", factors = FALSE)
|
x |
A vector or dataframe. |
col |
If x is a dataframe: The column containing the age values. |
month_separator |
sign that separates months from years. Defaults to ; as in 2;4.10 |
day_separator |
sign that separates days from months. Defaults to . as in 2;4.10 |
factors |
If TRUE, the original formatting will be retained but the factor levels will be rearranged to reflect the correct temporal order (if x is a vector). If x is a dataframe, the factor levels will be rearranged in the original column but an age_formatted column will still be appended to the original dataframe. To get rid of the column, use x[,!names(x)=="age_formatted"]. Default is FALSE. |
a dataframe or a vector, depending on the input.
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