I am also adding an answer that augments this a little bit because I came to this entry when searching for an answer, and this had almost what I needed, but I needed a bit more, which I got via @MrFlik 's answer and the R lazyeval vignettes.
I wanted to make a function that could take a dataframe and a vector of column names (as strings) that I want to be converted from a string to a Date object. I couldn't figure out how to make as.Date()
take an argument that is a string and convert it to a column, so I did it as shown below.
Below is how I did this via SE mutate (mutate_()
) and the .dots
argument. Criticisms that make this better are welcome.
library(dplyr)
dat <- data.frame(a="leave alone",
dt="2015-08-03 00:00:00",
dt2="2015-01-20 00:00:00")
# This function takes a dataframe and list of column names
# that have strings that need to be
# converted to dates in the data frame
convertSelectDates <- function(df, dtnames=character(0)) {
for (col in dtnames) {
varval <- sprintf("as.Date(%s)", col)
df <- df %>% mutate_(.dots= setNames(list(varval), col))
}
return(df)
}
dat <- convertSelectDates(dat, c("dt", "dt2"))
dat %>% str