In order to save a dataframe as a csv file, I had to convert a list-type column of the dataframe that was a list of vectors of different lengths and of different dates.
ex.
To be able to save the dataframe, I did DF$dates <- as.character(DF$dates) and then used write.csv() to save it. I want to be able to read in this file and use this column as a list column of vectors again, but the resulting column after reading in the file is a column of characters that is difficult to work with.
The rows are all variations of this: c("12/01/23", "12/02/23", "12/04/23", "12/05/23", "12/07/23", "12/13/23", "12/14/23", "12/15/23", "12/16/23", "12/17/23", "12/19/23", "12/20/23", "12/21/23", "12/22/23", "12/23/23", "12/24/23", "12/25/23", "12/30/23", "12/31/23")" which is just one long string that I am not able to use for much.
I’ve tried several things such as as.list() on a cell, using str_sub() to remove the parentheses, str_replace_all() to replace the backslashes, and strsplit() to separate the dates into different list objects and nothing has been simple or worked well enough.
>Solution :
You can re-parse the contents of "dates" (c() and all) and reattach it in the following way. I’m not sure if you’re using dplyr or not, so here is a base R solution:
DF <- data.frame(dates = 'c("12/01/23", "12/02/23", "12/04/23", "12/05/23", "12/07/23", "12/13/23", "12/14/23", "12/15/23", "12/16/23", "12/17/23", "12/19/23", "12/20/23", "12/21/23", "12/22/23", "12/23/23", "12/24/23", "12/25/23", "12/30/23", "12/31/23")')
DF$dates_list <- lapply(DF$dates, \(x) eval(str2expression(x)))