How can I strptime a character like "15/09/2016 20h26"?
I searched this on the strptime official manual and arround the Stack overflow, but I didn't find anything that could help me.
I tried this code:
u <- ("15/09/2016 20h26")
strptime(u, format = "%d/%m/%Y %H:%M")
and
strptime(u, format = "%d/%m/%Y %H'h'%M")
Anyone has any idea of what I need to do?
Edit:
Firstly, sorry for the clearness of the question.
I tried to use a sample to describe my problem, but it seems like a typo.
I have a txt file with 400 observations of time in the format "20/01/2016 14h00". I'm trying to convert this to date with strptime:
here is my code:
y <- read.table('materias_data.txt', sep = ";")
l <- strptime(y, format = "%d/%m/%Y %Hh%M")
I got this error when I run the code:
Error in strptime(y, format = "%d/%m/%Y %Hh%M") : input string is too long
What can I do to convert correctly the data.frame?
Edit 2: Solution
Reading the comments I realized what the problem is:
strptime doesn't work to data.frame but it works on characters.
The step to solve this is
y <- read.table('materias_data.txt', sep = ";")
l <- strptime(y[,1], format = "%d/%m/%Y %Hh%M")
df <- data.frame(l)
Thank You
The relatively new anytime package I added to CRAN parses this automatically without extra help:
R> anytime("15/09/2016 20h26")
[1] "2016-09-15 20:26:00 CDT"
R>
Another nice feature is that it also automatically converts from factor (or ordered) to character.
So if you have a data.frame df with a column of values like this in, say, a column called datestr then just pass the column:
R> anytime( df[, "datestr"] )
which you can assign to a new column in the data.frame as well:
R> df[, "parsedtime"] <- anytime( df[, "datestr"] )