On 9/23/2015 5:57 AM, Therneau, Terry M., Ph.D. wrote:
Thanks for all for the comments, I hadn't intended to start a war.My summary:1. Most important: I wasn't missing something obvious. This is always my first suspicion when I submit something to R-help, and it's true more often than not.2. Obviously (at least it is now), the CSV standard does not specify that quotes should force a character result. R is not "wrong". Wrt to using what Excel does as litmus test, I consider that to be totally uninformative about standards: neither pro (like Duncan) or anti (like Rolf), but simply irrelevant. (Like many MS choices.)3. I'll have to code in my own solution, either pre-scan the first few lines to create a colClasses, or use read_csv from the readr library (if there are leading zeros it keeps the string as character, which may suffice for my needs), or something else.4. The source of the data is a "text/csv" field coming from an http POST request. This is an internal service on an internal Mayo server and coded by our own IT department; this will not be the first case where I have found that their definition of "csv" is not quite standard.Terry T.On 23/09/15 10:00, Therneau, Terry M., Ph.D. wrote:I have a csv file from an automatic process (so this will happen thousands of times), for which the first row is a vector of variable names and the second row often starts something like this: 5724550,"000202075214",2005.02.17,2005.02.17,"F", ..... Notice the second variable which is a character string (note the quotation marks) a sequence of numeric digits leading zeros are significant The read.csv function insists on turning this into a numeric. Is there any simple set of options that will turn this behavior off? I'm looking for a way to tell it to "obey the bloody quotes" -- I still want the first, third, etc columns to become numeric. There can be more than one variable like this, and not always in the second position. This happens deep inside the httr library; there is an easy way for me to add more options to the read.csv call but it is not so easy to replace it with something else.______________________________________________ [email protected] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
A fairly simple workaround is to add two lines of code to the process, and then add the colClasses parameter as you suggested in item 2 above.
want <- read.csv('yourfile', quote='', stringsAsFactors= FALSE, nrows=1)
classes <- sapply(want, class)
want <- read.csv('yourfile', stringsAsFactors= FALSE, colClasses=classes)
I don't know if you want your final file to convert strings to factors,
so you can modify as needed. In addition, if your files aren't as
regular as I inferred, you can increase the number of rows to read in
the first line to ensure getting the classes right.
Hope this is helpful, Dan -- Daniel Nordlund Bothell, WA USA ______________________________________________ [email protected] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
