[R] read.csv interpreting numbers as factors
Jeff Newmiller
jdnewmil at dcn.davis.CA.us
Tue Dec 10 11:06:47 CET 2013
It is bad netiquette to hijack an existing thread for a new topic. Please start a new email thread when changing topics.
If your data really consists of what you show, then read.csv won't behave that way. I suggest that you open the file in a text editor and look for odd characters. They may be invisible.
Going out on a limb, you may be trying to read a tab separated file, and if so then you need to use the sep=”\t" argument to read.csv.
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Bill <william108 at gmail.com> wrote:
>Why does R interpret a column of numbers in a csv file as a factor when
>using read.csv() and how can I prevent that. The data looks like
>
>9928
>3502
>146
>404
>1831
>686
>249
>
>I tried kick=read.csv("kick.csv",stringsAsFactors =FALSE)
>as well as
>kick=read.csv("kick.csv")
>
>Thanks
>
>
>On Mon, Dec 2, 2013 at 5:16 PM, William Dunlap <wdunlap at tibco.com>
>wrote:
>
>> > It seems so inefficient.
>>
>> But ifelse knows nothing about the expressions given
>> as its second and third arguments -- it only sees their
>> values after they are evaluated. Even if it could see the
>> expressions, it would not be able to assume that f(x[i])
>> is the same as f(x)[i] or things like
>> ifelse(x>0, cumsum(x), cumsum(-x))
>> would not work.
>>
>> You can avoid the computing all of f(x) and then extracting
>> a few elements from it by doing something like
>> x <- c("Wednesday", "Monday", "Wednesday")
>> z1 <- character(length(x))
>> z1[x=="Monday"] <- "Mon"
>> z1[x=="Tuesday"] <- "Tue"
>> z1[x=="Wednesday"] <- "Wed"
>> or
>> LongDayNames <- c("Monday","Tuesday","Wednesday")
>> ShortDayNames <- c("Mon", "Tue", "Wed")
>> z2 <- character(length(x))
>> for(i in seq_along(LongDayNames)) {
>> z2[x==LongDayNames[i]] <- ShortDayNames[i]
>> }
>>
>> To avoid the repeated x==value[i] you can use match(x, values).
>> z3 <- ShortDayNames[match(x, LongDayNames)]
>>
>> z1, z2, and z3 are identical character vectors.
>>
>> Or, you can use factors.
>> > factor(x, levels=LongDayNames, labels=ShortDayNames)
>> [1] Wed Mon Wed
>> Levels: Mon Tue Wed
>>
>> Bill Dunlap
>> Spotfire, TIBCO Software
>> wdunlap tibco.com
>>
>>
>> > -----Original Message-----
>> > From: r-help-bounces at r-project.org
>[mailto:r-help-bounces at r-project.org]
>> On Behalf
>> > Of Bill
>> > Sent: Monday, December 02, 2013 4:50 PM
>> > To: Duncan Murdoch
>> > Cc: r-help at r-project.org
>> > Subject: Re: [R] ifelse -does it "manage the indexing"?
>> >
>> > It seems so inefficient. I mean the whole first vector will be
>evaluated.
>> > Then if the second if is run the whole vector will be evaluated
>again.
>> Then
>> > if the next if is run the whole vector will be evaluted again. And
>so on.
>> > And this could be only to test the first element (if it is false
>for each
>> > if statement). Then this would be repeated again and again. Is that
>> really
>> > the way it works? Or am I not thinking clearly?
>> >
>> >
>> > On Mon, Dec 2, 2013 at 4:48 PM, Duncan Murdoch
>> > <murdoch.duncan at gmail.com>wrote:
>> >
>> > > On 13-12-02 7:33 PM, Bill wrote:
>> > >
>> > >> ifelse ((day_of_week == "Monday"),1,
>> > >> ifelse ((day_of_week == "Tuesday"),2,
>> > >> ifelse ((day_of_week == "Wednesday"),3,
>> > >> ifelse ((day_of_week == "Thursday"),4,
>> > >> ifelse ((day_of_week == "Friday"),5,
>> > >> ifelse ((day_of_week == "Saturday"),6,7)))))))
>> > >>
>> > >>
>> > >> In code like the above, day_of_week is a vector and so
>day_of_week
>> ==
>> > >> "Monday" will result in a boolean vector. Suppose day_of_week is
>> Monday,
>> > >> Thursday, Friday, Tuesday. So day_of_week == "Monday" will be
>> > >> True,False,False,False. I think that ifelse will test the first
>> element
>> > >> and
>> > >> it will generate a 1. At this point it will not have run
>day_of_week
>> ==
>> > >> "Tuesday" yet. Then it will test the second element of
>day_of_week
>> and it
>> > >> will be false and this will cause it to evaluate day_of_week ==
>> "Tuesday".
>> > >> My question would be, does the evaluation of day_of_week ==
>"Tuesday"
>> > >> result in the generation of an entire boolean vector (which
>would be
>> in
>> > >> this case False,False,False,True) or does the ifelse "manage the
>> indexing"
>> > >> so that it only tests the second element of the original vector
>> (which is
>> > >> Thursday) and for that matter does it therefore not even bother
>to
>> > >> generate
>> > >> the first boolean vector I mentioned above
>(True,False,False,False)
>> but
>> > >> rather just checks the first element?
>> > >> Not sure if I have explained this well but if you understand
>I
>> would
>> > >> appreciate a reply.
>> > >>
>> > >
>> > > See the help for the function. If any element of the test is
>true, the
>> > > full first vector will be evaluated. If any element is false,
>the
>> second
>> > > one will be evaluated. There are no shortcuts of the kind you
>> describe.
>> > >
>> > > Duncan Murdoch
>> > >
>> > >
>> >
>> > [[alternative HTML version deleted]]
>> >
>> > ______________________________________________
>> > R-help at r-project.org mailing list
>> > 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.
>>
>
> [[alternative HTML version deleted]]
>
>______________________________________________
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