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Newbie question: Statistical functions (e.g., mean, sd) in a "transform" statement?
5 messages · Ben Fairbank, Charles C. Berry, Michael Kubovy +2 more
Ben, transform() is probably the wrong tool if what you want is to 'apply a function' to the corresponding elements of time1, time2, ... , and return a vector of results. If this is what you are after, the 'apply' family of functions is what you want. See ?apply and ?mapply and the 'See Also's on each page. Chuck Berry
On Fri, 19 Jan 2007, Ben Fairbank wrote:
Greetings listeRs -
Given a data frame such as
times
time1 time2 time3 time4
1 70.408543 48.92378 7.399605 95.93050
2 17.231940 27.48530 82.962916 10.20619
3 20.279220 10.33575 66.209290 30.71846
4 NA 53.31993 12.398237 35.65782
5 9.295965 NA 48.929201 NA
6 63.966518 42.16304 1.777342 NA
one can use "transform" to total all or some columns, thus,
times2 <- transform(times,totaltime=time1+time2+time3+time4)
times2
time1 time2 time3 time4 totaltime
1 70.408543 48.92378 7.399605 95.93050 222.6624
2 17.231940 27.48530 82.962916 10.20619 137.8863
3 20.279220 10.33575 66.209290 30.71846 127.5427
4 NA 53.31993 12.398237 35.65782 NA
5 9.295965 NA 48.929201 NA NA
6 63.966518 42.16304 1.777342 NA NA
I cannot, however, find a way, other than "for" looping,
to use statistical functions, such as mean or sd, to
compute the new column. For example,
times2<-transform(times,meantime=(mean(c(time1,time2,time3,time4),na.rm= TRUE)))
times2
time1 time2 time3 time4 meantime 1 70.408543 48.92378 7.399605 95.93050 45.54178 2 17.231940 27.48530 82.962916 10.20619 45.54178 3 20.279220 10.33575 66.209290 30.71846 45.54178 4 NA 53.31993 12.398237 35.65782 45.54178 5 9.295965 NA 48.929201 NA 45.54178 6 63.966518 42.16304 1.777342 NA 45.54178 How can this be done? And, generally, what is the recommended method for creating computed new columns in data frames when "for" loops take too long? With thanks for any suggestions, Ben Fairbank Using version 2.4.1 on a Windows XP professional operating system. [[alternative HTML version deleted]]
______________________________________________ R-help at stat.math.ethz.ch 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.
Charles C. Berry (858) 534-2098
Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu UC San Diego
http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0901
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Try this using the builtin data set anscombe: transform(anscombe, rowMeans = rowMeans(anscombe))
On 1/19/07, Ben Fairbank <BEN at ssanet.com> wrote:
Greetings listeRs -
Given a data frame such as
times
time1 time2 time3 time4
1 70.408543 48.92378 7.399605 95.93050
2 17.231940 27.48530 82.962916 10.20619
3 20.279220 10.33575 66.209290 30.71846
4 NA 53.31993 12.398237 35.65782
5 9.295965 NA 48.929201 NA
6 63.966518 42.16304 1.777342 NA
one can use "transform" to total all or some columns, thus,
times2 <- transform(times,totaltime=time1+time2+time3+time4)
times2
time1 time2 time3 time4 totaltime
1 70.408543 48.92378 7.399605 95.93050 222.6624
2 17.231940 27.48530 82.962916 10.20619 137.8863
3 20.279220 10.33575 66.209290 30.71846 127.5427
4 NA 53.31993 12.398237 35.65782 NA
5 9.295965 NA 48.929201 NA NA
6 63.966518 42.16304 1.777342 NA NA
I cannot, however, find a way, other than "for" looping,
to use statistical functions, such as mean or sd, to
compute the new column. For example,
times2<-transform(times,meantime=(mean(c(time1,time2,time3,time4),na.rm= TRUE)))
times2
time1 time2 time3 time4 meantime
1 70.408543 48.92378 7.399605 95.93050 45.54178
2 17.231940 27.48530 82.962916 10.20619 45.54178
3 20.279220 10.33575 66.209290 30.71846 45.54178
4 NA 53.31993 12.398237 35.65782 45.54178
5 9.295965 NA 48.929201 NA 45.54178
6 63.966518 42.16304 1.777342 NA 45.54178
How can this be done? And, generally, what is the recommended method
for creating computed new columns in data frames when "for" loops take
too long?
With thanks for any suggestions,
Ben Fairbank
Using version 2.4.1 on a Windows XP professional operating system.
[[alternative HTML version deleted]]
______________________________________________ R-help at stat.math.ethz.ch 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.
On Fri, 2007-01-19 at 11:54 -0600, Ben Fairbank wrote:
Greetings listeRs -
Here are two solutions, depending on whether you wanted the NA's or not, and I assume you wanted the row means:
times3 <- transform(times, meantime = rowMeans(times)) times3
time1 time2 time3 time4 meantime 1 70.408543 48.92378 7.399605 95.93050 55.66561 2 17.231940 27.48530 82.962916 10.20619 34.47159 3 20.279220 10.33575 66.209290 30.71846 31.88568 4 NA 53.31993 12.398237 35.65782 NA 5 9.295965 NA 48.929201 NA NA 6 63.966518 42.16304 1.777342 NA NA
times4 <- transform(times, meantime = rowMeans(times, na.rm = TRUE)) times4
time1 time2 time3 time4 meantime 1 70.408543 48.92378 7.399605 95.93050 55.66561 2 17.231940 27.48530 82.962916 10.20619 34.47159 3 20.279220 10.33575 66.209290 30.71846 31.88568 4 NA 53.31993 12.398237 35.65782 33.79200 5 9.295965 NA 48.929201 NA 29.11258 6 63.966518 42.16304 1.777342 NA 35.96897 HTH G
Given a data frame such as
times
time1 time2 time3 time4
1 70.408543 48.92378 7.399605 95.93050
2 17.231940 27.48530 82.962916 10.20619
3 20.279220 10.33575 66.209290 30.71846
4 NA 53.31993 12.398237 35.65782
5 9.295965 NA 48.929201 NA
6 63.966518 42.16304 1.777342 NA
one can use "transform" to total all or some columns, thus,
times2 <- transform(times,totaltime=time1+time2+time3+time4)
times2
time1 time2 time3 time4 totaltime
1 70.408543 48.92378 7.399605 95.93050 222.6624
2 17.231940 27.48530 82.962916 10.20619 137.8863
3 20.279220 10.33575 66.209290 30.71846 127.5427
4 NA 53.31993 12.398237 35.65782 NA
5 9.295965 NA 48.929201 NA NA
6 63.966518 42.16304 1.777342 NA NA
I cannot, however, find a way, other than "for" looping,
to use statistical functions, such as mean or sd, to
compute the new column. For example,
times2<-transform(times,meantime=(mean(c(time1,time2,time3,time4),na.rm= TRUE)))
times2
time1 time2 time3 time4 meantime 1 70.408543 48.92378 7.399605 95.93050 45.54178 2 17.231940 27.48530 82.962916 10.20619 45.54178 3 20.279220 10.33575 66.209290 30.71846 45.54178 4 NA 53.31993 12.398237 35.65782 45.54178 5 9.295965 NA 48.929201 NA 45.54178 6 63.966518 42.16304 1.777342 NA 45.54178 How can this be done? And, generally, what is the recommended method for creating computed new columns in data frames when "for" loops take too long? With thanks for any suggestions, Ben Fairbank Using version 2.4.1 on a Windows XP professional operating system. [[alternative HTML version deleted]]
______________________________________________ R-help at stat.math.ethz.ch 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.
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