Re: [R] Difference between times
On 18/04/2014 21:46, David Winsemius wrote:
On Apr 18, 2014, at 12:59 PM, Prof Brian Ripley wrote:
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've misspelled CET (not CEST)
Neither CET nor CEST are portable time-zone names. We have not been given the
'at a minimum' information required by the posting guide, so please read
?Sys.timezone on your system.
Dear Prof;
Thanks for the impetus to yet again read that page. Despite frequently reading help pages and in
particular reading that one many times, I still was not getting the 'tz' arguments correct on a
Mac. I do now see that I was spelling my TZ incorrectly (as Americas/Los_Angeles rather
than America/Los_Angeles.
Fellow Mac users may face a problem when using the Finder unless they set it up to
display hidden ('dot') files. The /usr/ folder is greyed out but it still
does open. If I restore my Finder defaults to not show system files and folders, I no
longer see that directory and would not have been able to resolve my spelling error on my
own:
dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
America/New_York)
dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
America/Los_Angeles)
dt1-dt2
Time difference of 3 hours
I don't suppose a warning could be issued by the as.POSIXct code when a tz
argument is not found in the database to let people know that 'UTC' will be the default?
No, as the underlying POSIX function does not report this. We could
perhaps do this on platforms which use --with-internal-tzcode but not
e.g. on Linux.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@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.
Re: [R] Difference between times
On 20/04/2014 08:50, Prof Brian Ripley wrote:
On 18/04/2014 21:46, David Winsemius wrote:
On Apr 18, 2014, at 12:59 PM, Prof Brian Ripley wrote:
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've misspelled CET (not CEST)
Neither CET nor CEST are portable time-zone names. We have not been
given the 'at a minimum' information required by the posting guide,
so please read ?Sys.timezone on your system.
Dear Prof;
Thanks for the impetus to yet again read that page. Despite frequently
reading help pages and in particular reading that one many times, I
still was not getting the 'tz' arguments correct on a Mac. I do now
see that I was spelling my TZ incorrectly (as Americas/Los_Angeles
rather than America/Los_Angeles.
Fellow Mac users may face a problem when using the Finder unless they
set it up to display hidden ('dot') files. The /usr/ folder is greyed
out but it still does open. If I restore my Finder defaults to not
show system files and folders, I no longer see that directory and
would not have been able to resolve my spelling error on my own:
You can always use the command-line or OlsonNames() in R.
dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
America/New_York)
dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
America/Los_Angeles)
dt1-dt2
Time difference of 3 hours
I don't suppose a warning could be issued by the as.POSIXct code when
a tz argument is not found in the database to let people know that
'UTC' will be the default?
No, as the underlying POSIX function does not report this. We could
perhaps do this on platforms which use --with-internal-tzcode but not
e.g. on Linux.
In fact we already do: R 3.1.0 on a Mac shows
as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
Americas/New_York)
[1] 2014-04-18 09:00:00 GMT
Warning messages:
1: In strptime(x, format, tz = tz) : unknown timezone 'Americas/New_York'
2: In as.POSIXct.POSIXlt(as.POSIXlt(x, tz, ...), tz, ...) :
unknown timezone 'Americas/New_York'
3: In as.POSIXlt.POSIXct(x, tz) : unknown timezone 'Americas/New_York'
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@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.
[R] to divide column cells by the mean of another column
Dear all,
I am getting data columnwise that I need to divide by the mean of another
column
If the column is the previous one this code works perfectly well:
fun1 - function(beginColumn, by, data) { indx - seq(beginColumn,
ncol(data), by = by) as.data.frame(t(100 - (t(data[, indx])/colMeans(data[,
indx - 1], na.rm = TRUE)) * 100))
}
(Arun helped me with this code, thank you again!...)
But, the things is now more complicated...
I need to program a function that allow me to divide for example cells from
column 3 on mean from column 2 and cells from column 4 on mean of column 2
and the 5 etc. Then column 6 is another column from whch I need to extract
the mean and to do the same with column 7 and 8, etc...
so if I have:
1 2 3 4 1 5
2 5 4 7 2 8
3 4 5 9 3 7
4 7 7 9 4 3
The serie 1,2,3,4 ar just enumerating so not useful at this timepoint.
the results should be (from excel...):
4,5 33,333 11,111 -11,111 11,111 -55,556 -77,778
-11,111 -100 -55,556 -55,556 -100
33,333
I tried to work on modyfying indx-1 by 2*indx-2, but this is not doing the
job... I tried many other things so that I am now stucked.
Does Anyone has a brilliant idea?
Many many thanks
André ZACHARIA
[[alternative HTML version deleted]]
__
R-help@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.
Re: [R] Extracting the names of coefficients of random effects
Brian Willis b.h.willis at bham.ac.uk writes: Hi All, I need to be able to manipulate the names of the coefficients from *ranef()*. If there is any missing data when fitting a mixed model using lmer, no estimate is returned for the associated level for that random effect. Thus if the data input for regions had levels *Region* Bolton Bradford Cambridge Durham and there was missing data on Bradford then * ranef(model)* gives (Intercept) Bolton: -0.0981763413 Cambridge0.0151102347 Durham 0.1837142259 I think you want to use rownames(): library(lme4) d - expand.grid(f=factor(LETTERS[1:10]),rep=1:10) d$y - rnorm(100) m - lmer(y~(1|f),data=d) rownames(ranef(m)[[1]]) This becomes a problem if I want to use *predict( )* on new data where there is no missing data on Bradford. In such an instance *predict (model, newdata = newInput) * gives the following error message ‘Error in (function (x, n) : new levels detected in newdata’ I could get round this by checking the Region field of the new data ‘newInput’ against the names of the levels of the intercept coefficients from* ranef().* However, I’m not sure how to access these since if *x- ranef(model) x * You should also check the allow.new.levels argument in ?predict.merMod, and send followups to r-sig-mixed-mod...@r-project.org. __ R-help@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.
Re: [R] How to use rainbow function without the gamma argument
Hi Boris, yes I tried this way and it worked. The fact is that I wanted to be compliant with the old code, I did not want to change anything. So I wanted to find a new way to rewrite the code. Thanks ~ Francesco Brundu On 19 April 2014 23:18, Boris Steipe boris.ste...@utoronto.ca wrote: Have you looked at ?rainbow ? Is there a reason why you don't simply leave the gamma parameter away? Try: pie(rep(1,100), col=rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75)) Cheers, B. On 2014-04-19, at 6:05 AM, Francesco Brundu wrote: Hi all, I am using an old code (probably written for R 2.5) and it stops when calling rainbow() with gamma argument. I saw that gamma argument is not present in newer version of R rainbow function. How can I translate this line of code: rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5) ? It fails with: Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5) : unused argument (gamma = 1.5) Calls: nmfconsensus ... matrix.abs.plot - image - image.default - rainbow Execution halted Thanks ~ Francesco Brundu [[alternative HTML version deleted]] __ R-help@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]] __ R-help@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.
Re: [R] How to use rainbow function without the gamma argument
Thanks Boris for the detailed answer. I thought to make it backward
compatible because I did not know if gamma parameter would change the
result if absent. As far as I can see changes in my results are not
relevant, so I think to follow your advice to update the old code.
Thanks
~ Francesco Brundu
On 20 April 2014 02:16, Boris Steipe boris.ste...@utoronto.ca wrote:
If it MUST be parameter-compatible with the old call, you could just add
... to your local version of rainbow. The unused parameter will then be
dropped.
Here's how:
# The original creates an error ...
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5)
Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5)
:
unused argument (gamma = 1.5)
# The code of the function is here:
rainbow
function (n, s = 1, v = 1, start = 0, end = max(1, n - 1)/n,
alpha = 1)
{
if ((n - as.integer(n[1L])) 0) {
if (start == end || any(c(start, end) 0) || any(c(start,
end) 1))
stop('start' and 'end' must be distinct and in [0, 1].)
hsv(h = seq.int(start, ifelse(start end, 1, 0) + end,
length.out = n)%%1, s, v, alpha)
}
else character()
}
bytecode: 0x101968950
environment: namespace:grDevices
# I add ... to the parameters and define a local version of rainbow
rainbow = function (n, s = 1, v = 1, start = 0, end = max(1, n - 1)/n,
alpha = 1, ...)
{
if ((n - as.integer(n[1L])) 0) {
if (start == end || any(c(start, end) 0) || any(c(start,
end) 1))
stop('start' and 'end' must be distinct and in [0, 1].)
hsv(h = seq.int(start, ifelse(start end, 1, 0) + end,
length.out = n)%%1, s, v, alpha)
}
else character()
}
# same code except for the dots...
# Now it accepts and discards unused arguments
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5,
gefingerpoken = TRUE)
[1] #BFFF #BF0900FF #BF1100FF #BF1A00FF #BF2300FF
#BF2B00FF #BF3400FF
[...]
[99] #5700BFFF #6000BFFF
# and if I want the original back, I just delete my local version ...
rm(rainbow)
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5)
Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5)
:
unused argument (gamma = 1.5)
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75)
[1] #BFFF #BF0900FF #BF1100FF #BF1A00FF #BF2300FF
#BF2B00FF #BF3400FF
[...]
[99] #5700BFFF #6000BFFF
You can read about the '...' argument in the Introduction to R. Here we
use it not to pass variables on, but to have them not cause an error when
present.
HOWEVER: I personally would consider this poor style. It's probably better
to review and update your old code. There may be other less obvious
problems.
YMMV
B.
On 2014-04-19, at 7:19 PM, Francesco Brundu wrote:
Hi Boris,
yes I tried this way and it worked. The fact is that I wanted to be
compliant with the old code, I did not want to change anything. So I wanted
to find a new way to rewrite the code.
Thanks
~ Francesco Brundu
On 19 April 2014 23:18, Boris Steipe boris.ste...@utoronto.ca wrote:
Have you looked at ?rainbow ?
Is there a reason why you don't simply leave the gamma parameter away?
Try:
pie(rep(1,100), col=rainbow(100, s = 1.0, v = 0.75, start = 0.0, end =
0.75))
Cheers,
B.
On 2014-04-19, at 6:05 AM, Francesco Brundu wrote:
Hi all,
I am using an old code (probably written for R 2.5) and it stops when
calling rainbow() with gamma argument. I saw that gamma argument is not
present in newer version of R rainbow function. How can I translate
this
line of code:
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5)
?
It fails with:
Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma =
1.5)
:
unused argument (gamma = 1.5)
Calls: nmfconsensus ... matrix.abs.plot - image - image.default -
rainbow
Execution halted
Thanks
~ Francesco Brundu
[[alternative HTML version deleted]]
__
R-help@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]]
__
R-help@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.
Re: [R] Need help to convert data frame to transaction set.
Hi Arul, Thanks for your reply. I am trying your approach now. I will reply once again with my result. Regards, Sathish On Sat, Apr 19, 2014 at 6:43 PM, arun smartpink...@yahoo.com wrote: Hi, Without a reproducible example using ?dput() or the package name, it is a bit difficult to comment. Assuming that you used: library(arules) data(AdultUCI) AdultUCI$ID - 1:nrow(AdultUCI) lst1 - split(AdultUCI[,ID], AdultUCI[,marital-status]) as(lst1, transactions) #transactions in sparse format with # 7 transactions (rows) and # 48842 items (columns) A.K. On Saturday, April 19, 2014 5:01 PM, Sathish Kumar tellsath...@gmail.com wrote: Hi, To convert coerce the data set to transaction data set I used the code trans4 - as(split(a[,Cust_ID], a[,Parts]), transactions) but I am getting the following error- Error in as(split(a[, Cust_ID], a[, Parts]), transactions) : nomethod or default for coercing âlistâ to âtransactionsâ Then I tried first converting the data set to matrix structure using the code c_m-as.matrix(c_df) c_m then entered the following code trans2 - as(c_m, transactions) but got the following error Error in as(c_m, transactions) : no method or default for coercing âmatrixâ to âtransactionsâ Please let me know how to correct the problem. Thanks Sathish __ R-help@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]] __ R-help@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.
Re: [R] Need help to convert data frame to transaction set.
Hi AK, First, your timely response is greatly appreciated. Contrary to what you are getting, I am getting the below when I enter the same code lines1 - readLines(spares.csv) head(lines1) [1] Cust_ID,Parts1,FENDERS2,FENDERS [4] 3,OIL FILTERS4,BMW SERVICE FLUIDS 4,OIL FILTER Why is that ? similarly, I get a different output when I have the same code you have given. part-read.csv(spares.csv,header=TRUE,sep=\t) str(part) 'data.frame': 838 obs. of 1 variable: $ Cust_ID.Parts: Factor w/ 838 levels 1,FENDERS,10,BUMPERS,..: 1 162 334 496 497 729 730 731 771 772 ... What is the mistake I am doing? Also, can you please explain below code? part$Parts - interaction(part[,2:5],sep= ,drop=TRUE) part - part[,1:2] Aru, I need to submit my class assignment tomorrow. I have to waiting for answers for last one week in different forums, you are the only person to reply. many thanks for that. Thanks Sathish On Sat, Apr 19, 2014 at 10:36 PM, arun smartpink...@yahoo.com wrote: Hi Satish, Using your code: part-read.csv(spares.csv) str(part) #'data.frame':838 obs. of 1 variable: # $ Cust_ID.Parts: Factor w/ 838 levels 100\tAIR\tFILTER\tHOUSING\t,..: 161 333 495 727 728 768 769 770 784 785 ... #I guess you have only two columns in the dataset lines1 - readLines(spares.csv) head(lines1) #[1] Cust_ID\tParts\t\t\t 1\tFENDERS\t\t\t #[3] 2\tFENDERS\t\t\t 3\tOIL\tFILTERS\t\t #[5] 4\tBMW\tSERVICE\tFLUIDS\t 4\tOIL\tFILTER\t\t part-read.csv(spares.csv,header=TRUE,sep=\t) str(part) #'data.frame':838 obs. of 5 variables: # $ Cust_ID: int 1 2 3 4 4 5 5 5 6 6 ... # $ Parts : Factor w/ 26 levels AIR,ALTERNATOR,..: 9 9 16 4 16 6 9 12 6 9 ... # $ X : Factor w/ 14 levels ,BLADES,CAR,..: 1 1 6 14 5 1 1 1 1 1 ... # $ X.1: Factor w/ 6 levels ,(AXLE,CARE,..: 1 1 1 4 1 1 1 1 1 1 ... # $ X.2: Factor w/ 2 levels ,BOOT): 1 1 1 1 1 1 1 1 1 1 ... part$Parts - interaction(part[,2:5],sep= ,drop=TRUE) part - part[,1:2] str(part) #'data.frame':838 obs. of 2 variables: # $ Cust_ID: int 1 2 3 4 4 5 5 5 6 6 ... # $ Parts : Factor w/ 31 levels ALTERNATOR ,..: 6 6 21 28 20 3 6 8 3 6 ... as(split(part[,Cust_ID],part[,Parts]),transactions) #transactions in sparse format with # 31 transactions (rows) and # 502 items (columns) A.K. On Saturday, April 19, 2014 11:08 PM, Sathish Kumar tellsath...@gmail.com wrote: Hi AK, I have attached the data set. And as you suggested, I coded as follows but still I am getting an error part-read.csv(spares.csv) part lst1-split(part[,Cust_ID],part[,Parts]) as(lst1, transactions) Error in as(lst1, transactions) : no method or default for coercing âlistâ to âtransactionsâ On Sat, Apr 19, 2014 at 6:43 PM, arun smartpink...@yahoo.com wrote: Hi, Without a reproducible example using ?dput() or the package name, it is a bit difficult to comment. Assuming that you used: library(arules) data(AdultUCI) AdultUCI$ID - 1:nrow(AdultUCI) lst1 - split(AdultUCI[,ID], AdultUCI[,marital-status]) as(lst1, transactions) #transactions in sparse format with # 7 transactions (rows) and # 48842 items (columns) A.K. On Saturday, April 19, 2014 5:01 PM, Sathish Kumar tellsath...@gmail.com wrote: Hi, To convert coerce the data set to transaction data set I used the code trans4 - as(split(a[,Cust_ID], a[,Parts]), transactions) but I am getting the following error- Error in as(split(a[, Cust_ID], a[, Parts]), transactions) : nomethod or default for coercing âlistâ to âtransactionsâ Then I tried first converting the data set to matrix structure using the code c_m-as.matrix(c_df) c_m then entered the following code trans2 - as(c_m, transactions) but got the following error Error in as(c_m, transactions) : no method or default for coercing âmatrixâ to âtransactionsâ Please let me know how to correct the problem. Thanks Sathish __ R-help@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]] __ R-help@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.
Re: [R] to divide column cells by the mean of another column
R has no cells.
You need to do your homework by reading An Introduction to R , which
ships with R, or one of the many R web tutorials of your choice. What
you describe is trivial once you have made a minimal effort to learn
R. In particular, ?[ explains how to index data frames; but a
tutorial is a better option for a learner.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Sun, Apr 20, 2014 at 2:52 AM, Andre Zacharia
andre.zacha...@gmail.com wrote:
Dear all,
I am getting data columnwise that I need to divide by the mean of another
column
If the column is the previous one this code works perfectly well:
fun1 - function(beginColumn, by, data) { indx - seq(beginColumn,
ncol(data), by = by) as.data.frame(t(100 - (t(data[, indx])/colMeans(data[,
indx - 1], na.rm = TRUE)) * 100))
}
(Arun helped me with this code, thank you again!...)
But, the things is now more complicated...
I need to program a function that allow me to divide for example cells from
column 3 on mean from column 2 and cells from column 4 on mean of column 2
and the 5 etc. Then column 6 is another column from whch I need to extract
the mean and to do the same with column 7 and 8, etc...
so if I have:
1 2 3 4 1 5
2 5 4 7 2 8
3 4 5 9 3 7
4 7 7 9 4 3
The serie 1,2,3,4 ar just enumerating so not useful at this timepoint.
the results should be (from excel...):
4,5 33,333 11,111 -11,111 11,111 -55,556 -77,778
-11,111 -100 -55,556 -55,556 -100
33,333
I tried to work on modyfying indx-1 by 2*indx-2, but this is not doing the
job... I tried many other things so that I am now stucked.
Does Anyone has a brilliant idea?
Many many thanks
André ZACHARIA
[[alternative HTML version deleted]]
__
R-help@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.
__
R-help@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.
Re: [R] to divide column cells by the mean of another column
Hi,
May be this helps:
fun1 - function(beginColumn, by, data) {
indx - seq(beginColumn, ncol(data), by = by)
dataNew - data[, indx[1]:ncol(data)]
indx1 - cumsum(seq(ncol(data)) %in% indx)
indx2 - indx1[indx1 != 0]
lst1 - lapply(split(seq_along(indx2), indx2), function(i) {
x1 - dataNew[, i, drop = FALSE]
if (ncol(x1) 1) {
x1[, -1]/mean(x1[, 1])
}
})
res - data.frame(lst1[sapply(lst1, length) 0])
colnames(res) - gsub(.*\\., , colnames(res))
res
}
set.seed(458)
dat1 - as.data.frame(matrix(sample(5,10*5,replace=TRUE),ncol=10))
set.seed(34)
dat2 - as.data.frame(matrix(sample(20,21*5,replace=TRUE),ncol=21))
fun1(2,4,dat1)
fun1(2,5,dat2)
A.K.
On Sunday, April 20, 2014 5:55 AM, Andre Zacharia andre.zacha...@gmail.com
wrote:
Dear all,
I am getting data columnwise that I need to divide by the mean of another
column
If the column is the previous one this code works perfectly well:
fun1 - function(beginColumn, by, data) { indx - seq(beginColumn,
ncol(data), by = by) as.data.frame(t(100 - (t(data[, indx])/colMeans(data[,
indx - 1], na.rm = TRUE)) * 100))
}
(Arun helped me with this code, thank you again!...)
But, the things is now more complicated...
I need to program a function that allow me to divide for example cells from
column 3 on mean from column 2 and cells from column 4 on mean of column 2
and the 5 etc. Then column 6 is another column from whch I need to extract
the mean and to do the same with column 7 and 8, etc...
so if I have:
1 2 3 4 1 5
2 5 4 7 2 8
3 4 5 9 3 7
4 7 7 9 4 3
The serie 1,2,3,4 ar just enumerating so not useful at this timepoint.
the results should be (from excel...):
4,5 33,333 11,111 -11,111 11,111 -55,556 -77,778
-11,111 -100 -55,556 -55,556 -100
33,333
I tried to work on modyfying indx-1 by 2*indx-2, but this is not doing the
job... I tried many other things so that I am now stucked.
Does Anyone has a brilliant idea?
Many many thanks
André ZACHARIA
[[alternative HTML version deleted]]
__
R-help@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.
__
R-help@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.
Re: [R] to divide column cells by the mean of another column
Hi Andre,
A slight correction:
fun1 - function(beginColumn, by, data) {
indx - seq(beginColumn, ncol(data), by = by)
dataNew - data[, indx[1]:ncol(data)]
indx1 - cumsum(seq(ncol(data)) %in% indx)
indx2 - indx1[indx1 != 0]
lst1 - lapply(split(seq_along(indx2), indx2), function(i) {
x1 - dataNew[, i, drop = FALSE]
if (ncol(x1) 1) {
x1[, -1, drop = FALSE]/mean(x1[, 1]) ###changed
}
})
res - data.frame(lst1[sapply(lst1, length) 0])
colnames(res) - gsub(.*\\., , colnames(res))
res
}
Also, you can try:
fun2 - function(beginColumn, by, data) {
indx - seq(beginColumn, ncol(data), by = by)
indx1 - head(indx + 1, -1)
indx2 - tail(indx - 1, -1)
vec1 - as.vector(sapply(seq_along(indx1), function(i) indx1[i]:indx2[i]))
if (ncol(data) tail(indx, 1)) {
vec2 - c(vec1, (tail(indx, 1) + 1):ncol(dat1))
} else {
vec2 - vec1
}
means1 - rep(colMeans(data[, indx]), each = by - 1, length.out =
length(vec2))
res - as.data.frame(t(t(data[, vec2])/means1))
res
}
set.seed(458)
dat1 - as.data.frame(matrix(sample(5,10*5,replace=TRUE),ncol=10))
identical(fun1(2,4,dat1),fun2(2,4,dat1))
#[1] TRUE
A.K.
Hi,
May be this helps:
fun1 - function(beginColumn, by, data) {
indx - seq(beginColumn, ncol(data), by = by)
dataNew - data[, indx[1]:ncol(data)]
indx1 - cumsum(seq(ncol(data)) %in% indx)
indx2 - indx1[indx1 != 0]
lst1 - lapply(split(seq_along(indx2), indx2), function(i) {
x1 - dataNew[, i, drop = FALSE]
if (ncol(x1) 1) {
x1[, -1]/mean(x1[, 1])
}
})
res - data.frame(lst1[sapply(lst1, length) 0])
colnames(res) - gsub(.*\\., , colnames(res))
res
}
set.seed(458)
dat1 - as.data.frame(matrix(sample(5,10*5,replace=TRUE),ncol=10))
set.seed(34)
dat2 - as.data.frame(matrix(sample(20,21*5,replace=TRUE),ncol=21))
fun1(2,4,dat1)
fun1(2,5,dat2)
A.K.
On Sunday, April 20, 2014 5:55 AM, Andre Zacharia andre.zacha...@gmail.com
wrote:
Dear all,
I am getting data columnwise that I need to divide by the mean of another
column
If the column is the previous one this code works perfectly well:
fun1 - function(beginColumn, by, data) { indx - seq(beginColumn,
ncol(data), by = by) as.data.frame(t(100 - (t(data[, indx])/colMeans(data[,
indx - 1], na.rm = TRUE)) * 100))
}
(Arun helped me with this code, thank you again!...)
But, the things is now more complicated...
I need to program a function that allow me to divide for example cells from
column 3 on mean from column 2 and cells from column 4 on mean of column 2
and the 5 etc. Then column 6 is another column from whch I need to extract
the mean and to do the same with column 7 and 8, etc...
so if I have:
1 2 3 4 1 5
2 5 4 7 2 8
3 4 5 9 3 7
4 7 7 9 4 3
The serie 1,2,3,4 ar just enumerating so not useful at this timepoint.
the results should be (from excel...):
4,5 33,333 11,111 -11,111 11,111 -55,556 -77,778
-11,111 -100 -55,556 -55,556 -100
33,333
I tried to work on modyfying indx-1 by 2*indx-2, but this is not doing the
job... I tried many other things so that I am now stucked.
Does Anyone has a brilliant idea?
Many many thanks
André ZACHARIA
[[alternative HTML version deleted]]
__
R-help@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.
__
R-help@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.
[R] What do the colours of the scatterplot3d actually show?
Hello, Please could anyone tell me the significance of the graded red-black color when using scatterplot3d? At first I thought it was just going from the very 'near' to the back of the chart area, but as you can see from the pictures some of the points in the back are bright red. The description from ?scatterplot mentions that hightlight3d points will be drawn in different colors related to y coordinates... but as you can tell from the plot is isn't true either; with come short points being black as well as red... http://r.789695.n4.nabble.com/file/n4689148/Rplot.png any help would be fantastic, I'm very very new to R Cheers, Jo -- View this message in context: http://r.789695.n4.nabble.com/What-do-the-colours-of-the-scatterplot3d-actually-show-tp4689148.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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.
Re: [R] What do the colours of the scatterplot3d actually show?
On 20.04.2014 17:51, mfbx9jhy wrote: Hello, Please could anyone tell me the significance of the graded red-black color when using scatterplot3d? At first I thought it was just going from the very 'near' to the back of the chart area, but as you can see from the pictures some of the points in the back are bright red. The description from ?scatterplot mentions that hightlight3d points will be drawn in different colors related to y coordinates... but as you can tell from the plot is isn't true either; with come short points being black as well as red... http://r.789695.n4.nabble.com/file/n4689148/Rplot.png any help would be fantastic, I'm very very new to R All points in bright red appear to be within [0,5] on the elong labelled axis. If you find a point that is not, please tell me which one and send me the data and the code so that I could check what happens. Best, Uwe Ligges Cheers, Jo -- View this message in context: http://r.789695.n4.nabble.com/What-do-the-colours-of-the-scatterplot3d-actually-show-tp4689148.html Sent from the R help mailing list archive at Nabble.com. __ R-help@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. __ R-help@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.
Re: [R] Difference between times
On Apr 20, 2014, at 2:16 AM, Prof Brian Ripley wrote:
On 20/04/2014 08:50, Prof Brian Ripley wrote:
On 18/04/2014 21:46, David Winsemius wrote:
On Apr 18, 2014, at 12:59 PM, Prof Brian Ripley wrote:
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've misspelled CET (not CEST)
Neither CET nor CEST are portable time-zone names. We have not been
given the 'at a minimum' information required by the posting guide,
so please read ?Sys.timezone on your system.
Dear Prof;
Thanks for the impetus to yet again read that page. Despite frequently
reading help pages and in particular reading that one many times, I
still was not getting the 'tz' arguments correct on a Mac. I do now
see that I was spelling my TZ incorrectly (as Americas/Los_Angeles
rather than America/Los_Angeles.
Fellow Mac users may face a problem when using the Finder unless they
set it up to display hidden ('dot') files. The /usr/ folder is greyed
out but it still does open. If I restore my Finder defaults to not
show system files and folders, I no longer see that directory and
would not have been able to resolve my spelling error on my own:
You can always use the command-line or OlsonNames() in R.
I'm not finding an OlsonNames function on a Mac (but is that because I haven't
updated?). Before seeing that the zone-checking feature had been added as a
feature, I was building an OlsonNames function that extracts the
sub-directories of the zoneinfo directory and appends the file names to them as
well as extracting the non-OlsonNames entries in zoneinfo. My plan had been to
make my own warnings in strptime, but that appears to be unnecessary.
OlsonNames - function(onlyOlson=FALSE) {
MacOlsonDirs - system('ls -p /usr/share/zoneinfo ', intern=TRUE)
OlsonNames - unlist( lapply( MacOlsonDirs[grep(/$, MacOlsonDirs)],
function(dir) paste0( dir,
system( paste0('ls -p
/usr/share/zoneinfo/', dir) ,
intern=TRUE) ) ) )
nonOlsonNames - MacOlsonDirs[grepl(^[A-Z], MacOlsonDirs) ! grepl(/$,
MacOlsonDirs) ]
if ( !onlyOlson){ c(OlsonNames, nonOlsonNames)} else {OlsonNames}
}
Yes. It is because I haven't updated. I now see this in the NEWS that was
posted in this list 10 days ago.
There is more support to explore the system's idea of time-zone
names. Sys.timezone() tries to give the current system setting
by name (and succeeds at least on Linux, OS X, Solaris and
Windows), and OlsonNames() lists the names in the system's Olson
database. Sys.timezone(location = FALSE) gives the previous
behaviour.
I guess I will have fun comparing my efforts with those of the masters.
dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
America/New_York)
dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
America/Los_Angeles)
dt1-dt2
Time difference of 3 hours
I don't suppose a warning could be issued by the as.POSIXct code when
a tz argument is not found in the database to let people know that
'UTC' will be the default?
No, as the underlying POSIX function does not report this. We could
perhaps do this on platforms which use --with-internal-tzcode but not
e.g. on Linux.
In fact we already do: R 3.1.0 on a Mac shows
My apologies. And thank you to whomever added the feature and to you, Prof, for
checking and letting us know. I was going to offer the code above as a patch
but that seems not needed now. I have not yet updated to 3.1.0. The
Mavericks/3.1.0 incompatibilities have been scaring me off from updating. Still
on Lion/3.0.2.
as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
Americas/New_York)
[1] 2014-04-18 09:00:00 GMT
Warning messages:
1: In strptime(x, format, tz = tz) : unknown timezone 'Americas/New_York'
2: In as.POSIXct.POSIXlt(as.POSIXlt(x, tz, ...), tz, ...) :
unknown timezone 'Americas/New_York'
3: In as.POSIXlt.POSIXct(x, tz) : unknown timezone 'Americas/New_York'
Thank you, all of R-Core.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
David Winsemius
Alameda, CA, USA
__
R-help@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.
Re: [R] R 3.0.3, Windows 7: Problem installing XML package
I keep on trying from various networks but I still get the same error. I
don't this this has anything to do with network or ability to download the
package (as I can install other packages fine). This must be something in
base R or dependencies issues (that R is not spelling out).
I know R is geared for Mac and Windows is kind of looked down upon but I
have no option but use windows and need this XML package running to
complete my education. Any help on this would be appreciated.
On Mon, Apr 14, 2014 at 2:24 PM, Alpesh Pandya alpeshpan...@gmail.comwrote:
Thank you for response Rui.
I still get the same error with this repository.
Installing package into âC:/Users/APandya/Documents/R/win-library/3.0â
(as âlibâ is unspecified)
trying URL '
http://cran.dcc.fc.up.pt/bin/windows/contrib/3.0/XML_3.98-1.1.zip'
Content type 'application/zip' length 4288136 bytes (4.1 Mb)
opened URL
downloaded 4.1 Mb
Error in read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type))
:
cannot open the connection
In addition: Warning messages:
1: In download.file(url, destfile, method, mode = wb, ...) :
downloaded length 4276224 != reported length 4288136
2: In unzip(zipname, exdir = dest) : error 1 in extracting from zip file
3: In read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) :
cannot open compressed file 'XML/DESCRIPTION', probable reason 'No such
file or directory'
On Mon, Apr 14, 2014 at 2:17 PM, Rui Barradas ruipbarra...@sapo.ptwrote:
Hello,
I have package XML installed on Windows 7, R 3.0.3 and I had no problem
at all. Can't you try (it worked with me)
install.packages(XML, repos = http://cran.dcc.fc.up.pt;)
Hope this helps,
Rui Barradas
Em 14-04-2014 16:24, Alpesh Pandya escreveu:
I have tried these sources (almost all US mirrors):
http://cran.cnr.Berkeley.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://cran.stat.ucla.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://streaming.stat.iastate.edu/CRAN/bin/windows/contrib/
3.0/XML_3.98-1.1.zip
http://ftp.ussg.iu.edu/CRAN/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://rweb.quant.ku.edu/cran/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://watson.nci.nih.gov/cran_mirror/bin/windows/
contrib/3.0/XML_3.98-1.1.zip
http://cran.mtu.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://cran.wustl.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://cran.case.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://ftp.osuosl.org/pub/cran/bin/windows/contrib/3.0/XML_3.98-1.1.zip
http://lib.stat.cmu.edu/R/CRAN/bin/windows/contrib/3.0/XML_3.98-1.1.zip
I have confirmed with IT that there is no restriction on downloading this
zip file from any of these sources. Also I am getting same error when I
try
from my home network as well.
On Sun, Apr 13, 2014 at 9:46 AM, Uwe Ligges
lig...@statistik.tu-dortmund.de
wrote:
On 13.04.2014 01:30, Alpesh Pandya wrote:
@Uwe I tried the same steps from office as well as home network with
same
results. Are you using windows 7 with R 3.0.3?
I have seen same question being asked by others without any
resolution. Is
anything special about XML package? I am OK use older version of
package
but in archives there are no zip files (only gz files). Is windows
platform
not recommended for R?
Right, and you can try to install these from sources.
But I doubt you need it. You still have not told us if you tried another
mirror to download the XML file from and what you local IT support tells
you while your downloads are incomplete.
Best,
Uwe Ligges
On Sat, Apr 12, 2014 at 7:22 PM, Uwe Ligges
lig...@statistik.tu-dortmund.de
wrote:
On 12.04.2014 22:39, Alpesh Pandya wrote:
Thank you for response Uwe. I tried multiple times by downloading
the
zip
file from many sources but still the same error. This is a major road
block
for me in using R. Appreciate any help on this.
Please ask your local IT staff.
I get, using the same mirror:
options(repos=c(CRAN=http://watson.nci.nih.gov/cran_mirror;))
install.packages(XML, lib=d:/temp)
trying URL 'http://watson.nci.nih.gov/cran_mirror/bin/windows/
contrib/3.0/XML_3.98-1.1.zip'
Content type 'application/zip' length 4288136 bytes (4.1 Mb)
opened URL
downloaded 4.1 Mb
package 'XML' successfully unpacked and MD5 sums checked
The downloaded binary packages are in
d:\temp\RtmpqMqL8L\downloaded_packages
Best,
Uwe Ligges
On Fri, Apr 11, 2014 at 6:53 PM, Uwe Ligges
lig...@statistik.tu-dortmund.de
wrote:
Works for me.
Best,
Uwe Ligges
On 11.04.2014 17:10, Alpesh Pandya wrote:
Using install.package('XML') command produces this error:
trying URL
'
http://watson.nci.nih.gov/cran_mirror/bin/windows/
contrib/3.0/XML_3.98-1.1.zip
'
Content type 'application/zip' length 4288136 bytes (4.1 Mb)
opened URL
downloaded 4.1 Mb
Error in read.dcf(file.path(pkgname, DESCRIPTION), c(Package,
Type)) :
cannot open the
Re: [R] R 3.0.3, Windows 7: Problem installing XML package
On 20/04/2014, 2:45 PM, Alpesh Pandya wrote: I keep on trying from various networks but I still get the same error. I don't this this has anything to do with network or ability to download the package (as I can install other packages fine). This must be something in base R or dependencies issues (that R is not spelling out). This is very clearly a problem with your setup, not with R: nobody else is reporting it. What you should do is download the .zip file using some other means (e.g. Firefox, etc.) and check whether the size comes out properly (4,288,694 bytes; not sure why you saw the numbers you saw). If it does, then install the package from the .zip file. If it doesn't, then tell your IT people to fix your system. Duncan Murdoch I know R is geared for Mac and Windows is kind of looked down upon but I have no option but use windows and need this XML package running to complete my education. Any help on this would be appreciated. On Mon, Apr 14, 2014 at 2:24 PM, Alpesh Pandya alpeshpan...@gmail.comwrote: Thank you for response Rui. I still get the same error with this repository. Installing package into ‘C:/Users/APandya/Documents/R/win-library/3.0’ (as ‘lib’ is unspecified) trying URL ' http://cran.dcc.fc.up.pt/bin/windows/contrib/3.0/XML_3.98-1.1.zip' Content type 'application/zip' length 4288136 bytes (4.1 Mb) opened URL downloaded 4.1 Mb Error in read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open the connection In addition: Warning messages: 1: In download.file(url, destfile, method, mode = wb, ...) : downloaded length 4276224 != reported length 4288136 2: In unzip(zipname, exdir = dest) : error 1 in extracting from zip file 3: In read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open compressed file 'XML/DESCRIPTION', probable reason 'No such file or directory' On Mon, Apr 14, 2014 at 2:17 PM, Rui Barradas ruipbarra...@sapo.ptwrote: Hello, I have package XML installed on Windows 7, R 3.0.3 and I had no problem at all. Can't you try (it worked with me) install.packages(XML, repos = http://cran.dcc.fc.up.pt;) Hope this helps, Rui Barradas Em 14-04-2014 16:24, Alpesh Pandya escreveu: I have tried these sources (almost all US mirrors): http://cran.cnr.Berkeley.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://cran.stat.ucla.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://streaming.stat.iastate.edu/CRAN/bin/windows/contrib/ 3.0/XML_3.98-1.1.zip http://ftp.ussg.iu.edu/CRAN/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://rweb.quant.ku.edu/cran/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://watson.nci.nih.gov/cran_mirror/bin/windows/ contrib/3.0/XML_3.98-1.1.zip http://cran.mtu.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://cran.wustl.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://cran.case.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://ftp.osuosl.org/pub/cran/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://lib.stat.cmu.edu/R/CRAN/bin/windows/contrib/3.0/XML_3.98-1.1.zip I have confirmed with IT that there is no restriction on downloading this zip file from any of these sources. Also I am getting same error when I try from my home network as well. On Sun, Apr 13, 2014 at 9:46 AM, Uwe Ligges lig...@statistik.tu-dortmund.de wrote: On 13.04.2014 01:30, Alpesh Pandya wrote: @Uwe I tried the same steps from office as well as home network with same results. Are you using windows 7 with R 3.0.3? I have seen same question being asked by others without any resolution. Is anything special about XML package? I am OK use older version of package but in archives there are no zip files (only gz files). Is windows platform not recommended for R? Right, and you can try to install these from sources. But I doubt you need it. You still have not told us if you tried another mirror to download the XML file from and what you local IT support tells you while your downloads are incomplete. Best, Uwe Ligges On Sat, Apr 12, 2014 at 7:22 PM, Uwe Ligges lig...@statistik.tu-dortmund.de wrote: On 12.04.2014 22:39, Alpesh Pandya wrote: Thank you for response Uwe. I tried multiple times by downloading the zip file from many sources but still the same error. This is a major road block for me in using R. Appreciate any help on this. Please ask your local IT staff. I get, using the same mirror: options(repos=c(CRAN=http://watson.nci.nih.gov/cran_mirror;)) install.packages(XML, lib=d:/temp) trying URL 'http://watson.nci.nih.gov/cran_mirror/bin/windows/ contrib/3.0/XML_3.98-1.1.zip' Content type 'application/zip' length 4288136 bytes (4.1 Mb) opened URL downloaded 4.1 Mb package 'XML' successfully unpacked and MD5 sums checked The downloaded binary packages are in d:\temp\RtmpqMqL8L\downloaded_packages Best, Uwe Ligges On Fri, Apr 11, 2014 at 6:53 PM, Uwe Ligges lig...@statistik.tu-dortmund.de wrote: Works for me.
Re: [R] R 3.0.3, Windows 7: Problem installing XML package
Hard to help you when the problem simply does not happen for others. As for Windows being not a focus, that is not at all true. I use it regularly on Windows at work. That being said, there are thousands of packages and those each involve their own subset of R users. There are also many operating system configurations that may not all be fully tested. Blaming R or Windows, or blaming us for preventing you from getting your education (isn't that something between you and your educational institution?) are not going to be effective strategies for problem solving. Are you able to use other aspects of R beyond the XML package? Have you tried communicating with the maintainers of that package? --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On April 20, 2014 11:45:44 AM PDT, Alpesh Pandya alpeshpan...@gmail.com wrote: I keep on trying from various networks but I still get the same error. I don't this this has anything to do with network or ability to download the package (as I can install other packages fine). This must be something in base R or dependencies issues (that R is not spelling out). I know R is geared for Mac and Windows is kind of looked down upon but I have no option but use windows and need this XML package running to complete my education. Any help on this would be appreciated. On Mon, Apr 14, 2014 at 2:24 PM, Alpesh Pandya alpeshpan...@gmail.comwrote: Thank you for response Rui. I still get the same error with this repository. Installing package into ���C:/Users/APandya/Documents/R/win-library/3.0��� (as ���lib��� is unspecified) trying URL ' http://cran.dcc.fc.up.pt/bin/windows/contrib/3.0/XML_3.98-1.1.zip' Content type 'application/zip' length 4288136 bytes (4.1 Mb) opened URL downloaded 4.1 Mb Error in read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open the connection In addition: Warning messages: 1: In download.file(url, destfile, method, mode = wb, ...) : downloaded length 4276224 != reported length 4288136 2: In unzip(zipname, exdir = dest) : error 1 in extracting from zip file 3: In read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open compressed file 'XML/DESCRIPTION', probable reason 'No such file or directory' On Mon, Apr 14, 2014 at 2:17 PM, Rui Barradas ruipbarra...@sapo.ptwrote: Hello, I have package XML installed on Windows 7, R 3.0.3 and I had no problem at all. Can't you try (it worked with me) install.packages(XML, repos = http://cran.dcc.fc.up.pt;) Hope this helps, Rui Barradas Em 14-04-2014 16:24, Alpesh Pandya escreveu: I have tried these sources (almost all US mirrors): http://cran.cnr.Berkeley.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://cran.stat.ucla.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://streaming.stat.iastate.edu/CRAN/bin/windows/contrib/ 3.0/XML_3.98-1.1.zip http://ftp.ussg.iu.edu/CRAN/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://rweb.quant.ku.edu/cran/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://watson.nci.nih.gov/cran_mirror/bin/windows/ contrib/3.0/XML_3.98-1.1.zip http://cran.mtu.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://cran.wustl.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://cran.case.edu/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://ftp.osuosl.org/pub/cran/bin/windows/contrib/3.0/XML_3.98-1.1.zip http://lib.stat.cmu.edu/R/CRAN/bin/windows/contrib/3.0/XML_3.98-1.1.zip I have confirmed with IT that there is no restriction on downloading this zip file from any of these sources. Also I am getting same error when I try from my home network as well. On Sun, Apr 13, 2014 at 9:46 AM, Uwe Ligges lig...@statistik.tu-dortmund.de wrote: On 13.04.2014 01:30, Alpesh Pandya wrote: @Uwe I tried the same steps from office as well as home network with same results. Are you using windows 7 with R 3.0.3? I have seen same question being asked by others without any resolution. Is anything special about XML package? I am OK use older version of package but in archives there are no zip files (only gz files). Is windows platform not recommended for R? Right, and you can try to install these from sources. But I doubt you need it. You still have not told us if you tried another mirror to download the XML file from and what you local IT support tells you while your downloads are incomplete. Best, Uwe Ligges On Sat, Apr 12, 2014 at 7:22 PM, Uwe Ligges
