[R] DOUBT
Respected sir/madam can you please suggest what is an unexpected symbol in the below code for running a multinomial logistic regression model <- multinom(adoption ~ age + education + HH size + landholding + Farmincome + nonfarmincome + creditaccesibility + LHI, data=newdata) [[alternative HTML version deleted]] __ R-help@r-project.org 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.
Re: [R] C stack usage 7970372 is too close to the limit
Hi John, Only the particular users getting error john. Please help me On Mon, 10 Aug 2020, 11:43 pm Raj kapoor, wrote: > Hi John, > > I have 10 user in the instance, 9 user is working and access the R studio > app, but while access the 10th user it's getting stack usage limit issues, > then we create the new users its working fine. > > On Mon, 10 Aug 2020, 10:21 pm John Harrold, > wrote: > >> Hello Raj, >> >> I've gotten this type of error in the past when I've done things like use >> while loops that didn't end. Basically, I think this means you're running >> out of memory. If you want more users, possibly increase the amount of ram >> in your machine. >> >> John >> >> On Mon, Aug 10, 2020 at 6:43 AM Raj kapoor >> wrote: >> >>> Hi Team, >>> >>> I have one production instance in aws, in CentoOs linux environment, i >>> have >>> 5 user to access the instance for using RStudio, In case R-studio >>> working 4 >>> users running good, while we access 5th users its getting error, >>> >>> First issue : C stack usage 7970372 is too close to the limit >>> >>> Second Issue : no stack overflow >>> >>> Please provide the solution. >>> >>> [[alternative HTML version deleted]] >>> >>> __ >>> R-help@r-project.org 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. >>> >> >> >> -- >> John >> :wq >> > [[alternative HTML version deleted]] __ R-help@r-project.org 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.
Re: [R] C stack usage 7970372 is too close to the limit
Hi John, I have 10 user in the instance, 9 user is working and access the R studio app, but while access the 10th user it's getting stack usage limit issues, then we create the new users its working fine. On Mon, 10 Aug 2020, 10:21 pm John Harrold, wrote: > Hello Raj, > > I've gotten this type of error in the past when I've done things like use > while loops that didn't end. Basically, I think this means you're running > out of memory. If you want more users, possibly increase the amount of ram > in your machine. > > John > > On Mon, Aug 10, 2020 at 6:43 AM Raj kapoor > wrote: > >> Hi Team, >> >> I have one production instance in aws, in CentoOs linux environment, i >> have >> 5 user to access the instance for using RStudio, In case R-studio working >> 4 >> users running good, while we access 5th users its getting error, >> >> First issue : C stack usage 7970372 is too close to the limit >> >> Second Issue : no stack overflow >> >> Please provide the solution. >> >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org 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. >> > > > -- > John > :wq > [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] C stack usage 7970372 is too close to the limit
Hi Team, I have one production instance in aws, in CentoOs linux environment, i have 5 user to access the instance for using RStudio, In case R-studio working 4 users running good, while we access 5th users its getting error, First issue : C stack usage 7970372 is too close to the limit Second Issue : no stack overflow Please provide the solution. [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] (no subject)
hello Team R, i have been using R for statistical analysis of phylogeny and i have installed the required packages phangorn and phytools but whenever i give the command "pml.fit" the program stops and it appears thatb r for windows GUI has stopped etc.. previously i thought it was a fault in my computer but it appears to happen the same when i give the command in any other computer. can i get a solution for this i have tried it on windows 10 and older versions can it be fixed? and also can you please send me the the commands for performing SH test of phylogeny using phangorn if possible. please do reply thank you [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] regarding a problem in R
hello Team R, i have been using R for statistical analysis of phylogeny and i have installed the required packages phangorn and phytools but whenever i give the command "pml.fit" the program stops and it appears thatb r for windows GUI has stopped etc.. previously i thought it was a fault in my computer but it appears to happen the same when i give the command in any other computer. can i get a solution for this i have tried it on windows 10 and older versions can it be fixed? and also can you please send me the the commands for performing SH test of phylogeny using phangorn if possible. please do reply thank you [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] help
hello Team R, i have been using R for statistical analysis of phylogeny and i have installed the required packages phangorn and phytools but whenever i give the command "pml.fit" the program stops and it appears thatb r for windows GUI has stopped etc.. previously i thought it was a fault in my computer but it appears to happen the same when i give the command in any other computer. can i get a solution for this i have tried it on windows 10 and older versions can it be fixed? and also can you please send me the the commands for performing SH test of phylogeny using phangorn if possible. please do reply thank you [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] How to take a multidimensional data set to higher dimensions using R?
Hi, I have a multidimensional data-set( multiple 'x' variables with a target variable). I want to take it to a higher dimensional space so that I can apply classification technique with ease . Is there a package in R which would allow me to take these data points from lower dimensional space to higher dimensions? so that is that higher dimension I can apply classification techniques to be effective. My datapoints are highly inseperable in lower dimension so i want to take the data points to higher dimension where they can be separable( something similar to kernel trick) -- Warm regards, Devesh. [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] Replace a column value on condition
I want to replace column c3 with values from column c2 whenever values of column Id are 2. In stata I could use replace c3 = c2 if id ==2. How could I do that in R? Thanks Sample data found below: dput(df4) structure(list(c2 = c(42L, 42L, 47L, 47L, 55L, 55L, 36L, 36L, 61L, 61L), c3 = c(68L, 59L, 68L, 50L, 62L, 50L, 63L, 45L, 65L, 45L), id = c(1, 2, 1, 2, 1, 2, 1, 2, 1, 2)), datalabel = Written by R. , time.stamp = 23 Mar 2015 13:54, .Names = c(c2, c3, id), formats = c(%9.0g, %9.0g, %9.0g), types = c(253L, 253L, 255L), val.labels = c(, , ), var.labels = c(, , ), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), version = 12L, class = data.frame) __ R-help@r-project.org 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.
[R] Aggegate minutes data to hourly data
I have a measurement that was taken in 15 minutes or more and want to aggregate it by hour. How could I do that? Sample data is found below date_time concentration 26/11/2013 15:46 529.25 26/11/2013 16:03 1596 26/11/2013 16:23 1027.111 26/11/2013 16:39 1001.9 26/11/2013 16:54 -80.25 26/11/2013 17:12 1064.125 26/11/2013 11:14 7969.7 26/11/2013 11:32 522 26/11/2013 11:58 845.111 26/11/2013 12:12 1166.875 26/11/2013 12:30 473.375 26/11/2013 12:42 466.2 26/11/2013 07:47 4358.833 26/11/2013 08:05 1257.545 26/11/2013 08:24 828.6 26/11/2013 08:45 942 26/11/2013 08:58 758.111 26/11/2013 09:13 832.333 26/11/2013 15:45 1876.909 26/11/2013 16:07 574.25 26/11/2013 16:27 1736.846 26/11/2013 16:43 1024.857 26/11/2013 16:59 858.538 26/11/2013 17:15 912.455 26/11/2013 11:18 2086.143 26/11/2013 11:39 2078.667 26/11/2013 12:03 1619.072 26/11/2013 12:16 1197.583 26/11/2013 12:35 619.308 26/11/2013 12:51 1222.571 26/11/2013 07:49 1357.929 26/11/2013 08:08 1120 26/11/2013 08:29 1381.6 26/11/2013 08:48 1493.429 26/11/2013 09:03 1113.786 26/11/2013 09:18 1217.143 __ R-help@r-project.org 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.
[R] rename columns with pattern
I want to rename columns 1 to 6 in the sample data set as bp_1 to bp_6. How could I do that in R? Thanks dput(dff) structure(list(one = c(1.00027378507871, 0.982313483915127, 1.1531279945243, 1.07400410677618, 1.22710472279261, 1.19762271047046, 1.10904859685147, 1.32060232717317), two = c(1.04707392197125, 1.00998288843258, 1.17598904859685, 1.09595482546201, 1.28599589322382, 1.26632675564591, 1.12986995208761, 1.30704654346338), three = c(1.06301619895049, 1.02743782797171, 1.1977093315081, 1.11466803559206, 1.2949441022131, 1.28365657768591, 1.1305452886151, 1.32089436459046), four = c(1.06994010951403, 1.03489904175222, 1.19799452429843, 1.1172022587269, 1.28742984257358, 1.27650013346977, 1.12265058179329, 1.30723134839151), five = c(1.07019712525667, 1.03722792607803, 1.19174811772758, 1.11514168377823, 1.26594387405886, 1.25720010677582, 1.11339630390144, 1.29178507871321), six = c(1.1909650924, 1.08407027150354, 1.24785877253023, 1.16373032169747, 1.31150581793292, 1.31042514031455, 1.16205338809035, 1.37122975131189), idd = 1:8), .Names = c(one, two, three, four, five, six, idd), row.names = c(NA, -8L), class = c(tbl_df, data.frame)) __ R-help@r-project.org 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.
[R] Create previous dates from date with consideration of leap year
Dear R Community, I wish to create 5 preceding dates from the date variable by ID. How could I create such dates? The code should consider leap year. Thanks Sample data follows: structure(list(id = 1:12, date = structure(c(9L, 6L, 11L, 8L, 7L, 5L, 4L, 3L, 12L, 1L, 10L, 2L), .Label = c(01feb2003, 03mar2008, 04feb2008, 07jul1991, 07jun2010, 13feb2005, 18dec1991, 22sep2005, 27apr1993, 29jan2009, 29may2002, 31jan2005 ), class = factor), case = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L)), .Names = c(id, date, case), class = data.frame, row.names = c(NA, -12L)) __ R-help@r-project.org 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.
[R] How to update R without losing packages
A solution on the link below provides the steps of updating R without losing packages in Unix. http://zvfak.blogspot.se/2012/06/updating-r-but-keeping-your-installed.html How could I do that on windows 7 platform? Thanks __ 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] How can I merge data with differing length?
How can I merge data frame df and tem shown below by filling the head of tem with missing values? a- rnorm(1825, 20) b- rnorm(1825, 30) date-seq(as.Date(2000/1/1), by = day, length.out = 1825) df-data.frame(date,a,b) tem- rpois(1095, lambda=21) Thanks __ 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] How to sum some columns based on their names
I want to sum columns based on their names. As an exampel how could I sum columns which contain 6574, 7584 and 85 as column names? In addition, how could I sum those which contain 6574, 7584 and 85 in ther names and have a prefix f. My data contains several variables with I want to sum columns based on their names. As an exampel how could I sum columns which contain 6574, 7584 and 85 as column names? In addition, how could I sum those which contain 6574, 7584 and 85 in ther names and have a prefix f. My data contains several variables with dput(df1) structure(list(date = structure(c(1230768000, 1230854400, 1230940800, 1231027200, 1231113600, 123120, 1231286400, 1231372800, 1231459200, 1231545600, 1231632000), class = c(POSIXct, POSIXt), tzone = UTC), f014card = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), f1534card = c(0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1), f3564card = c(1, 6, 1, 5, 5, 4, 4, 7, 6, 4, 6), f6574card = c(3, 6, 4, 5, 5, 2, 10, 3, 4, 2, 4), f7584card = c(13, 6, 1, 4, 10, 6, 8, 12, 10, 4, 3), f85card = c(5, 3, 1, 0, 2, 10, 7, 9, 1, 7, 3), m014card = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), m1534card = c(0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0), m3564card = c(12, 7, 4, 7, 12, 13, 12, 7, 12, 2, 11), m6574card = c(3, 4, 8, 8, 8, 10, 7, 6, 7, 7, 5), m7584card = c(8, 10, 5, 4, 12, 7, 14, 11, 9, 1, 11), m85card = c(1, 4, 3, 0, 3, 4, 5, 5, 4, 5, 0)), .Names = c(date, f014card, f1534card, f3564card, f6574card, f7584card, f85card, m014card, m1534card, m3564card, m6574card, m7584card, m85card), class = data.frame, row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11)) __ 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] Generate sequence of date based on a group ID
I want to generate a sequence of date based on a group id(similar IDs should have same date). The id variable contains unequal observations and the length of the data set also varies. How could I create a sequence that starts on specific date (say January 1, 2000 onwards) and continues until the end without specifying length? Sample data follows: df-structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L), out1 = c(0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 1L)), .Names = c(id, out1), class = data.frame, row.names = c(NA, -23L)) __ 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] How to combine character month and year columns into one column
Dear R users, I have a data with month and year columns which are both characters and wanted to create a new column like Jan-1999 with the following code. The result is all NA for the month part. What is wrong with the and what is the right way to combine the two? ddf$MonthDay - paste(month.abb[ddf$month], ddf$Year, sep=- ) Thanks dput(ddf) structure(list(month = c(01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12), Year = c(1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999), views = c(42, 49, 44, 38, 37, 35, 38, 39, 38, 39, 38, 46), MonthDay = c(NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999)), .Names = c(month, Year, views, MonthDay), row.names = 109:120, class = data.frame) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to combine character month and year columns into one column
Many thanks for your quick answer which has created what I wished. May I ask followup question on the same issue. I failed to convert the new column into date format with this code. The class of MonthDay is still character df$MonthDay - format(df$MonthDay, format=c(%b %Y)) I would appreciate if you could suggest a working solution Thanks On 23 September 2014 18:03, Marc Schwartz marc_schwa...@me.com wrote: On Sep 23, 2014, at 10:41 AM, Kuma Raj pollar...@gmail.com wrote: Dear R users, I have a data with month and year columns which are both characters and wanted to create a new column like Jan-1999 with the following code. The result is all NA for the month part. What is wrong with the and what is the right way to combine the two? ddf$MonthDay - paste(month.abb[ddf$month], ddf$Year, sep=- ) Thanks dput(ddf) structure(list(month = c(01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12), Year = c(1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999, 1999), views = c(42, 49, 44, 38, 37, 35, 38, 39, 38, 39, 38, 46), MonthDay = c(NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999, NA-1999)), .Names = c(month, Year, views, MonthDay), row.names = 109:120, class = data.frame) Since you are trying to use ddf$month as an index into month.abb, you will either need to coerce ddf$month to numeric in your code, or adjust how the data frame is created. In the case of the former approach: paste(month.abb[as.numeric(ddf$month)], ddf$Year, sep=- ) [1] Jan-1999 Feb-1999 Mar-1999 Apr-1999 May-1999 Jun-1999 [7] Jul-1999 Aug-1999 Sep-1999 Oct-1999 Nov-1999 Dec-1999 Regards, Marc Schwartz __ 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] How to identify outliers with values five times 99th percentile
I have a data frame with some extreme values which I wish to identify and repeat an analysis without these extreme values. How could I identify several columns with values which are 5 times higher than the 99th percentile? Sample data is pasted below. dput(df) structure(list(ad1 = c(98, 6.9, 8.1, 56, 3.9, 6.9, 6.9, 5.8, 7.2, 20.5, 9.4, 7.6, 5.3, 7.9, 62.2, 9.2, 11.9, 8.8, 23.1, 5.4, 9.4, 56, 8.6, 20.7, 21, 10.5, 5.5, 4.3, 15.8, 6.8, 10.4, 5.1), ad2 = c(14.9, 19.7, 1, 17.7, 14.9, 13.6, 18.8, 20.9, 46, 16.5, 11.7, 1, 9.2, 23.6, 19.7, 1, 11.4, 11, 23.1, 1, 1, 8.9, 11.3, 6.4, 15.2, 1, 17.3, 10.1, 13.3, 21.3, 12.3, 15.4 ), ad3 = c(0.91, 0.95, 10.7, 4.4, 0.43, 0.8, 3.1, 1.9, 2.3, 5.6, 3.9, 7.3, 0.37, 4.1, 15.1, 21.8, 3, 0.79, 1, 4.6, 0.61, 0.46, 0.87, 23.5, 3.8, 3.1, 0.33, 1.9, 3.2, 1.7, 0.53, 62.5 ), ad4 = c(225.5, 269.7, 326, 485.4, 193.2, 274.1, 553.2, 166.8, 435.9, 433.2, 187.1, 660.4, 235.4, 356.5, 378.8, 500.5, 323.5, 327.1, 289.5, 301.2, 291.7, 333.5, 351.7, 384.1, 347, 1354, 440.4, 189.2, 381, 252.7, 391.1, 255.1), ad5 = c(337.9, 355.6, 419.5, 798.5, 225, 355.9, 394.4, 340.6, 463.9, 291.9, 312.3, 491, 290.5, 231.9, 358, 386.4, 306.7, 440.6, 297.9, 339.3, 341.1, 366.2, 325.4, 357, 412.2, 370.2, 421.3, 346.3, 289.1, 257.4, 368, 322.6), ad6 = c(64.5, 130.6, 76, 167.8, 47.3, 117, 60.7, 91.9, 221.9, 91.1, 105.1, 110.8, 64.5, 184.5, 191.6, 259.4, 879.5, 142.1, 55.3, 123.1, 62.2, 75.2, 154.6, 100.7, 93.1, 136.7, 74.3, 41.8, 110.1, 109.1, 172.5, 87.7 ), ad7 = c(128L, 987L, 158L, 124L, 137L, 215L, 141L, 98L, 291L, 261L, 106L, 137L, 141L, 159L, 221L, 108L, 123L, 107L, 137L, 175L, 257L, 97L, 168L, 145L, 147L, 188L, 145L, 128L, 153L, 187L, 123L, 354L), ad8 = c(3.26, 3.98, 2.88, 2.85, 4.17, 3.16, 3.09, 4.35, 3.46, 3.81, 3.78, 3.81, 4.17, 4.27, 4.27, 2.97, 3.43, 3.48, 3.78, 3.86, 3.11, 3.12, 3.16, 4.24, 3.81, 3.11, 5.31, 3.75, 3.78, 3.55, 4.08, 3.5), ad9 = c(433L, 211L, 66L, 173L, 224L, 466L, 224L, 273L, 94L, 321L, 160L, 107L, 121L, 186L, 455L, 80L, 897L, 186L, 285L, 134L, 107L, 355L, 261L, 249L, 332L, 107L, 273L, 107L, 160L, 535L, 160L, 121L)), .Names = c(ad1, ad2, ad3, ad4, ad5, ad6, ad7, ad8, ad9), class = data.frame, row.names = c(NA, -32L)) __ 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] Obtain coefficients of several nlme objects
I have several lme objects like the ones shown below and I wish to combine the coefficients and confidence intervals of fixed effects of several models. Is there a function that could do that job? m1 - lme(mark1 ~ pm10 + temp + + age + gender + bmi + statin + smoke + dow + season , data = df , random = ~ 1 | id,na.action=na.exclude, method=ML) m2 - lme(mark2 ~ pm10 + temp + + age + gender + bmi + statin + smoke + dow + season , data = df , random = ~ 1 | id,na.action=na.exclude, method=ML) __ 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] Subset a column with specific characters
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I would like to subset a column based on the contents of a column with
specific character. In the sample data I wish to do the following:
First keep the data based on column prog if prog contains ca, and
secondly to drop if race contains ic
Thanks
library(foreign)
hsb2 - read.dta('http://www.ats.ucla.edu/stat/stata/notes/hsb2.dta')
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[R] Loop through columns of outcomes
I have asked this question on SO, but it attracted no response, thus I am
cross- posting it here with the hope that someone would help.
I want to estimate the effect of pm10 and o3 on three outcome(death, cvd
and resp). What I want to do is run one model for each of the main
predictors (pm10 and o3) and each outcome(death, cvd and resp). Thus I
expect to obtain 6 models. The script below works for one outcome (death)
and I wish to use it for more dependent variables.
library(quantmod)
library(mgcv)
library(dlnm)
df - chicagoNMMAPS
outcomes- c(death, cvd, resp )
varlist0 - c(pm10, o3)
m1 - lapply(varlist0,function(v) {
f - sprintf(death~ s(time,bs='cr',k=200)+s(temp,bs='cr') +
Lag(%s,0:6),v)
gam(as.formula(f),family=quasipoisson,na.action=na.omit,data=df)
})
Thanks
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop through columns of outcomes
Thanks for the script which works perfectly. I am interested to do
model checking and also interested to extract the coefficients for
linear and spline terms. For model checkup I could run this script
which will give different plots to test model fit: gam.check(m2[[1]]).
Thanks to mnel from SO I could also extract the linear terms with the
following script:
m2 - unlist(m1, recursive = FALSE) ## unlist
First extract the model elements:
mod1-m2[[1]]
mod2-m2[[2]]
mod3-m2[[3]]
mod4-m2[[4]]
mod5-m2[[5]]
mod6-m2[[6]]
And run the following:
mlist - list(mod1, mod2, mod3,mod4,mod5,mod6) ## Creates a list of models
names(mlist) - list(mod1, mod2, mod3,mod4,mod5,mod6)
slist - lapply(mlist, summary) ## obtain summaries
plist - lapply(slist, `[[`, 'p.table') ## list of the coefficients
linear terms
For 6 models this is relatively easy to do, but how could I shorten
the process if I have large number of models?
Thanks
On 12 November 2013 12:32, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Use nested lapply(). Like this:
m1 - lapply(varlist0,function(v) {
lapply(outcomes, function(o){
f - sprintf(%s~ s(time,bs='cr',k=200)+s(temp,bs='cr') +
Lag(%s,0:6), o, v)
gam(as.formula(f),family=quasipoisson,na.action=na.omit,data=df)
})})
m1 - unlist(m1, recursive = FALSE)
m1
Hope this helps,
Rui Barradas
Em 12-11-2013 09:53, Kuma Raj escreveu:
I have asked this question on SO, but it attracted no response, thus I am
cross- posting it here with the hope that someone would help.
I want to estimate the effect of pm10 and o3 on three outcome(death, cvd
and resp). What I want to do is run one model for each of the main
predictors (pm10 and o3) and each outcome(death, cvd and resp). Thus I
expect to obtain 6 models. The script below works for one outcome (death)
and I wish to use it for more dependent variables.
library(quantmod)
library(mgcv)
library(dlnm)
df - chicagoNMMAPS
outcomes- c(death, cvd, resp )
varlist0 - c(pm10, o3)
m1 - lapply(varlist0,function(v) {
f - sprintf(death~ s(time,bs='cr',k=200)+s(temp,bs='cr') +
Lag(%s,0:6),v)
gam(as.formula(f),family=quasipoisson,na.action=na.omit,data=df)
})
Thanks
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop through columns of outcomes
Very helpful, many thanks.
On 12 November 2013 16:09, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Once again, use lapply.
mlist - lapply(seq_along(m2), function(i) m2[[i]])
names(mlist) - paste0(mod, seq_along(mlist))
slist - lapply(mlist, summary)
plist - lapply(slist, `[[`, 'p.table')
Hope this helps,
Rui Barradas
Em 12-11-2013 13:28, Kuma Raj escreveu:
Thanks for the script which works perfectly. I am interested to do
model checking and also interested to extract the coefficients for
linear and spline terms. For model checkup I could run this script
which will give different plots to test model fit: gam.check(m2[[1]]).
Thanks to mnel from SO I could also extract the linear terms with the
following script:
m2 - unlist(m1, recursive = FALSE) ## unlist
First extract the model elements:
mod1-m2[[1]]
mod2-m2[[2]]
mod3-m2[[3]]
mod4-m2[[4]]
mod5-m2[[5]]
mod6-m2[[6]]
And run the following:
mlist - list(mod1, mod2, mod3,mod4,mod5,mod6) ## Creates a list of
models
names(mlist) - list(mod1, mod2, mod3,mod4,mod5,mod6)
slist - lapply(mlist, summary) ## obtain summaries
plist - lapply(slist, `[[`, 'p.table') ## list of the coefficients
linear terms
For 6 models this is relatively easy to do, but how could I shorten
the process if I have large number of models?
Thanks
On 12 November 2013 12:32, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Use nested lapply(). Like this:
m1 - lapply(varlist0,function(v) {
lapply(outcomes, function(o){
f - sprintf(%s~ s(time,bs='cr',k=200)+s(temp,bs='cr')
+
Lag(%s,0:6), o, v)
gam(as.formula(f),family=quasipoisson,na.action=na.omit,data=df)
})})
m1 - unlist(m1, recursive = FALSE)
m1
Hope this helps,
Rui Barradas
Em 12-11-2013 09:53, Kuma Raj escreveu:
I have asked this question on SO, but it attracted no response, thus I
am
cross- posting it here with the hope that someone would help.
I want to estimate the effect of pm10 and o3 on three outcome(death,
cvd
and resp). What I want to do is run one model for each of the main
predictors (pm10 and o3) and each outcome(death, cvd and resp). Thus I
expect to obtain 6 models. The script below works for one outcome
(death)
and I wish to use it for more dependent variables.
library(quantmod)
library(mgcv)
library(dlnm)
df - chicagoNMMAPS
outcomes- c(death, cvd, resp )
varlist0 - c(pm10, o3)
m1 - lapply(varlist0,function(v) {
f - sprintf(death~ s(time,bs='cr',k=200)+s(temp,bs='cr') +
Lag(%s,0:6),v)
gam(as.formula(f),family=quasipoisson,na.action=na.omit,data=df)
})
Thanks
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Remove a column of a matrix with unnamed column header
I have a matrix names test which I want to convert to a data frame. When I use a command test2-as.data.frame(test) it is executed without a problem. But when I want to browse the content I receive an error message Error in data.frame(outcome = c(cardva, respir, cereb, neoplasm, : duplicate row.names: Estimate . The problem is clearly due to a duplicate in row name . But I am unable to remove this column. I need help on how to remove this specific column that has essentially no column header name. dput of the matrix is here: dput(test) structure(c(cardva, respir, cereb, neoplasm, ami, ischem, heartf, pneumo, copd, asthma, dysrhy, diabet, 0.00259492159959046, 0.00979775441709427, 0.00103414632535868, 0.00486468139227382, 0.0164825543879707, 0.0116647168053943, -0.0012137908515233, 0.00730433232907741, 0.00355583994565985, 0.000712387285735019, -0.00103763671307935, 0.00981500221106926, 0.00325476724733837, 0.0049232113728293, 0.00520118026087645, 0.00386848394426742, 0.00688121694253705, 0.00585772614064902, 0.00564983058883797, 0.0061328202328586, 0.0108212194251692, 0.0173804438930357, 0.00867931407250442, 0.0106638104533486, 0.425323120845664, 0.0466180768654915, 0.842402292743715, 0.208609687427072, 0.0166336682608816, 0.0464833846710956, 0.8299010611324, 0.233685747699204, 0.742469001175026, 0.967306766450795, 0.904840885401235, 0.357394700741248), .Dim = c(12L, 4L), .Dimnames = list( c(Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate), c(outcome, beta, se, pval ))) test2-as.data.frame(test) test2 Error in data.frame(outcome = c(cardva, respir, cereb, neoplasm, : duplicate row.names: Estimate [[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] Remove a column of a matrix with unnamed column header
Beend, Thanks for that. Conversion of test to a data frame resulted in a factor. Is there a possibility to selectively convert to numeric? I have tried this code and that has not produced the intended result. test[, c(2:4)] - sapply(test[, c(2:4)], as.numeric) On 8 November 2013 11:31, Berend Hasselman b...@xs4all.nl wrote: On 08-11-2013, at 10:40, Kuma Raj pollar...@gmail.com wrote: I have a matrix names test which I want to convert to a data frame. When I use a command test2-as.data.frame(test) it is executed without a problem. But when I want to browse the content I receive an error message Error in data.frame(outcome = c(cardva, respir, cereb, neoplasm, : duplicate row.names: Estimate . The problem is clearly due to a duplicate in row name . But I am unable to remove this column. I need help on how to remove this specific column that has essentially no column header name. dput of the matrix is here: dput(test) structure(c(cardva, respir, cereb, neoplasm, ami, ischem, heartf, pneumo, copd, asthma, dysrhy, diabet, 0.00259492159959046, 0.00979775441709427, 0.00103414632535868, 0.00486468139227382, 0.0164825543879707, 0.0116647168053943, -0.0012137908515233, 0.00730433232907741, 0.00355583994565985, 0.000712387285735019, -0.00103763671307935, 0.00981500221106926, 0.00325476724733837, 0.0049232113728293, 0.00520118026087645, 0.00386848394426742, 0.00688121694253705, 0.00585772614064902, 0.00564983058883797, 0.0061328202328586, 0.0108212194251692, 0.0173804438930357, 0.00867931407250442, 0.0106638104533486, 0.425323120845664, 0.0466180768654915, 0.842402292743715, 0.208609687427072, 0.0166336682608816, 0.0464833846710956, 0.8299010611324, 0.233685747699204, 0.742469001175026, 0.967306766450795, 0.904840885401235, 0.357394700741248), .Dim = c(12L, 4L), .Dimnames = list( c(Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate, Estimate), c(outcome, beta, se, pval ))) test2-as.data.frame(test) test2 Error in data.frame(outcome = c(cardva, respir, cereb, neoplasm, : duplicate row.names: Estimate rownames(test) - NULL Berend [[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] Subject: Regress multiple independent variables on multiple dependent variables
I want to estimate the effect of several independent variables on several dependent variables. In the example below I wanted to estimate the effect of three independent variables on ozone and temperature. My aim is to create a list of dependent and independent variables and automate the process rather than writing every dependent and independent variable in each model as I have done below. Example data is provided by the following library: library(faraway) data(ozone) mo3 - glm(O3 ~ humidity + ibh + ibt, data=ozone) mtemp- glm(temp ~ humidity + ibh + ibt, data=ozone) Thanks [[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] Logistic Regression - Variable Selection Methods With Prediction
Can I atleast get help with what pacakge to use for logistic
regression with all possible models and do prediction. I know i can
use regsubsets but i am not sure if it has any prediction functions to
go with it.
Thanks
On Oct 25, 6:54 pm, RAJ dheerajathr...@gmail.com wrote:
Hello,
I am pretty new to R, I have always used SAS and SAS products. My
target variable is binary ('Y' and 'N') and i have about 14 predictor
variables. My goal is to compare different variable selection methods
like Forward, Backward, All possible subsests. I am using
misclassification rate to pick the winner method.
This is what i have as of now,
Reg - glm (Graduation ~., DFtrain,family=binomial(link=logit))
step - extractAIC(Reg, direction=forward)
pred - predict(Reg, DFtest,type=response)
mis - mean({pred 0.5} != {DFtest[,Graduation] == Y})
This program actually works but I needed to check to make sure am
doing this right. Also, I am getting the same misclassification rates
for all different methods.
I also tried to use
Reg - leaps(Graduation ~., DFtrain)
pred - predict(Reg, DFtest,type=response)
mis - mean({pred 0.5} != {DFtest[,Graduation] == Y})
#print(summary(mis))
which doesnt work
and
Reg - regsubsets(Graduation ~., DFtrain)
pred - predict(Reg, DFtest,type=response)
mis - mean({pred 0.5} != {DFtest[,Graduation] == Y})
#print(summary(mis))
The Regsubsets will work but the 'predict' function does not work with
it. Is there any other way to do predictions when using regsubsets
Any help is appreciated.
Thanks,
__
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PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Logistic Regression - Variable Selection Methods With Prediction
Hello,
I am pretty new to R, I have always used SAS and SAS products. My
target variable is binary ('Y' and 'N') and i have about 14 predictor
variables. My goal is to compare different variable selection methods
like Forward, Backward, All possible subsests. I am using
misclassification rate to pick the winner method.
This is what i have as of now,
Reg - glm (Graduation ~., DFtrain,family=binomial(link=logit))
step - extractAIC(Reg, direction=forward)
pred - predict(Reg, DFtest,type=response)
mis - mean({pred 0.5} != {DFtest[,Graduation] == Y})
This program actually works but I needed to check to make sure am
doing this right. Also, I am getting the same misclassification rates
for all different methods.
I also tried to use
Reg - leaps(Graduation ~., DFtrain)
pred - predict(Reg, DFtest,type=response)
mis - mean({pred 0.5} != {DFtest[,Graduation] == Y})
#print(summary(mis))
which doesnt work
and
Reg - regsubsets(Graduation ~., DFtrain)
pred - predict(Reg, DFtest,type=response)
mis - mean({pred 0.5} != {DFtest[,Graduation] == Y})
#print(summary(mis))
The Regsubsets will work but the 'predict' function does not work with
it. Is there any other way to do predictions when using regsubsets
Any help is appreciated.
Thanks,
__
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Re: [R] Contingency table in R
Hi Laura and R users, I would like to know whether we can do siginificance test between Column Yes and Column No. Any one tried? I have seen it in Tabulaiton software packages from our vendors and in SPSS Custom Table. Thanks, On Wed, Mar 2, 2011 at 7:43 PM, Laura Clasemann violagirl...@msn.comwrote: Hi, I have a table in R with data I needed and need to create a contingency table out of it. The table I have so far looks like this: Binger r DietType No Yes Dangerous 15 12 Healthy52 9 None 134 24 Unhealthy 72 23 These are the error messages that I keep getting whenever I try to get a contingency table. I'm not sure why it won't work for me, any help would be appreciated! nametable-table(excat,recat) Error in table(excat, recat) : object 'excat' not found [[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.htmlhttp://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] sapply, lattice functions
You're right. It's necessary for xyplot though to prevent grouping. On Mar 20, 2010 10:43 AM, Dieter Menne dieter.me...@menne-biomed.de wrote: Sundar Dorai-Raj-2 wrote: Or perhaps more clearly, histogram(~a1 + b1 + c1, data = aa, o... Why outer=TRUE? Looks same for me without: Dieter library(lattice) aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5)) histogram(~a1 + b1 + c1, data = aa) -- View this message in context: http://n4.nabble.com/sapply-lattice-functions-tp1618134p1676043.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz [[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] sapply, lattice functions
Or perhaps more clearly, histogram(~a1 + b1 + c1, data = aa, outer = TRUE) --sundar On Fri, Mar 19, 2010 at 3:50 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try this: histogram(~ values | ind, stack(aa)) On Fri, Mar 19, 2010 at 5:44 PM, Santosh santosh2...@gmail.com wrote: Dear R-gurus aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5)) sapply(aa,function(x) histogram(x,breaks=NULL)) or px - sapply(aa,function(x) histogram(x,breaks=NULL)) print(px,split=c(1,1,1,1),more=F) The above code does not seem to work. am I missing something? Thanks, Santosh [[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. [[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] Source code for the t-distribution
Here it is. https://svn.r-project.org/R/trunk/src/nmath/pt.c --sundar On Tue, Mar 9, 2010 at 4:24 AM, Ravi Kulkarni ravi.k...@gmail.com wrote: I have tried looking for the source code for the pt() function in https://svn.r-project.org/R/trunk/src/library/stats/ and am unable to find it there. Can someone please tell me where to find it? Thanks, Ravi Kulkarni -- View this message in context: http://n4.nabble.com/Source-code-for-the-t-distribution-tp1585875p1585875.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. [[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] capturing errors in Sweave
Thanks, Berwin. That works just great!
--sundar
On Tue, Mar 2, 2010 at 12:57 AM, Berwin A Turlach
ber...@maths.uwa.edu.auwrote:
G'day Sundar,
On Mon, 1 Mar 2010 23:46:55 -0800
Sundar Dorai-Raj sdorai...@gmail.com wrote:
Thanks for the input, but I don't want try in the Sweave output. I
want the output to look just like it does in the console, as if an
uncaptured error really did occur.
I don't think that you will get around using try; and you will have
to work moderately hard to make the output appear as it does on the
console. Probably somewhere along the lines:
Sweave code start ++
Function-4a=
MySqrt - function(x) {
if (missing(x)) {
stop('x' is missing with no default)
}
if (!is.numeric(x)) {
stop('x' should only be numeric)
}
if (x 0) {
stop('x' should be non-negative)
}
return(sqrt(x))
}
@
echo=FALSE=
tmp - try(MySqrt())
@
eval=FALSE=
MySqrt()
@
echo=FALSE=
cat(tmp[1])
@
echo=FALSE=
tmp - try(MySqrt(a))
@
eval=FALSE=
MySqrt(a)
@
echo=FALSE=
cat(tmp[1])
@
echo=FALSE=
tmp - try(MySqrt(-2))
@
eval=FALSE=
MySqrt(-2)
@
echo=FALSE=
cat(tmp[1])
@
=
MySqrt(4)
@
+++ Sweave code end ++
Now what I would like to know is how to include easily warning messages
in my Sweave output without having to try whether Jean Lobry's [1] hack
still works. :)
HTH.
Cheers,
Berwin
[1]
https://www.stat.math.ethz.ch/pipermail/r-help/2006-December/121975.html
== Full address
Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr)
School of Maths and Stats (M019)+61 (8) 6488 3383 (self)
The University of Western Australia FAX : +61 (8) 6488 1028
35 Stirling Highway
Crawley WA 6009e-mail: ber...@maths.uwa.edu.au
Australiahttp://www.maths.uwa.edu.au/~berwin
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Re: [R] capturing errors in Sweave
What I ended up using was: cat(unclass(tmp)) --sundar On Tue, Mar 2, 2010 at 8:58 AM, Berwin A Turlach ber...@maths.uwa.edu.auwrote: G'day Sundar, On Tue, 2 Mar 2010 01:03:54 -0800 Sundar Dorai-Raj sdorai...@gmail.com wrote: Thanks, Berwin. That works just great! You are welcome. I noticed by now that cat(tmp) is sufficient; the tmp[1] in cat(tmp[1]) was a left over from earlier attempts to get the output to look correct. Cheers, Berwin [[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] capturing errors in Sweave
Hi,
I'm writing a manual using Sweave and I want to be able to print errors from
bad code. Here's an example:
Function-4a=
MySqrt - function(x) {
if (missing(x)) {
stop('x' is missing with no default)
}
if (!is.numeric(x)) {
stop('x' should only be numeric)
}
if (x 0) {
stop('x' should be non-negative)
}
return(sqrt(x))
}
MySqrt()
MySqrt(a)
MySqrt(-2)
MySqrt(4)
@
And I would like the output to be:
MySqrt - function(x) {
...
}
MySqrt()
Error in MySqrt() : 'x' is missing with no default
MySqrt(a)
Error in MySqrt(a) : 'x' should only be numeric
MySqrt(-2)
Error in MySqrt(-2) : 'x' should be non-negative
MySqrt(2)
[1] 1.414214
I.e. I want the Error statements to print in the tex file and not just make
Sweave to bomb.
Thanks,
--sundar
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Re: [R] capturing errors in Sweave
Thanks for the input, but I don't want try in the Sweave output. I want
the output to look just like it does in the console, as if an uncaptured
error really did occur.
--sundar
On Mon, Mar 1, 2010 at 11:42 PM, Sharpie ch...@sharpsteen.net wrote:
Sundar Dorai-Raj-2 wrote:
Hi,
I'm writing a manual using Sweave and I want to be able to print errors
from
bad code. Here's an example:
Function-4a=
MySqrt - function(x) {
if (missing(x)) {
stop('x' is missing with no default)
}
if (!is.numeric(x)) {
stop('x' should only be numeric)
}
if (x 0) {
stop('x' should be non-negative)
}
return(sqrt(x))
}
MySqrt()
MySqrt(a)
MySqrt(-2)
MySqrt(4)
@
And I would like the output to be:
MySqrt - function(x) {
...
}
MySqrt()
Error in MySqrt() : 'x' is missing with no default
MySqrt(a)
Error in MySqrt(a) : 'x' should only be numeric
MySqrt(-2)
Error in MySqrt(-2) : 'x' should be non-negative
MySqrt(2)
[1] 1.414214
I.e. I want the Error statements to print in the tex file and not just
make
Sweave to bomb.
Thanks,
--sundar
You can catch errors in R using the try() function:
foo - try( log(A), silent = TRUE )
if( class( foo ) == 'try-error' ){
print( unclass( foo ) )
}
[1] Error in log(\A\) : Non-numeric argument to mathematical function\n
There might be a more clever way to override the default error handling
that
would save you from wrapping everything in try()-- but this method should
work.
Hope it helps!
-Charlie
-
Charlie Sharpsteen
Undergraduate-- Environmental Resources Engineering
Humboldt State University
--
View this message in context:
http://n4.nabble.com/capturing-errors-in-Sweave-tp1574642p1574686.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] How do I juxtapose two lattice graphs with common X axes such that the X axes line up?
Try googling latticeExtra x.same for some examples. Here's one:
http://www.mail-archive.com/r-help@r-project.org/msg39048.html
On Wed, Jan 20, 2010 at 9:44 AM, George Chen glc...@stanford.edu wrote:
Hello,
I would like to juxtapose two lattice graphs with common X axes such that
the X axes line up. I am using plot right now but the edges are not neat
and it would be nice if I could just draw 1 X axis and not both of them.
Here is my code:
upper-bwplot(SignalUsed~as.factor(AllNormalHitsNamesCount),data=NmlOverviewArray2,
xlab=,
ylab=Intensity of Individual Antibody Responses,
main=Intensity, Frequency, Distribution, Quantity of Normal
Antibody Responses,
box.ratio=1,
panel = function (AllNormalHitsNamesCount,...) {
panel.bwplot(...)
}
)
lower-barchart(as.vector(table(NmlOverviewArray2$AllNormalHitsNamesCount))
~as.factor(as.numeric(names(table(NmlOverviewArray2$AllNormalHitsNamesCount,
data=NmlOverviewArray2,
ylab=Number of Individual Antibody Responses,
xlab=Occurrence of Individual Antibody Responses
(Out of 45 Normals),
box.ratio=1)
plot (upper, newpage=TRUE, more=TRUE, position = c(0,.15,1,1))
plot (lower, newpage=FALSE, more=TRUE, position = c(0,0,1,.3))
George
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Re: [R] Dynamic arguments in rbind function
Use a list instead of assign then do.call(rbind, thelist).
import.files - c(a.txt, b.txt, c.txt, d.txt, e.txt)
imp - vector(list, length(import.files))
for (i in 1:length(import.files)) {
imp[[i]] - read.delim(import.files[i], sep = , header = TRUE)
}
combined - do.call(rbind, imp)
HTH,
--sundar
On Mon, Jan 4, 2010 at 4:31 PM, Steven Kang stochastick...@gmail.comwrote:
Hi, all
Basically, I have unknown number of data that need to be imported and
collapsed row-wisely.
The code below works fine, however the rbind function may require 50
arguments if there are 50 data files...
Thus, I would like to explore whether there are any methods in using
dynamic
objects (i.e from the resulting objects in the for loop) as an argument in
the *rbind* function.
setwd(.)
import.files - c(a.txt, b.txt, c.txt, d.txt, e.txt)
for (i in 1:length(import.files)) {
assign(paste(imp, i, sep = .), read.delim(eval(paste(.\\,
import.files[i], sep = )), header = TRUE))
}
combined - rbind(*imp.1, imp.2, imp.3, imp.4, imp.5, imp.6*)
Your expertise in resolving this issue would be greatly appreciated.
Steve
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Re: [R] makefile for sweave
Is texi2dvi in your PATH? What happens if you open a CMD window and
type texi2dvi at the prompt?
--sundar
On Thu, Nov 26, 2009 at 6:14 AM, Wolfgang Raffelsberger wr...@igbmc.fr wrote:
Dear all,
I can't get texi2dvi working right. Basically I'd like to convert a .lex to
.pdf without having to fiddle with the issue Sweave.sty not being in my
current directory (as this was sugested in other posts on this list).
When I'm in the R-Gui I can get the help via
?texi2dvi
(So I conclude its installed.)
However, when I try to use it to concert a .tex to .pdf I get trouble ...
For example :
A)
The file test02.r contains :
Sweave(Sweave_test01.rnw)
library(tools)
texi2dvi(Sweave_test01.tex, pdf =T)
Now, when I run on the linux command line :
R --vanilla -q test02.r
I get :
Sweave(Sweave_test01.rnw)
Writing to file Sweave_test01.tex
Processing code chunks ...
1 : term hide (label=chunk_ini)
2 : term verbatim eps pdf (label=Fig01)
3 : term tex (label=packageVersionInfo)
Loading required package: xtable
You can now run LaTeX on 'Sweave_test01.tex'
library(tools)
texi2dvi(Sweave_test01.tex, pdf =T)
Error in texi2dvi(Sweave_test01.tex, pdf = T) :
Running 'texi2dvi' on 'Sweave_test01.tex' failed.
Messages:
sh: texi2dvi: command not found
Execution halted
B)
In a previous message on this list I found the following command line(s)
suggested, but I my case it won't work
star5_R_test_ R CMD texi2dvi --help
/usr/local/lib64/R/bin/Rcmd: line 62: exec: texi2dvi: not found
similarly, when execute (as sugested) I get the same error message
star5_R_test_ R CMD texi2dvi -p Sweave_test01.tex
/usr/local/lib64/R/bin/Rcmd: line 62: exec: texi2dvi: not found
I don't understand how can a command can be present (= installed) and still
not being found as the error messages suggest ?
For completeness :
sessionInfo()
R version 2.10.0 (2009-10-26)
x86_64-unknown-linux-gnu
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] xtable_1.5-6 mouse4302probe_2.5.0 AnnotationDbi_1.8.1
[4] mouse4302cdf_2.5.0 MASS_7.3-3 fdrtool_1.2.5 [7]
limma_3.2.1 affyPLM_1.22.0 preprocessCore_1.8.0
[10] gcrma_2.18.0 affy_1.24.2 Biobase_2.6.0
loaded via a namespace (and not attached):
[1] Biostrings_2.14.3 DBI_0.2-4 IRanges_1.4.4 RSQLite_0.7-3
[5] affyio_1.14.0 splines_2.10.0 tools_2.10.0
Thank's in advance,
Wolfgang
Charles C. Berry a écrit :
On Tue, 8 Sep 2009, Welma Pereira wrote:
Hello, I have the following makefile. The problem is that the
bibliography
doesn t work. Any help would be appreciated! I really don t don t what to
do..:-(
# The sources of the report (tex, Rnw and other files (e.g. bib, idx))
TEX_CMPS = Report problem
RNW_CMPS = prop1 prop2 ExeExps
OTHER = Report.bib
# The name of the report to produce
all: Report.pdf
code: $(RNW_CMPS:=.R)
clean:
rm -f *.log *.dvi *~
# On what does the report depends?
Report.pdf: $(TEX_CMPS:=.tex) $(RNW_CMPS:=.tex) ${OTHER} makefile
TEXINPUTS=${TPUTS} pdflatex $
TEXINPUTS=${TPUTS} pdflatex $
IIRC
R CMD texi2dvi -p target.tex
takes care of finding sweave.sty and running latex thru all the iterations
needed to build cross-references and a usable pdf.
Try
R CMD texi2dvi --help
at the shell prompt.
HTH,
Chuck
rm *.log
# mv *.aux $(dir $)
# How to build the tex files from the Rnw (Sweave) files
%.tex: %.Rnw
echo library(utils); options(width=60); Sweave('$') | ${R_PRG}
--no-save --vanilla
mv $(notdir $*.tex) $(dir $)
# How to build the R code files from the Rnw (Sweave) files
%.R: %.Rnw
echo library(utils); Stangle('$') | ${R_PRG} --no-save --vanilla
%.bib:
TEXINPUTS=${TPUTS} pdflatex $
bibtex $
%.aux:
TEXINPUTS=${TPUTS} pdflatex $
bibtex $
%.idx:
TEXINPUTS=${TPUTS} pdflatex $
makeindex $
cheers!
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Charles C. Berry (858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
__
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wolfgang Raffelsberger, PhD
Laboratoire de BioInformatique et Génomique Intégratives
IGBMC, 1 rue Laurent Fries, 67404 Illkirch Strasbourg, France
Tel (+33) 388 65 3300 Fax (+33) 388 65 3276
wolfgang.raffelsberger (at) igbmc.fr
Re: [R] Why F value and Pr are not show in summary() of an aov() result?
It's hard to read your code, so I won't comment on your specific
example. So when all else fails read the documentation for
?summary.aov:
They have columns ‘Df’, ‘Sum Sq’, ‘Mean
Sq’, as well as ‘F value’ and ‘Pr(F)’ if there are non-zero
residual degrees of freedom.
So if you do df.residual(afit), is it 0?
--sundar
On Sun, Nov 22, 2009 at 7:19 AM, Peng Yu pengyu...@gmail.com wrote:
I have the following code. I'm wondering why summary() doesn't show F
value and Pr?
Rscript multi_factor.R
a=3
b=4
c=5
d=6
e=7
A=1:a
B=1:b
C=1:c
D=1:d
E=1:e
X=matrix(nr=a*b*c*d*e,nc=5)
colnames(X)=LETTERS[1:5]
for(i_a in 1:a-1) {
+ for(i_b in 1:b-1) {
+ for(i_c in 1:c-1) {
+ for(i_d in 1:d-1) {
+ for(i_e in 1:e-1) {
+ X[(((i_a * b + i_b) * c + i_c) * d + i_d) * e + i_e + 1, ]
= c(i_a+1, i_b+1, i_c+1, i_d+1, i_e+1)
+ }
+ }
+ }
+ }
+ }
Y=matrix(nr=a*b*c*d*e,nc=1)
for(i in 1:(a*b*c*d*e)) {
+ fa=X[i,'A']
+ fb=X[i,'B']
+ fc=X[i,'C']
+ fd=X[i,'D']
+ fe=X[i,'E']
+
+ Y[i,1]= fa +fb +fc +fe +fa*fb +fa*fc +fb*fc +fa*fe +fc*fe
+fa*fb*fc +fa*fc*fe + rnorm(1)
+ }
aframe = data.frame(
+ A=as.factor(X[,'A'])
+ , B=as.factor(X[,'B'])
+ , C=as.factor(X[,'C'])
+ , D=as.factor(X[,'D'])
+ , E=as.factor(X[,'E'])
+ ,Y)
afit=aov(Y ~ A * B * C * D * E, aframe)
summary(afit)
Df Sum Sq Mean Sq
A 2 1512240 756120
B 3 453324 151108
C 4 2549895 637474
D 5 2 0.3693
E 6 1451057 241843
A:B 6 33875 5646
A:C 8 189839 23730
B:C 12 56024 4669
A:D 10 7 1
B:D 15 25 2
C:D 20 18 1
A:E 12 107574 8964
B:E 18 21 1
C:E 24 180413 7517
D:E 30 16 1
A:B:C 24 4167 174
A:B:D 30 37 1
A:C:D 40 42 1
B:C:D 60 63 1
A:B:E 36 30 1
A:C:E 48 13298 277
B:C:E 72 62 1
A:D:E 60 79 1
B:D:E 90 87 1
C:D:E 120 122 1
A:B:C:D 120 140 1
A:B:C:E 144 131 1
A:B:D:E 180 145 1
A:C:D:E 240 225 1
B:C:D:E 360 398 1
A:B:C:D:E 720 713 1
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Re: [R] denoting max value in ylim
ylim = c(0, max(log10(D10$Part.P))) Make sure you remove any 0s or NAs before computing the max though. --sundar On Fri, Nov 20, 2009 at 6:12 AM, helene frigstad helenefrigs...@hotmail.com wrote: Hi, is there any way to set the ylim range from zero to whatever is the max value in that dataset? I am plotting many similar plots to the one below, and would like to avoid having to find the max value each time. plot (D10$Part.P ~ D10$Klorofyll,pch=16,log = xy, xlab = (Chla), ylab = (POP), ylim = c (0, ???)) abline(m3, untf=TRUE, lty=1,col=blue) text(5, 0.05,paste(round(glm(D10$Part.P ~ D10$Klorofyll, data = D10, family = Gamma(link = identity))$coef, 2),collapse = )) thank you very much for your time and help. Best regards, Helene Frigstad -- View this message in context: http://old.nabble.com/denoting-max-value-in-ylim-tp26441590p26441590.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] lapply() not converting columns to factors (no error message)
Works for me: x - read.csv(url(http://dc170.4shared.com/download/153147281/a5c78386/Testvcomp10.csv?tsid=20091116-075223-c3093ab0;)) names(x) x[2:13] - lapply(x[2:13], factor) levels(x$P1L55) [1] 0 1 is.factor(x$P1L96) [1] TRUE sessionInfo() R version 2.10.0 (2009-10-26) i386-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.0 tools_2.10.0 On Mon, Nov 16, 2009 at 4:50 AM, A Singh aditi.si...@bristol.ac.uk wrote: Sorry, my file is at: http://www.4shared.com/file/153147281/a5c78386/Testvcomp10.html -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ 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] lapply() not converting columns to factors (no error message)
Could it be you have factor redefined in your workspace? Have you tried it in a clean directory? I.e. a directory where no .RData exists? On Mon, Nov 16, 2009 at 5:07 AM, A Singh aditi.si...@bristol.ac.uk wrote: Oh, strange! I thought it might be a problem with the 'base' package installation, because the same thing's worked for me too before but won't do now. I tried to reinstall it (base), but R says its there already which I expected it to be anyway. I don't quite know where the issue is. Very odd. --On 16 November 2009 04:59 -0800 Sundar Dorai-Raj sdorai...@gmail.com wrote: Works for me: x - read.csv(url(http://dc170.4shared.com/download/153147281/a5c78386/Testvc omp10.csv?tsid=20091116-075223-c3093ab0)) names(x) x[2:13] - lapply(x[2:13], factor) levels(x$P1L55) [1] 0 1 is.factor(x$P1L96) [1] TRUE sessionInfo() R version 2.10.0 (2009-10-26) i386-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.0 tools_2.10.0 On Mon, Nov 16, 2009 at 4:50 AM, A Singh aditi.si...@bristol.ac.uk wrote: Sorry, my file is at: http://www.4shared.com/file/153147281/a5c78386/Testvcomp10.html -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ 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. -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ 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] Trellis settings get lost when printing to pdf
Did you make the changes before or after starting the device: library(lattice) ## before doesn't change the settings on the device: trellis.par.set(plot.symbol = list(col = red)) trellis.device(pdf, file = tmp.pdf) xyplot(1 ~ 1) dev.off() ## after does trellis.device(pdf, file = tmp.pdf) trellis.par.set(plot.symbol = list(col = red)) xyplot(1 ~ 1) dev.off() I never do things like this, though. I would suggest creating a theme instead and supplying it to xyplot (or whatever plot you're using) using par.settings: my.theme - list(plot.symbol = list(col = red)) trellis.device(pdf, file = tmp.pdf) xyplot(1 ~ 1, par.settings = my.theme) dev.off() HTH, --sundar 2009/11/13 Joel Fürstenberg-Hägg joel_furstenberg_h...@hotmail.com: Hi all, I've got some problems when changing the trellis settings for the lattice plots. The plots look exactly as I want them to when calling show.settings() as well as when plotting them in the graphical window. But when printing to a pdf file, none of the settings are used!? Does anyone know what might have happened? Because the when changing the trellis settings, these should remain in the new state until you close R right..? # Change settings for the boxplot appearance new.dot=trellis.par.get(box.dot) new.rectangle=trellis.par.get(box.rectangle) new.umbrella=trellis.par.get(box.umbrella) new.symbol=trellis.par.get(plot.symbol) new.strip.background=trellis.par.get(strip.background) new.strip.shingle=trellis.par.get(strip.shingle) new.dot$pch=| new.dot$col=black new.rectangle$col=black new.rectangle$fill=grey65 new.umbrella$col=black new.umbrella$lty=1 # Continous line, not dotted new.symbol$col=black new.strip.background$col=grey87 # Background colour in the upper label new.strip.shingle$col=black # Border colour around the upper label trellis.par.set(box.dot=new.dot, box.rectangle=new.rectangle, box.umbrella=new.umbrella, plot.symbol=new.symbol, strip.background=new.strip.background, strip.shingle=new.strip.shingle) Best regards, Joel _ Nya Windows 7 - Hitta en dator som passar dig! Mer information. http://windows.microsoft.com/shop [[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] tapply function
you must have missing values in data. Try tapply(data, group, mean, na.rm = TRUE) If that's not the case, read the bottom of this email about the posting guide. HTH, --sundar On Tue, Nov 3, 2009 at 5:28 AM, FMH kagba2...@yahoo.com wrote: Hi, I tried to use tapply function to find the mean of the data in each group as the following command, but the result are NA, as there are several missing values in each group. tapply(data,group,mean) Could someone please advice me the way to ignore the missing data in order for the fucntion to run successfully? Thanks Fir __ 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] Issue with %in% - not matching identical rows in data frames
?%in% says x and table must be vectors. You supplied data.frames. So %in% is coercing your today.sequence to a vector using as.character(today.sequence) Perhaps you should paste the columns together first: x - do.call(paste, c(sequence, sep = ::)) table - do.call(paste, c(today.sequence, sep = ::)) x[7] %in% table I'm not sure if this is what you want/need, but it does match your example. HTH, --sundar On Tue, Nov 3, 2009 at 7:53 AM, Kaushik Krishnan kaushik.s.krish...@gmail.com wrote: Hi folks I have two data frames. I know that the nth (let's say the 7th) row in the first data frame (sequence) is there in the second (today.sequence). When I try to check that by doing 'sequence[7,] %in% today.sequence', I get all FALSE when it should be all TRUE. I'm certain I'm making some trivial mistake. Any solutions? The code to recreate the data frames and see for yourself is: sequence - structure(list(DATE = structure(c(14549, 14549, 14553, 14550, 14557, 14550, 14551, 14550), class = Date), DATASET = c(1L, 2L, 1L, 2L, 2L, 3L, 3L, 4L), REP = c(1L, 0L, 2L, 2L, 3L, 0L, 1L, 0L), WRONGS_ABS = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), WRONGS_RATIO = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), DONE = c(1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), class = data.frame, row.names = c(NA, -8L)) today.sequence - structure(list(DATE = structure(c(14551, 14550), class = Date), DATASET = 3:4, REP = c(1L, 0L), WRONGS_ABS = c(0L, 0L), WRONGS_RATIO = c(0L, 0L), DONE = c(0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), row.names = 7:8, class = data.frame) sequence[7,] #You should see '2009-11-03 3 1 0 0 0' today.sequence #You can clearly see that sequence [7,] is the first row in today.sequence sequence[7,] %in% today.sequence #This should show 'TRUE TRUE TRUE TRUE TRUE TRUE'. Instead # it shows 'FALSE FALSE FALSE FALSE FALSE FALSE' Thanks -- Kaushik Krishnan (kaushik.s.krish...@gmail.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] How can I run a function to a piece of text?
Based solely on what you told us, this can be done using eval(parse(text=...)) cmd - sprintf(mean(%s), script) eval(parse(text = cmd)) However, with more context, there may be a better solution. See, for example, install.packages(fortunes) library(fortunes) fortune(parse()) HTH, --sundar On Fri, Oct 16, 2009 at 11:39 AM, Javier PB j.perez-barbe...@macaulay.ac.uk wrote: Dear users, I got really stuck trying to apply a function to a piece of code that I created using different string functions. To make things really easy, this is a wee example: x-c(1:10) script-x, trim = 0, na.rm = FALSE ##script created by a number of paste() and rep() steps mean(script) ##function that I want to apply: doesn't work Is there any way to convert the script class so that the function mean() can read it as if it were text typed in the console? Thanks and have a superb day Javier -- View this message in context: http://www.nabble.com/How-can-I-run-a-function-to-a-piece-of-text--tp25930315p25930315.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] Error in family$family : $ operator is invalid for atomic vectors
Check to see if you have an old workspace being loaded. You might have an object called 'family' which you might need to remove. --sundar On Oct 11, 2009 12:15 PM, romunov romu...@gmail.com wrote: Thank you Jorge and Barry for your input. I've fiddled around a bit and as a result, am even more confused. If I start R console via Notepad++ (I use Npp2R) and execute the model1, it goes through just fine. Here is the sessionInfo() for this working session: sessionInfo() R version 2.9.2 (2009-08-24) i386-pc-mingw32 locale: LC_COLLATE=Slovenian_Slovenia.1250;LC_CTYPE=Slovenian_Slovenia.1250;LC_MONETARY=Slovenian_Slovenia.1250;LC_NUMERIC=C;LC_TIME=Slovenian_Slovenia.1250 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.9.2 And if I run R normally, via an icon from the desktop (Rgui.exe) it gives the aforementioned error. Here is the sessionInfo() for non-working session. Is it possible that grid, reshape, plyr, ggplot2 and proto could be causing this? If so, how can I prevent them from loading automatically or unloading from a live session? sessionInfo() R version 2.9.2 (2009-08-24) i386-pc-mingw32 locale: LC_COLLATE=Slovenian_Slovenia.1250;LC_CTYPE=Slovenian_Slovenia.1250;LC_MONETARY=Slovenian_Slovenia.1250;LC_NUMERIC=C;LC_TIME=Slovenian_Slovenia.1250 attached base packages: [1] stats graphics grDevices utils datasets grid methods [8] base other attached packages: [1] reshape_0.8.3 plyr_0.1.9proto_0.3-8 loaded via a namespace (and not attached): [1] ggplot2_0.8.3 Cheers, Roman On Sun, Oct 11, 2009 at 6:32 PM, Jorge Ivan Velez jorgeivanve...@gmail.comwrote: Hi Romain, It works for me: model1 - glm(as.vector(x) ~dept*sex*admit,poisson) model1 ... [[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 speed up R with version 2.9.2?
Another possibility is a very large .RData file in the directory where you're starting R. You can try Rgui --no-restore (I don't have windows, so I'm not sure if this an option with RGui, though I know it is with R.) --sundar On Fri, Oct 2, 2009 at 8:50 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Its under 5 seconds on my Vista laptop. Do you have any startup files? If Rgui --vanilla is much faster then your startup files are the problem. On Fri, Oct 2, 2009 at 11:45 AM, FMH kagba2...@yahoo.com wrote: Thank you for your answer. I'm using Win XP with 2GB RAM in memory. Cheers Fir - Original Message From: stephen sefick ssef...@gmail.com To: FMH kagba2...@yahoo.com Cc: r-help@r-project.org Sent: Fri, October 2, 2009 4:38:10 PM Subject: Re: [R] How to speed up R with version 2.9.2? You're fine, but please do read the posting guide. What OS etc. Where you doing anything else on the computer? Is this a RAM limitation? I have 2.9.2 running on two flavours of linux, mac os x and windows all 2.9.2 and there doesn't seem to be a problem. regards, Stephen On Fri, Oct 2, 2009 at 10:35 AM, FMH kagba2...@yahoo.com wrote: Dear All, I'm sorry if my question does not suit with this R group. I have recently installed R software with version 2.9.2, but i found the program took almost 1 minute as soon as it was opened, before it can be used. However, the previous version 2.9.1 only take few seconds after the menu bar was clicked. This circumstance has caused me to wait for couple of minutes as several R windows were opened simultaneously. Are there any ways to speed up on this 2.9.2 version? Could someone give some hints, please? Thank you Fir __ 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. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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-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] Skipping missing files when importing data
Try ?file.exists.
if (file.exists(fxxx)) {
read.table(fxxx)
} else {
cat(\, fxxx, \ is missing\n, sep = )
}
HTH,
--sundar
On Sun, Sep 20, 2009 at 9:28 PM, jiangrm jian...@gmail.com wrote:
Trying to import a bunch of data files named like f001, f002, f999. Some
of the files may be
missing and the missing files vary from time to time.
Used for loop and read.table. When it reaches the missing file (say f100), it
shows:
Error in file(file, r) : cannot open the connection
In addition: Warning message:
In file(file, r) :
cannot open file 'f100': No such file or directory
and the program stops.
How can I skip the missing ones and keep the program running? Guess either
checking the validity of
filenames, or ignore the error message may work. Which functions should be
used or any better ideas?
-RJ
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Re: [R] xyplot: Can I identify groups but not use them for regression?
I think this ought to work for you:
library(lattice)
set.seed(42)
d - data.frame(year = c(rep(2007,12), rep(2008,12)),
treatment = rep(LETTERS[1:3], each = 4, times = 2))
d$cover - rnorm(nrow(d))
d$variable - rnorm(nrow(d))
xyplot(variable ~ cover | year, d,
panel = function(x, y, ...) {
panel.superpose(x, y, ...)
panel.lmline(x, y, ...)
},
groups = treatment)
HTH,
--sundar
On Fri, Sep 18, 2009 at 3:42 PM, Seth W Bigelow sbige...@fs.fed.us wrote:
I wish to identify groups representing different treatments, but to plot
them and do a regression using a continuous variable (cover)
ignoring the groupings.
d$year - NA
d$year -c(rep(2007,12), rep(2008,12))
d$treatment - c(rep(A,4),rep(B,4),rep(C,4), rep(A,4), rep(B,4),
rep(C,4))
d$cover - rnorm(24)
d$variable - rnorm(24)
xyplot(variable ~ cover | year, d,
type=c(p,r),
groups=treatment
)
As it stands, a different regression line is plotted for each treatment.
Oh, and how do I display the actual numeric value of year (e.g., 2007)
in the strip, rather than the word year?
--Seth
Dr. Seth W. Bigelow
Biologist, USDA-FS Pacific Southwest Research Station
1731 Research Park Drive, Davis California
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice graph tweaking
A reproducible example would be nice. Try grid = FALSE for the first question, though I'm unaware which lattice plot you are using where the default is TRUE. So I can't guarantee that will even work. For your second question, add par.settings = list(strip.background = list(col = white)) to your call (e.g. xyplot(..., par.settings = ...)). To see what other parameters you can set in par.settings, try: str(trellis.par.get()) HTH, --sundar On Tue, Aug 25, 2009 at 6:14 AM, Wallis, Daviddavid.walli...@imperial.ac.uk wrote: To: silwood-r Subject: Removing lattice graph gridlines and editing label box colour Hi, Is it possible to remove the background gridlines from a lattice graph (ie graph made up of multiple individual graphs with annoying blue grid in the backgroun)? Also, Is it possible to change the colour of the individual graph label boxes? - ie the default pink boxes above the individual graphs Thanks [[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] Help with nls and error messages singular gradient
Hi, Michael,
I think the SPSS answer is wrong. Your starting values are way off.
Look at this plot for verification:
con - textConnection(time bod
11 0.47
22 0.74
33 1.17
44 1.42
55 1.60
67 1.84
79 2.19
8 11 2.17)
mydata - read.table(con, header = TRUE)
close(con)
beta - c(3, -0.1) # your initial values
beta - c(2.4979, -2.02456) # SPSS answer
mycurve - function(x) {
beta[1]/(1 - exp(beta[1] * x))
}
curve(mycurve, from = 1, to = 11,
ylim = range(mydata$bod, mycurve(mydata$time)))
points(mydata$time, mydata$bod)
You might want to look at package nls2 which allows a brute force grid
search to find some starting values. Or rethink the equation you're
trying to fit.
HTH,
--sundar
On Tue, Aug 25, 2009 at 9:10 AM, Michael Pearmainmpearm...@google.com wrote:
Hi All,
I'm trying to run nls on the data from the study by Marske (Biochemical
Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
(1988).
Data is as follows, (stored as mydata)
time bod
1 1 0.47
2 2 0.74
3 3 1.17
4 4 1.42
5 5 1.60
6 7 1.84
7 9 2.19
8 11 2.17
I then run the following;
#Plot initial curve
plot(mydata$time, mydata$bod,xlab=Time (in days),ylab=biochemical oxygen
demand (mg/l) )
model - nls(bod ~ beta1/(1 - exp(beta2*time)), data =
mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T)
The start values are recommended, (I have used these values in SPSS without
any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456)
but return the error message,
Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start =
list(beta1 = 3, : singular gradient
Can anyone offer any advice?
Thanks in advance
Mike
--
Michael Pearmain
Senior Analytics Research Specialist
“Statistics are like women; mirrors of purest virtue and truth, or like
whores to use as one pleases”
Google UK Ltd
Belgrave House
76 Buckingham Palace Road
London SW1W 9TQ
United Kingdom
t +44 (0) 2032191684
mpearm...@google.com
If you received this communication by mistake, please don't forward it to
anyone else (it may contain confidential or privileged information), please
erase all copies of it, including all attachments, and please let the sender
know it went to the wrong person. Thanks.
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with nls and error messages singular gradient
Another alternative is to use SSlogis which is very similar to the
model you're fitting except with one additional parameter:
Asym - 3
xmid - 0
scal - 10
model - nls(bod ~ SSlogis(time, Asym, xmid, scal), data = mydata)
summary(model)
plot(bod ~ time, mydata)
newdata - data.frame(time = seq(1, 11, length = 100))
lines(newdata$time, predict(model, newdata))
HTH,
--sundar
On Tue, Aug 25, 2009 at 10:05 AM, Sundar Dorai-Rajsdorai...@gmail.com wrote:
Hi, Michael,
I think the SPSS answer is wrong. Your starting values are way off.
Look at this plot for verification:
con - textConnection(time bod
1 1 0.47
2 2 0.74
3 3 1.17
4 4 1.42
5 5 1.60
6 7 1.84
7 9 2.19
8 11 2.17)
mydata - read.table(con, header = TRUE)
close(con)
beta - c(3, -0.1) # your initial values
beta - c(2.4979, -2.02456) # SPSS answer
mycurve - function(x) {
beta[1]/(1 - exp(beta[1] * x))
}
curve(mycurve, from = 1, to = 11,
ylim = range(mydata$bod, mycurve(mydata$time)))
points(mydata$time, mydata$bod)
You might want to look at package nls2 which allows a brute force grid
search to find some starting values. Or rethink the equation you're
trying to fit.
HTH,
--sundar
On Tue, Aug 25, 2009 at 9:10 AM, Michael Pearmainmpearm...@google.com wrote:
Hi All,
I'm trying to run nls on the data from the study by Marske (Biochemical
Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
(1988).
Data is as follows, (stored as mydata)
time bod
1 1 0.47
2 2 0.74
3 3 1.17
4 4 1.42
5 5 1.60
6 7 1.84
7 9 2.19
8 11 2.17
I then run the following;
#Plot initial curve
plot(mydata$time, mydata$bod,xlab=Time (in days),ylab=biochemical oxygen
demand (mg/l) )
model - nls(bod ~ beta1/(1 - exp(beta2*time)), data =
mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T)
The start values are recommended, (I have used these values in SPSS without
any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456)
but return the error message,
Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start =
list(beta1 = 3, : singular gradient
Can anyone offer any advice?
Thanks in advance
Mike
--
Michael Pearmain
Senior Analytics Research Specialist
“Statistics are like women; mirrors of purest virtue and truth, or like
whores to use as one pleases”
Google UK Ltd
Belgrave House
76 Buckingham Palace Road
London SW1W 9TQ
United Kingdom
t +44 (0) 2032191684
mpearm...@google.com
If you received this communication by mistake, please don't forward it to
anyone else (it may contain confidential or privileged information), please
erase all copies of it, including all attachments, and please let the sender
know it went to the wrong person. Thanks.
[[alternative HTML version deleted]]
__
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Reg: Creation of own Package and Permanent Dataset in R
Hi, This is Raj from ClinAsia and we have a small query with respect to R Statistical Package. Our Query: Actually while opening R console and R commander we see some packages like car and datasets. In these packages we have default datasets. For example: Women and Prestige so on. Now we created a Sales dataset importing either from excel, xml or text file. Now we are trying to store that dataset permanently in any one of the packages mentioned above (car or datasets). I am able to create them temporarily untill that pirticular session. But once we close the session and try to log into R Console and R Commander. We are not able to find the earlier created datasets Sales in the packages (Car and Datasets). Kindly suggest how to create permanent datasets in packages and also suggest how to create our own packages. If possible please send us the code it will be very helpful for us. Thanks and Regards, Raj ClinAsia 91-40-20010112 [[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 check if ... is empty
Try
dots - list(...)
if (length(dots) == 0) {
## do something
}
On Thu, Jul 16, 2009 at 6:46 AM, Thomas Roth (geb.
Kaliwe)hamstersqu...@web.de wrote:
Hi,
I was wondering what would be the best way to check if the three dots
argument contains any arguments (i.e. does ... contain any arguments or not?
)
#Example
test = function(x,y, ...)
{
#Wanted R-Code
# if(empty(...))
# do some calculation
plot(x,y,...)
}
Thanks
Thomas Roth
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Re: [R] Lattice group colors?
Look at show.settings() and str(trellis.par.get()). This will show you what the default settings are. The group colors are set by the superpose.* elements (e.g. superpose.line is for group lines). To set them, I usually create a list and pass it to par.settings. For example, my.theme - list(superpose.line = list(col = c(darkred, darkblue), lty = 1:2)) xyplot(y ~ x, data = mydata, groups = g, par.settings = my.theme, auto.key = list(points = FALSE, lines = TRUE)) HTH, --sundar On Mon, Jun 22, 2009 at 6:30 AM, Fredrik Karlssondargo...@gmail.com wrote: Dear list, I have been struggling to find how I would go about changing the bakground colors of groups in a lattice barchart in a way so that the auto.key generated also does the right thing and pick it up for the key. I have used the col argument (which I guess is sent to par()) in a way so that the colors are like I would want them, but now I guess I need to know which part part of the theme I need to change in order to make this change permanent for all the plots, and all the keys. I am of course thankful for all the help I can get. (Also, how does one know these things about the theme variables? Is there some documentation somewhere?) /Fredrik -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ 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] reading in multiple files
You could try: do.call(rbind, lapply(list.files(path/to/files, full = TRUE), read.csv)) And add more arguments to lapply if the files are not csv, have no header, etc. --sundar On Tue, Jun 9, 2009 at 11:18 AM, Erin Hodgesserinm.hodg...@gmail.com wrote: Dear R People: I have about 6000 files to be read in that I'd like to go to one matrix. There are two columns, 1 line in each file. Is there a way to bring them in to produce one matrix or data frame, please? Thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.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] find a sequence of characters in a vector
use gregexpr and paste aze - paste(c(a, z, e), collapse = ) sequence - paste(c(a,z,e,r,t,a,z,a,z,e,c), collapse = ) gregexpr(aze, sequence, fixed = TRUE) [[1]] [1] 1 8 attr(,match.length) [1] 3 3 HTH, --sundar On Fri, Jun 5, 2009 at 6:22 AM, Ptit_Bleuptit_b...@yahoo.fr wrote: Hello, I'm just looking for an easy way to find the positions of a complete sequence in a bigger vector. For example : c(a,z,e) in c(a,z,e,r,t,a,z,a,z,e,c) and the result should be 1 8 that is the positions of the beginning of the complete sequence. I tried with %in%, match, is.element but all I get is, for example which(c(a,z,e) in c(a,z,e,r,t,a,z,a,z,e,c)) 1 2 3 meaning that each character is in the bigger vector. It must be easy, except for me. Sorry. If you have a solution, thanks in advance to share it. Have a good week-end, Ptit Bleu. -- View this message in context: http://www.nabble.com/find-a-sequence-of-characters-in-a-vector-tp23888063p23888063.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] error message re: max(i), but code and output seen O.K.
This error is thrown if the argument to max is either NULL or length zero: [~] Rscript -e max(NULL) [1] -Inf Warning message: In max(NULL) : no non-missing arguments to max; returning -Inf [~] Rscript -e max(numeric(0)) [1] -Inf Warning message: In max(numeric(0)) : no non-missing arguments to max; returning -Inf HTH, --sundar On Wed, May 20, 2009 at 11:23 AM, Kirsten Miles sirole@gmail.com wrote: I have a researcher who is consistently get the warning message: In max(i) : no non-missing arguments to max; returning -Inf Best as I can tell the code is working properly and the output is as expected. I would like some help in understanding why he is getting this error message and what its implications are. I have his code. Sincerely, Kirsten Miles Support Specialist Research Computing Lab Charles L. Brown Science and Engineering Library kd...@virginia.edu [[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] what is wrong with this code?
You're missing a ) off end of the first line. You should consider using an editor (e.g. ESS/Emacs) that does parentheses matching. I found this in less than 5 sec (less time than I'm taking to write you a note) by cut and pasting in Emacs. --sundar On Tue, May 19, 2009 at 12:52 PM, deanj2k dl...@le.ac.uk wrote: dlogl - -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta))) d2logl - (n/theta^2) - sum((-2y/theta^3)*(1-exp(y/theta))/(1+exp(y/theta))) - sum(((2*y/theta^4)*exp(y/theta))/((1+exp(y/theta))^2)) returns the error message: Error: unexpected symbol in: dlogl - -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta))) d2logl do you know what i have done wrong -- View this message in context: http://www.nabble.com/what-is-wrong-with-this-code--tp23623227p23623227.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] convert large integers to hex
Thanks for both answers. In the end I decided to use Gabor's bc package. Thanks, --sundar On Thu, May 7, 2009 at 5:10 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: There is an interface between R and bc -- not on CRAN but available from its home page here: http://r-bc.googlecode.com source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;) bc(obase = 16; 123456789123456789, retclass = character) [1] 1B69B4BACD05F15 On Wed, May 6, 2009 at 9:59 PM, jim holtman jholt...@gmail.com wrote: You can use the 'bc' command (use Cygwin if on Windows); /cygdrive/c: bc bc 1.06 Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc. This is free software with ABSOLUTELY NO WARRANTY. For details type `warranty'. x=6595137340052185552 obase=16 x 5B86A277DEB9A1D0 You can call this from R. On Wed, May 6, 2009 at 3:26 PM, Sundar Dorai-Raj sdorai...@gmail.comwrote: Hi, I'm wondering if someone has solved the problem of converting very large integers to hex. I know about format.hexmode and as.hexmode, but these rely on integers. The numbers I'm working with are overflowing and losing precision. Here's an example: x - 6595137340052185552 # stored as character as.integer(x) # warning about inaccurate conversion format.hexmode(as.numeric(x)) # warnings about loss of precision as.hexmode(x) # more warnings and does not do what I expected I'm planning on writing a function that will do this, but would like to know if anybody already has a solution. Basically, I would like the functionality of format.hexmode on arbitrarily large integers. Thanks, --sundar __ 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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] convert large integers to hex
Hi, I'm wondering if someone has solved the problem of converting very large integers to hex. I know about format.hexmode and as.hexmode, but these rely on integers. The numbers I'm working with are overflowing and losing precision. Here's an example: x - 6595137340052185552 # stored as character as.integer(x) # warning about inaccurate conversion format.hexmode(as.numeric(x)) # warnings about loss of precision as.hexmode(x) # more warnings and does not do what I expected I'm planning on writing a function that will do this, but would like to know if anybody already has a solution. Basically, I would like the functionality of format.hexmode on arbitrarily large integers. Thanks, --sundar __ 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] setting trellis auto.key color values
Set the colors in graph.sets and not auto.key. graph.sets - list(axis.text = list(cex = 0.65), par.ylab.text = list(cex = 1.25), par.xlab.text = list(cex = 1.25), superpose.polygon = list(col = 3:5)) Then remove the col = 3:5 from auto.key and barchart. Also, you can simplify your code by removing gator_IR$ and including data = gator_IR. I.e. barchart(MEAN ~ Hydro | as.factor(IR_ID), data = gator_IR, layout = c(4, 1), groups = Rain, ylim = c(0, 1), ...) HTH, --sundar On Tue, May 5, 2009 at 8:32 AM, steve_fried...@nps.gov wrote: I'm working with Lattice graphics and I would like very much to color code the auto.key fill color with the same corresponding colors that I use in the panels. I've looked on the web for clues, and on the CRAN-R help sites searching on trellis auto.key color and variations, unfortunately the responses there are not very specific. Would someone please explain I have the Book, if you can point me to the pages that explain this I'd appreciate that too graph.sets - list(axis.text = list(cex = 0.65), par.ylab.text = list(cex = 1.25), par.xlab.text = list(cex = 1.25)) barchart(gator_IR$MEAN ~ gator_IR$Hydro | as.factor(gator_IR$IR_ID),layout=c(4,1),col = c(3:5), groups=gator_IR$Rain, ylim=c(0,1), par.settings = graph.sets, main = Alligator Nesting Performance, ylab= Mean HSI, auto.key = list(top, columns=3, col=c(3:5)) this script creates the graph nicely, with the qualification that the color for the key titles are the correct, but the filling colors are default pastels. Where do I change them ? Windows XP with R 2.8.1 Thank you. Steve Steve Friedman Ph. D. Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ 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] evaluate a expression
Hi, Ning, Try: eval(parse(text = expr)) HTH, --sundar On Sat, May 2, 2009 at 5:39 AM, Ning Ma pnin...@gmail.com wrote: Hi, I am new to R. Can anyone tell me how to evaluate an expression stored in a string? such as: expr - 3*5 I want to get the result 15. Thanks in advance. __ 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] Flipping axes of qqnorm
Try (re)reading ?qqnorm. Use datax = TRUE. --sundar On Sun, Apr 26, 2009 at 4:37 PM, Chris_d dewhurstch...@gmail.com wrote: Hi all, I have just started using R to produce qqnorm plots. I am trying to switch the x and y axes so that the theoretical values are plotted on the y axis and my data on the x axis. Can anyone help me with this? Thanks -- View this message in context: http://www.nabble.com/Flipping-axes-of-qqnorm-tp23248007p23248007.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] Cannot clean infinite values
Use ?is.infinite inf - is.infinite(data) data[inf] - 0.3 * sign(data[inf]) On Sun, Apr 26, 2009 at 5:44 PM, Nigel Birney na...@cam.ac.uk wrote: Hello all, I have to import numeric data from file but found it contains Infinite values which need to be eliminated. I tried to replace them in this way: data[which(data==-Inf)] - -0.3 data[which(data==+Inf)] - 0.3 But, somehow, the Infinite values stayed there. Any suggestions? regards, N. -- View this message in context: http://www.nabble.com/Cannot-clean-infinite-values-tp23248409p23248409.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] Setting lattice par parameters
Because you're not calling trellis.par.set correctly. It should be: trellis.par.set(par.ylab.text = list(cex = 0.65), par.xlab.text = list(cex = 0.65)) However, I usually do things like this: my.theme - list(par.ylab.text = list(cex = 0.65), par.xlab.text = list(cex = 0.65)) barchart(..., par.settings = my.theme) so the settings are only changed for the current plot and not globally. HTH, --sundar On Thu, Apr 23, 2009 at 6:25 AM, steve_fried...@nps.gov wrote: Hello I'm plotting a large suite of barcharts and need to modify the size of the text for both the yaxis and xaxis labels. I've tried using the following: trellis.par.set(list(par.ylab.text = list(cex = 0.65)), trellis.par.set(list = par.xlab.text = list(cex = 0.65 On inspection, however after I invoke this line, trellis.par.get(par.ylab.text) trellis.par.get(par.xlab.text) It looks like the parameters have not been modified. Do I need to do something different than this approach ? Thank you very much in advance. Steve Steve Friedman Ph. D. Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ 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] question on using lattice panel plots
Try: z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30), each = 50),2)) z - as.data.frame(z) z - cbind(z, runif(300)) names(z) - c(Method, sigma, Error) z$sigma - factor(z$sigma, c(5, 10, 30)) library(lattice) sigma - as.numeric(levels(z$sigma)) sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s))) sigmaExpr - as.expression(sigmaExprList) bwplot(Error~Method | sigma, data = z, horiz = F, xlab = Method, strip = strip.custom(var.name = sigmaExpr, strip.levels = FALSE, strip.names = TRUE), layout = c(3,1)) HTH, --sundar On Thu, Apr 16, 2009 at 12:43 PM, Ranjan Maitra mai...@iastate.edu wrote: Hi, I think this question is best explained using the following self-contained toy example: ## cut code here and paste to R window z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30), each = 50),2)) z - as.data.frame(z) z - cbind(z, runif(300)) names(z) - c(Method, sigma, Error) library(lattice) bwplot(Error~Method | sigma, data = z, horiz = F, xlab = Method, layout = c(3,1)) ## end code Now the question: I would like the panels to be in the order of sigma, i. e. 5, 10, 30 and not 10, 30 and 5 as is currently the case. Is this possible? Not to seek too much indulgence, but to ask anyway, I wonder if it is possible to have a Greek sigma = 5, a Greek sigma = 10 and a Greek sigma = 30. (Sort of what we would get using expression(sigma == 5), expression(sigma == 10), expression(sigma == 10) on base R figures). Please let me know if my question is not clear. Many thanks for any suggestions and help and best wishes! Ranjan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question on using lattice panel plots
Sorry, that should be:
sigma - as.numeric(levels(z$sigma))
sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr - as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
horiz = F, xlab = Method,
strip = function(which.given, which.panel, var.name,
strip.levels = FALSE,
strip.names = TRUE, ...) {
strip.default(which.given, which.panel,
var.name = sigmaExpr[which.panel],
strip.levels = FALSE,
strip.names = TRUE, ...)
},
layout = c(3,1))
Not sure how to do this with strip.custom.
--sundar
On Thu, Apr 16, 2009 at 1:20 PM, Sundar Dorai-Raj sdorai...@gmail.com wrote:
Try:
z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30),
each = 50),2))
z - as.data.frame(z)
z - cbind(z, runif(300))
names(z) - c(Method, sigma, Error)
z$sigma - factor(z$sigma, c(5, 10, 30))
library(lattice)
sigma - as.numeric(levels(z$sigma))
sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr - as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
horiz = F, xlab = Method,
strip = strip.custom(var.name = sigmaExpr,
strip.levels = FALSE, strip.names = TRUE),
layout = c(3,1))
HTH,
--sundar
On Thu, Apr 16, 2009 at 12:43 PM, Ranjan Maitra mai...@iastate.edu wrote:
Hi,
I think this question is best explained using the following
self-contained toy example:
## cut code here and paste to R window
z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30),
each = 50),2))
z - as.data.frame(z)
z - cbind(z, runif(300))
names(z) - c(Method, sigma, Error)
library(lattice)
bwplot(Error~Method | sigma, data = z, horiz = F, xlab = Method,
layout = c(3,1))
## end code
Now the question:
I would like the panels to be in the order of sigma, i. e. 5, 10, 30
and not 10, 30 and 5 as is currently the case. Is this possible?
Not to seek too much indulgence, but to ask anyway, I wonder if it is
possible to have a Greek sigma = 5, a Greek sigma = 10 and a Greek
sigma = 30. (Sort of what we would get using expression(sigma == 5),
expression(sigma == 10), expression(sigma == 10) on base R figures).
Please let me know if my question is not clear.
Many thanks for any suggestions and help and best wishes!
Ranjan
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Re: [R] Changing the y-axis units of a lattice histogram
Try: library(lattice) histogram( ~ height | voice.part, data = singer, type = c, scales = list(y = list(at = seq(0, 20, 5), labels = seq(0, 200, 50 HTH, --sundar On Fri, Apr 3, 2009 at 2:01 PM, Judith Flores jur...@yahoo.com wrote: Hello, I need to multiply the number of counts in the y-axis of a lattice histogram by a constant factor, such that the plot would represent a different type of variable plotted in the y-axis, not counts. Can this be done? Thank you, Judith __ 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] labeling panels in lattice plots
Try converting year to a factor xyplot(min + max + ave ~ month | factor(year), data = rain.stats, ...) Also, notice the inclusion of the data argument. HTH, --sundar On Tue, Mar 31, 2009 at 6:28 AM, steve_fried...@nps.gov wrote: I am using windows XP with R 2.8.1 I am generating a lattice plot of annual rain patterns using the following function: xyplot(rain.stats$min+ rain.stats$max + rain.stats$ave ~ rain.stats$month |rain.stats$year, lty = 1, data = rain.stats, type = c(l,l, l), col = c(red, blue, green), distribute.type = TRUE, main = Annual Monthly Minimum, Average and Maximum PPT) Each panel is labels with rain.stats$year instead of the actual value referenced by this variable. the data.frame I'm using has the following form (year values range from 1965 to 2000 and month from 1 to 12). dim(rain.stats) 432 6 rain.stats[1:10,] year month X2 min max ave 1 1965 1 1 0. 1.9196 0.3650112 2 1966 1 1 2.1483 4.9615 3.5247034 3 1967 1 1 0.4038 3.9145 1.7133045 4 1968 1 1 0.2033 3.2119 1.1844769 5 1969 1 1 1.2533 5.6226 2.9505545 6 1970 1 1 0.9142 3.8861 2.6248453 7 1971 1 1 0.1191 1.6109 0.6570289 8 1972 1 1 0.2309 2.9380 0.9259674 9 1973 1 1 0.9471 3.6342 1.9019848 10 1974 1 1 0.1739 9.0225 1.0672980 The plot illustrates exactly what I'm after with the exception of the panel labels. I'm following an example in the Lattice Book by Deepayan Sarkar (which I find very informative - thanks), but I'm not getting the results as per the example. Can anyone offer a solution? Thanks in advance. Steve Steve Friedman Ph. D. Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can not get a prediction interval from Predict
?predict.glm has no interval argument. Perhaps you're thinking of ?predict.lm, which is different. To get intervals in glm, I've used: example(predict.glm) pr - predict(budworm.lg, se.fit = TRUE) family - family(budworm.lg) lower - family$linkinv(pr$fit - qnorm(0.95) * pr$se.fit) upper - family$linkinv(pr$fit + qnorm(0.95) * pr$se.fit) Note that these are confidence limits and not prediction limits. The latter would require more thought. You could also try RSiteSearch(glm interval). HTH, --sundar On Tue, Mar 31, 2009 at 9:58 AM, Taylor Davis taylor.da...@klasresearch.com wrote: I am trying to get a prediction interval from a glm regression. With newdat being my set of values to be fitted, and glmreg the name of my regression, I am using the following code. predict(glmreg, newdat, se.fit = TRUE, interval = confidence, level = 0.90) The problem is that I am only getting the standard error and the fitted value, not a prediction interval. Any help would be great! Thanks so much. ___ TAYLOR DAVIS | KLAS taylor.da...@klasresearch.com | 801.734.6279 __ 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] use R Group SFBA April meeting reminder; video of Feb k
Could be that you have some sort of ad filter in your browser that's blocking the video? It appears just fine for me in Firefox 3. On Mon, Mar 30, 2009 at 3:55 PM, Ted Harding ted.hard...@manchester.ac.uk wrote: On 30-Mar-09 22:13:04, Jim Porzak wrote: Next week Wednesday evening, April 8th, Mike Driscoll will be talking about Building Web Dashboards using R see: http://www.meetup.com/R-Users/calendar/9718968/ for details to RSVP. Also of interest, our member Ron Fredericks has just posted a well edited video of the February kickoff panel discussion at Predictive Analytics World The R and Science of Predictive Analytics: Four Case Studies in R with * Bo Cowgill, Google * Itamar Rosenn, Facebook * David Smith, Revolution Computing * Jim Porzak, The Generations Network and chaired by Michael Driscoll, Dataspora LLC see: http://www.lecturemaker.com/2009/02/r-kickoff-video/ Best, Jim Porzak It could be very interesting to watch that video! However, I have had a close look at the web page you cite: http://www.lecturemaker.com/2009/02/r-kickoff-video/ and cannot find a link to a video. Lots of links to non-video things, but none that I could see to a video. There is a link on that page at: How Google and Facebook are using R by Michael E. Driscoll | February 19, 2009 http://dataspora.com/blog/predictive-analytics-using-r/ Following that link leads to a page, on which the first link, in: (March 26th Update: Video now available) Last night, I moderated our Bay Area R Users Group kick-off event with a panel discussion entitled The R and Science of Predictive Analytics, co-located with the Predictive Analytics World conference here in SF. leads you back to where you came from, and likewise the link at the bottom of the page: A video of the event is now available courtesy of Ron Fredericks and LectureMaker. Could you help by describing where on that web page it can be found? With thanks, Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 30-Mar-09 Time: 23:55:07 -- XFMail -- __ 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] Requesting help with lattice again
For the first question, add a groups argument. E.g. barchart(HSI ~ Scenario | Region, Wbirdsm, groups = HydroState) Also note that using Wbirdsm$HSI makes your call less readable, so I added the data argument. For your second question, setting the key does not set the color theme. You want to set the colors using par.setting. E.g. dotplot(HSI ~ Scenario | Region, Wbirdsm, groups = HydroState, par.settings = list(superpose.symbol = list(col = c(red, green, blue))), auto.key = list(space = right)) Then use auto.key instead of key. HTH, --sundar On Wed, Mar 25, 2009 at 7:02 AM, steve_fried...@nps.gov wrote: Hello, this is a request for assistance that I submitted earlier, this time with the dataset. My mistake for taking up bandwidth. I've also rephrased the question to address an additional concern. I'm working on a windows XP machine with R 2.8.1 1). I'd like a barchart (or other lattice type display) HSI ~ of the three factors (Region, Scenario and HydroState). However barchart complains library(reshape) library(lattice) barchart(Wbirdsm$HSI ~ WBirdsm$Scenario | WBridsm$Region) # This works, but only marginally. I'd like specify the interaction between the variables ( Scenario and HydroState) What is the best way to combine these factors so they can be treated as a single variable and rewrite the barchart call? barchart(Wbirdsm$HSI ~ WBirdsm$interaction between Scenario and HydroState) ~ Wbirdsm$Region) The second question refers back to my original posting. 2) # using dotplot instead of barchart I use the following code. key.variable - list(space = right, text = list(levels(wbirdm$variable)), points = list(pch = 1:3, col=c(red,green, blue))) dotplot(wbirdm$value ~ wbirdm$variable | wbirdm$Region, col=c(1:9), pch= rep(c(1:3), key = key.variable, groups=wbirdm$variable, ylab= Mean HSI)) # However the key legend doesn't appear in the plot with this sequence and the labels are too along the panels. Is there a way to address this? thank you very much for the assistance. Steve This is part of the larger data base, after I passed it thru melt. Region Species Scenario HydroState HSI 1 Eastern Panhandle WBLong NSM Ave 0.165945170 2 Eastern Panhandle WBLong NSM Dry 0.056244263 3 Eastern Panhandle WBLong NSM Wet 0.290692607 4 Eastern Panhandle WBLong ECB Ave 0.165945170 5 Eastern Panhandle WBLong ECB Dry 0.056244263 6 Eastern Panhandle WBLong ECB Wet 0.290692607 7 Eastern Panhandle WBLong CERP Ave 0.165945170 8 Eastern Panhandle WBLong CERP Dry 0.056244263 9 Eastern Panhandle WBLong CERP Wet 0.290692607 10 Long Pine Key / South Taylor Slough WBLong NSM Ave 0.151159734 11 Long Pine Key / South Taylor Slough WBLong NSM Dry 0.067348863 12 Long Pine Key / South Taylor Slough WBLong NSM Wet 0.20738 13 Long Pine Key / South Taylor Slough WBLong ECB Ave 0.151159734 14 Long Pine Key / South Taylor Slough WBLong ECB Dry 0.067348863 15 Long Pine Key / South Taylor Slough WBLong ECB Wet 0.20738 16 Long Pine Key / South Taylor Slough WBLong CERP Ave 0.151159734 17 Long Pine Key / South Taylor Slough WBLong CERP Dry 0.067348863 18 Long Pine Key / South Taylor Slough WBLong CERP Wet 0.20738 19 Northern Taylor Slough WBLong NSM Ave 0.115503291 20 Northern Taylor Slough WBLong NSM Dry 0.005617136 21 Northern Taylor Slough WBLong NSM Wet 0.252428530 22 Northern Taylor Slough WBLong ECB Ave 0.115503291 23 Northern Taylor Slough WBLong ECB Dry 0.005617136 24 Northern Taylor Slough WBLong ECB Wet 0.252428530 25 Northern Taylor Slough WBLong CERP Ave 0.115503291 26 Northern Taylor Slough WBLong CERP Dry 0.005617136 27 Northern Taylor Slough WBLong CERP Wet 0.252428530 28 East Slough WBLong NSM Ave 0.313215457 29 East Slough WBLong NSM Dry 0.046917053 30 East Slough WBLong NSM Wet 0.447002596 31 East Slough WBLong ECB Ave 0.313215457 32 East Slough WBLong ECB Dry 0.046917053 33 East Slough WBLong ECB Wet 0.447002596 34 East Slough WBLong CERP
Re: [R] XYplot simple question
Convert Plot to factor:
xyplot(AbvBioAnnProd ~ Year | Plot, type = c(b, r), pch = 16)
Also note that using the type argument with multiple values prevents
the necessity of a custom panel function.
HTH,
--sundar
On Mon, Mar 16, 2009 at 5:54 AM, AllenL allen.laroc...@gmail.com wrote:
Hello R friends,
Simple question today: I am desiring to do an xyplot with the below code,
which graphs time series across different experimental Plots-
xyplot(AbvBioAnnProd~Year|Plot) ### Plots each monoculture biomass vs
time
xyplot(AbvBioAnnProd~Year|Plot,panel=function(x,y){
panel.xyplot(x,y,type=b,pch=16)
panel.abline(lm(y~x))
})
What I want to add are unique labels for each panel, where instead of all
labeled plot with the slide-bar visual (although that is okay) I want
the Plot number to appear (i.e. the value of Plot for that panel).
This should be easy, right?
-Al
--
View this message in context:
http://www.nabble.com/XYplot-simple-question-tp22537350p22537350.html
Sent from the R help mailing list archive at Nabble.com.
__
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] XYplot simple question
Sorry, I should have
xyplot(AbvBioAnnProd ~ Year | factor(Plot), type = c(b, r), pch = 16)
On Mon, Mar 16, 2009 at 6:17 AM, Sundar Dorai-Raj sdorai...@gmail.com wrote:
Convert Plot to factor:
xyplot(AbvBioAnnProd ~ Year | Plot, type = c(b, r), pch = 16)
Also note that using the type argument with multiple values prevents
the necessity of a custom panel function.
HTH,
--sundar
On Mon, Mar 16, 2009 at 5:54 AM, AllenL allen.laroc...@gmail.com wrote:
Hello R friends,
Simple question today: I am desiring to do an xyplot with the below code,
which graphs time series across different experimental Plots-
xyplot(AbvBioAnnProd~Year|Plot) ### Plots each monoculture biomass vs
time
xyplot(AbvBioAnnProd~Year|Plot,panel=function(x,y){
panel.xyplot(x,y,type=b,pch=16)
panel.abline(lm(y~x))
})
What I want to add are unique labels for each panel, where instead of all
labeled plot with the slide-bar visual (although that is okay) I want
the Plot number to appear (i.e. the value of Plot for that panel).
This should be easy, right?
-Al
--
View this message in context:
http://www.nabble.com/XYplot-simple-question-tp22537350p22537350.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.
__
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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] Get user system name
Assuming USER is defined on your system then Sys.getenv(USER) ought to work. --sundar On Mon, Mar 16, 2009 at 3:04 PM, Etienne Bellemare Racine etienn...@gmail.com wrote: I would like to get the name of the user form the system. Is it possible ? Something like system.user() returning something like [1] etber12 Thanks, Etienne [[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] Get user system name
If that's the case, it's best to post the output from version to this thread. Type version from the R command prompt and paste the results here. On Mon, Mar 16, 2009 at 3:14 PM, Etienne Bellemare Racine etienn...@gmail.com wrote: It seems like it is not defined. I get an empty string. Is there a workaround or another solution ? -- Etienne Sundar Dorai-Raj a écrit : Assuming USER is defined on your system then Sys.getenv(USER) ought to work. --sundar On Mon, Mar 16, 2009 at 3:04 PM, Etienne Bellemare Racine etienn...@gmail.com wrote: I would like to get the name of the user form the system. Is it possible ? Something like system.user() returning something like [1] etber12 Thanks, Etienne [[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] Matrix Construction; Subdiagonal
Does this help? A - matrix(0, 6, 6) vec - 1:5 A[row(A) == col(A) + 1] - vec --sundar On Wed, Mar 11, 2009 at 4:42 PM, Stu Field s...@colostate.edu wrote: I'm trying to enter a vector into the subdiagonal of a matrix but cannot find a command in R which corresponds to the MatLab version of diag(vec, k), where vec = the vector of interest, and k = the diagonal (k=0 for the diagonal; k=-1 for the subdiagonal; k=1 for superdiagonal, etc.) Is there an equivalent command in R? I'm looking for something like this: vec = seq(1, 5, 1) # vector of interest A = xyz(vec,-1) # creates a 6x6 matrix with vec on the subdiagonal where xyz is some function similar to diag, but with differing arguments. I can't believe there is not a simple way to do this... Thanks for your help, ~~ Stu Field, PhD Postdoctoral Fellow Department of Biology Colorado State University 1878 Campus Delivery Fort Collins, CO 80523-1878 Office: E208 Anatomy/Zoology Phone: (970) 491-5744 ~~ [[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] Matrix Construction; Subdiagonal
You can always write your own function:
myDiag - function(x, vec, k) {
x[row(x) == col(x) - k] - vec
x
}
myDiag(A, vec, -1)
Of course, you should probably do some input checking too.
--sundar
On Wed, Mar 11, 2009 at 4:57 PM, Stu Field s...@colostate.edu wrote:
Sure, that'll work fine, thanks.
But I guess I was looking for something more similar to MatLab, I'm really
surprised R doesn't have a preset command for this (?)
Thanks again,
Stu
On 11 • Mar • 2009, at 5:49 PM, Sundar Dorai-Raj wrote:
Does this help?
A - matrix(0, 6, 6)
vec - 1:5
A[row(A) == col(A) + 1] - vec
--sundar
On Wed, Mar 11, 2009 at 4:42 PM, Stu Field s...@colostate.edu wrote:
I'm trying to enter a vector into the subdiagonal of a matrix but
cannot find a command in R which corresponds to the MatLab version of
diag(vec, k), where vec = the vector of interest, and k = the diagonal
(k=0 for the diagonal; k=-1 for the subdiagonal; k=1 for
superdiagonal, etc.)
Is there an equivalent command in R?
I'm looking for something like this:
vec = seq(1, 5, 1) # vector of interest
A = xyz(vec,-1) # creates a 6x6 matrix with vec on the
subdiagonal
where xyz is some function similar to diag, but with differing
arguments.
I can't believe there is not a simple way to do this...
Thanks for your help,
~~
Stu Field, PhD
Postdoctoral Fellow
Department of Biology
Colorado State University
1878 Campus Delivery
Fort Collins, CO 80523-1878
Office: E208 Anatomy/Zoology
Phone: (970) 491-5744
~~
[[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.
~~
Stu Field, PhD
Postdoctoral Fellow
Department of Biology
Colorado State University
1878 Campus Delivery
Fort Collins, CO 80523-1878
Office: E208 Anatomy/Zoology
Phone: (970) 491-5744
~~
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice: Customizing point-sizes with groups
Try this: xyplot(y ~ x, temp, groups = groups, par.settings = list( superpose.symbol = list( cex = c(1, 3), pch = 19, col = c(blue, red See: str(trellis.par.get()) for other settings you might want to change. Also, you should drop the ; from all your scripts. HTH, --sundar On Mon, Mar 9, 2009 at 6:49 PM, Paul Boutros paul.bout...@utoronto.ca wrote: Hello, I am creating a scatter-plot in lattice, and I would like to customize the size of each point so that some points are larger and others smaller. Here's a toy example: library(lattice); temp - data.frame( x = 1:10, y = 1:10, cex = rep( c(1,3), 5), groups = c( rep(A, 5), rep(B, 5) ) ); xyplot(y ~ x, temp, cex = temp$cex, pch = 19); This works just fine if I create a straight xy-plot, without groups. However when I introduce groupings the cex argument specifies the point-size for the entire group. For example: xyplot(y ~ x, temp, cex = temp$cex, pch = 19, group = groups); Is it possible to combine per-spot sizing with groups in some way? One work-around is to manually specify all graphical parameters, but I thought there might be a better way than this: temp$col - rep(blue, 10); temp$col[temp$groups == B] - red; xyplot(y ~ x, temp, cex = temp$cex, pch = 19, col = temp$col); Any suggestions/advice is much appreciated! Paul __ 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] Lattice: Customizing point-sizes with groups
Sorry, I missed your point the first time. Why not create a group for each subset then? xyplot(y ~ x, temp, groups = interaction(cex, groups), par.settings = list( superpose.symbol = list( cex = c(1, 2, 3, 4), pch = 19, col = c(blue, red, green, purple On Tue, Mar 10, 2009 at 8:11 AM, Paul C. Boutros paul.bout...@utoronto.ca wrote: Hi Sundar, Thanks for your help! Unfortunately your code seems to give the same result. Compare this: temp - data.frame( x = 1:10, y = 1:10, cex = rep( c(1,3), 5), col = c( rep(blue, 5), rep(red, 5) ), groups = c( rep(A, 5), rep(B, 5) ) ); xyplot(y ~ x, temp, groups = groups, par.settings = list( superpose.symbol = list( cex = c(1, 3), pch = 19, col = c(blue, red And this: xyplot(y ~ x, temp, cex = temp$cex, col = temp$col, pch = 19); Once I introduce groups, I lose the ability to customize individual data-points and seem only to be able to customize entire groups. Paul -Original Message- From: Sundar Dorai-Raj [mailto:sdorai...@gmail.com] Sent: Tuesday, March 10, 2009 5:49 AM To: paul.bout...@utoronto.ca Cc: r-help@r-project.org Subject: Re: [R] Lattice: Customizing point-sizes with groups Try this: xyplot(y ~ x, temp, groups = groups, par.settings = list( superpose.symbol = list( cex = c(1, 3), pch = 19, col = c(blue, red See: str(trellis.par.get()) for other settings you might want to change. Also, you should drop the ; from all your scripts. HTH, --sundar On Mon, Mar 9, 2009 at 6:49 PM, Paul Boutros paul.bout...@utoronto.ca wrote: Hello, I am creating a scatter-plot in lattice, and I would like to customize the size of each point so that some points are larger and others smaller. Here's a toy example: library(lattice); temp - data.frame( x = 1:10, y = 1:10, cex = rep( c(1,3), 5), groups = c( rep(A, 5), rep(B, 5) ) ); xyplot(y ~ x, temp, cex = temp$cex, pch = 19); This works just fine if I create a straight xy-plot, without groups. However when I introduce groupings the cex argument specifies the point-size for the entire group. For example: xyplot(y ~ x, temp, cex = temp$cex, pch = 19, group = groups); Is it possible to combine per-spot sizing with groups in some way? One work-around is to manually specify all graphical parameters, but I thought there might be a better way than this: temp$col - rep(blue, 10); temp$col[temp$groups == B] - red; xyplot(y ~ x, temp, cex = temp$cex, pch = 19, col = temp$col); Any suggestions/advice is much appreciated! Paul __ 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] 2 Simple Lattice Plot Questions
Convert Year to a factor and both problems will be solved. --sundar On Tue, Mar 10, 2009 at 3:48 PM, jimdare jamesdar...@gmail.com wrote: Hi, I have created the plot below and have a few questions about changes. 1) How do I change the Year title of each plot so it reads from the top 2006,2007,2008,2009. 2) How do I get rid of those vertical grey bars in the title bar of each plot? I apologise for my ignorance... one of those days :( James http://www.nabble.com/file/p22445242/PLOT.jpg -- View this message in context: http://www.nabble.com/2-Simple-Lattice-Plot-Questions-tp22445242p22445242.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] 2 Simple Lattice Plot Questions
I don't believe Elena's suggestion will work. However, the following will: xyplot(..., scales = list(y = list(at = seq(5, 25, 5 though you may need to extend the limits a little as well: xyplot(..., ylim = lattice:::extend.limits(c(0, 30))) and add the scales argument from the first example to place explicit tick marks rather than let xyplot do so. HTH, --sundar On Tue, Mar 10, 2009 at 7:04 PM, Elena Wilson ewil...@dbmcons.com.au wrote: Have you tried specifying the levels of y's you want to display, e.g. ylim=c(0,5,10,15,20,30)? -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of jimdare Sent: Wednesday, 11 March 2009 12:50 PM To: r-help@r-project.org Subject: Re: [R] 2 Simple Lattice Plot Questions Thanks very much. One other thing... see how the Y axis begins before 0. At first glance it appears that the 0's are actually worth something. When I use ylim=c(0,30) I get the graph I want, but the tick marks show only 5,10,15,20,25. I want them to show 0,5,10,15,20,30. Does anyone know how to change this setting? Cheers, jimdare wrote: Hi, I have created the plot below and have a few questions about changes. 1) How do I change the Year title of each plot so it reads from the top 2006,2007,2008,2009. 2) How do I get rid of those vertical grey bars in the title bar of each plot? I apologise for my ignorance... one of those days :( James http://www.nabble.com/file/p22445242/PLOT.jpg -- View this message in context: http://www.nabble.com/2-Simple-Lattice-Plot-Questions-tp22445242p22447415.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. -- Message protected by MailGuard: e-mail anti-virus, anti-spam and content filtering. http://www.mailguard.com.au/mg https://login.mailguard.com.au/report/1x1TEaABIw/4soNUVnBh04s1coMEHC4LA/7.998 __ 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] Date conversion
Hi, There are possibly several ways to do this. My approach would be: dates - strptime(as.character(DATE), %d%b%Y) year - dates$year + 1900 week - floor(dates$yday/365 * 52) HTH, --sundar On Thu, Mar 5, 2009 at 8:58 AM, Pele drdi...@yahoo.com wrote: Hi R users, I have a factor variable called date as shown below: Can anyone share the best / most efficient way to extract year and week (e.g. year = 2006, week = 52 for first record, etc..)? My data set has 1 million records. DATE 11DEC2006 11SEP2006 01APR2007 02DEC2007 Thanks in advance for any help! -- View this message in context: http://www.nabble.com/Date-conversion-tp22355788p22355788.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] lattice: remove box around a wireframe
(Sorry for the repeat. Forgot to copy R-help) Try, test = data.frame(expand.grid(c(1:10), c(1:10))) z = test[,1] + test[,2] test = cbind(test, z) names(test) = c(x, y, z) require(lattice) wireframe(z ~ x*y, data = test, par.settings = list(axis.line = list(col = transparent)), par.box = c(col = transparent) ) --sundar On Wed, Mar 4, 2009 at 8:17 AM, Thomas Roth (geb. Kaliwe) hamstersqu...@web.de wrote: #Hi, # #somebody knows how to remove the outer box around a wireframe and reduce the height # # test = data.frame(expand.grid(c(1:10), c(1:10))) z = test[,1] + test[,2] test = cbind(test, z) names(test) = c(x, y, z) require(lattice) wireframe(z ~ x*y, data = test, par.box = c(col = transparent) ) #not this one but the remaining outer box. Thanks in advance Thomas Roth __ 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] levelplot help needed
To reorder the y-labels, simply reorder the factor levels:
df - data.frame(x_label = factor(x_label),
y_label = factor(y_label, rev(y_label)),
values = as.vector(my.data))
Not sure about putting the strips at the bottom. A quick scan of
?xyplot and ?strip.default suggests that this is not possible, but I'm
sure Deepayan will correct me if I'm wrong (he often does).
--sundar
On Fri, Feb 27, 2009 at 5:51 AM, Antje niederlein-rs...@yahoo.de wrote:
Hi there,
I'm looking for someone who can give me some hints how to make a nice
levelplot. As an example, I have the following code:
# create some example data
# --
xl - 4
yl - 10
my.data - sapply(1:xl, FUN = function(x) { rnorm( yl, mean = x) })
x_label - rep(c(X Label 1, X Label 2, X Label 3, X Label 4), each =
yl)
y_label - rep(paste(Y Label , 1:yl, sep=), xl)
df - data.frame(x_label = factor(x_label),y_label = factor(y_label), values
= as.vector(my.data))
df1 - data.frame(df, group = rep(Group 1, xl*yl))
df2 - data.frame(df, group = rep(Group 2, xl*yl))
df3 - data.frame(df, group = rep(Group 3, xl*yl))
mdf - rbind(df1,df2,df3)
# plot
# --
graph - levelplot(mdf$values ~ mdf$x_label * mdf$y_label | mdf$group,
aspect = xy, layout = c(3,1),
scales = list(x = list(labels =
substr(levels(factor(mdf$x_label)),0,5), rot = 45)))
print(graph)
# --
(I need to put this strange x-labels, because in my real data the values of
the x-labels are too long and I just want to display the first 10 characters
as label)
My questions:
* I'd like to start with Y Label 1 in the upper row (that's a more general
issue, how can I have influence on the order of x,y, and groups?)
* I'd like to put the groups at the bottom
Can anybody give me some help?
Antje
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Re: [R] lattice contourplot line types
The only way I can figure out to do this is to use two calls to
panel.contourplot:
library(lattice)
x - seq(-2, 2, length = 20)
y - seq(-2, 2, length = 20)
grid - expand.grid(x=x, y=y)
grid$z - dnorm(grid$x) * dnorm(grid$y)
contourplot(z ~ x * y, grid,
panel = function(at, lty, col, ...) {
at.o - at[seq(1, length(at), 2)]
at.e - at[seq(2, length(at), 2)]
panel.contourplot(at = at.o, lty = 1, col = blue, ...)
panel.contourplot(at = at.e, lty = 4, col = red, ...)
},
at = pretty(grid$z, 10))
HTH,
--sundar
On Sun, Feb 22, 2009 at 12:45 PM, Andrew Beckerman
a.becker...@sheffield.ac.uk wrote:
Dear all -
I would like to adjust the line type of specific contours in contourplot
from the lattice package, but it seems like lty does not take a list in the
call.
Here is my call to contourplot:
contourplot(preds~size+trt|Size.Name,
data=pred.dat,layout=c(2,4),
at=c(0.025,0.5,0.975),
par.strip.text=list(cex=1.2),
scales=list(cex=0.5),
xlab=list(Size,cex=1.2),
ylab=list(Treatment,cex=1.2),
panel=function(x,y,z,...){
panel.contourplot(x,y,z,lwd=1,...)
panel.grid(h=-1,v=-1,col=grey,...)})
I would like to specify lty=c(2,1,2) corresponding to the
at=c(0.025,0.5,0.975), and have tried this in both the core part of the
call, and in panel.countourplot. However, it only recognises the first
type.
If there is no straightforward answer, I can provide the data.
Best wishes,
Andrew
R version 2.8.1 (2008-12-22)
i386-apple-darwin8.11.1
locale:
en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] MASS_7.2-45 lattice_0.17-17
loaded via a namespace (and not attached):
[1] grid_2.8.1
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Re: [R] Uninstall question
This is on the Mac FAQ: http://cran.cnr.berkeley.edu/bin/macosx/RMacOSX-FAQ.html#How-can-R-for-Mac-OS-X-be-uninstalled_003f HTH, --sundar On Tue, Feb 17, 2009 at 7:17 AM, ANJAN PURKAYASTHA anjan.purkayas...@gmail.com wrote: I need to uninstall R 2.7.1 from my Mac. What is the best way to uninstall it? Simply delete the R icon in the Applications folder? Or is it more involved? TIA, Anjan -- = anjan purkayastha, phd bioinformatics analyst whitehead institute for biomedical research nine cambridge center cambridge, ma 02142 purkayas [at] wi [dot] mit [dot] edu 703.740.6939 [[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] Adding abline in Lattice graph
Try:
coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
panel= function(...) {
panel.smooth(...)
panel.abline(h = 126, col = red)
panel.abline(v = 6.5, col = blue)
},
xlab=ABC, ylab=FBG)
Also note that you removed your with call and give coplot a data argument.
HTH,
--sundar
On Thu, Feb 12, 2009 at 3:41 PM, Veerappa Chetty chett...@gmail.com wrote:
Hi,I would like add a horizontal line at 126 (col=red) and a vertical line
at 6.5 ( col= blue) in each panel .How should I use the panel.abline
function in the following code I am using:
--
library(lattice)
with(reg.dat.5,coplot(lbxglu~lbxgh|eth,panel=panel.smooth,xlab=ABC,
ylab=FBG))
Thanks a lot.
Professor of Family Medicine
Boston University
Tel: 617-414-6221, Fax:617-414-3345
emails: chett...@gmail.com,vche...@bu.edu
[[alternative HTML version deleted]]
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Re: [R] Adding abline in Lattice graph
Sorry, that was my lack of understanding on how coplot works. Try the following:
coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
panel= function(...) {
panel.smooth(...)
abline(h = 126, col = red)
abline(v = 6.5, col = blue)
},
xlab=ABC, ylab=FBG)
On Thu, Feb 12, 2009 at 3:49 PM, Sundar Dorai-Raj sdorai...@gmail.com wrote:
Try:
coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
panel= function(...) {
panel.smooth(...)
panel.abline(h = 126, col = red)
panel.abline(v = 6.5, col = blue)
},
xlab=ABC, ylab=FBG)
Also note that you removed your with call and give coplot a data argument.
HTH,
--sundar
On Thu, Feb 12, 2009 at 3:41 PM, Veerappa Chetty chett...@gmail.com wrote:
Hi,I would like add a horizontal line at 126 (col=red) and a vertical line
at 6.5 ( col= blue) in each panel .How should I use the panel.abline
function in the following code I am using:
--
library(lattice)
with(reg.dat.5,coplot(lbxglu~lbxgh|eth,panel=panel.smooth,xlab=ABC,
ylab=FBG))
Thanks a lot.
Professor of Family Medicine
Boston University
Tel: 617-414-6221, Fax:617-414-3345
emails: chett...@gmail.com,vche...@bu.edu
[[alternative HTML version deleted]]
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Re: [R] changing settings on a barchart (lattice)
Add the adj argument to panel.text to left (adj = 0) or right(adj =
1) justify the text. Add the font argument to change the font. See
?text.
--sundar
On Wed, Feb 11, 2009 at 1:25 PM, Dimitri Liakhovitski ld7...@gmail.com wrote:
Thanks a lot, Sundar. I experimented somewhat and here is the code
that works well - it allows me to modify most of the stuff I want to
modify:
p-as.vector(c(0.1, 0.2, 0.3, 0.4))
names(p)-c(A,BB,,)
barchart(~sort(p), main=list(Chart Title,cex=1),xlab=list(X axis
title,cex=1),xlim=c(0,0.42),
layout = c(1,1),
stack = TRUE,
auto.key = list(points = FALSE, rectangles = TRUE, space =
top),
panel = function(y,x,...){
panel.grid(h = 0, v = -1, col = gray60, lty
=dotted)
panel.barchart(x,y,col=brown)
panel.text(x,y,label = round(x,2),cex=1)
}
)
One last question: How can I modify the way the value labels (those
that are at the end of the bars) appear? Can I make them bold? Make
them appear a bit to the right or to the left of where they currently
are?
Thanks a lot!
Dimitri
On Wed, Feb 11, 2009 at 12:46 PM, Sundar Dorai-Raj sdorai...@gmail.com
wrote:
Pass a list to xlab and main for the font sizes:
barchart(..., xlab = list(x-axis, cex = 2), main = list(title, cex = 2))
For value labels and a grid you'll need a custom panel function:
barchart(..., panel = function(x, y, ...) {
panel.barchart(x, y, ...)
panel.text(x, y, format(y), cex = 1.2)
panel.grid(h = -1, v = -1)
})
This is untested, but I think it should get you started.
--sundar
On Wed, Feb 11, 2009 at 9:10 AM, Dimitri Liakhovitski ld7...@gmail.com
wrote:
Hello!
I apologize - I never used lattice before, so my question is probably
very basic - but I just can't find the answer in the archive nor in
the documentation:
I have a named numeric vector p of 6 numbers (of the type 6 numbers
with people's names to whom those numbers belong). I want a simple bar
chart.
I am doing:
library(lattice)
trellis.par.set(fontsize=list(text=12)) # Changes only axes font
barchart(~sort(p),main=Text for main,xlab=Text for X axis,
col=PrimaryColors[3])
It works just fine.
Question: Where how can I change such things as font size for X axis
label (below the numbers), font size for Title, value labels (to label
bars with actual numbers), add grids, etc.?
Thanks a lot!
--
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com
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--
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com
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Re: [R] Efficent way to create an nxn upper triangular matrix of one's
Try
x - diag(n)
x[upper.tri(x)] - 1
On Wed, Feb 11, 2009 at 1:22 PM, Dale Steele dale.w.ste...@gmail.com wrote:
The code below create an nxn upper triangular matrix of one's. I'm
stuck on finding a more efficient vectorized way - Thanks. --Dale
n - 9
data - matrix(data=NA, nrow=n, ncol=n)
data
for (i in 1:n) {
data[,i] - c(rep(1,i), rep(0,n-i))
}
data
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Re: [R] sapply
I'm not sure what you really want, so perhaps a simple example would help (i.e. what a sample of the input looks like and what the output you need looks like). My guess would be sapply(df, diff) but again, I'm not sure. --sundar On Sun, Feb 8, 2009 at 4:24 PM, glenn g1enn.robe...@btinternet.com wrote: Newbie question sorry (have tried the help pages I promise) I have a dataframe (date,stockprice) say and looking how I might get the return of: dataframe (difference in days, change in stock price) using sapply - I require a very simple function and don't really want to go down the zoo and quant mod route Regards glenn [[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] subsets problem
you can try lapply(lapply(uniques, function(x) subset(df, date == x)), myfun) or possibly more accurate (subset may be finicky due to scoping): lapply(lapply(uniques, function(x) df[df$date == x, ]), myfun) or use ?split lapply(split(df, df$date), myfun) HTH, --sundar On Sun, Feb 8, 2009 at 5:00 PM, glenn g1enn.robe...@btinternet.com wrote: Help with this much appreciated I have a large dataframe that I would like to subset where the constraint Test1 - subset(df, date == uniques[[1]]), where uniques is a list of dates that must be matched to create Test1. I would like to perform an operation on Test1 that results in a single column of data. So far so good. How do loop through all values in the uniques list (say there is 50), perform an operationon Test1Test50, and then bolt all the lists together in a single list please ? Regards Glenn [[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] Foreign function call
You're missing that R_TSConv is an R object. You can use stats:::R_TSConv to see the value. Not sure how this helps you though. On Wed, Feb 4, 2009 at 10:08 AM, rkevinbur...@charter.net wrote: Let me get more specific. I think it this can be answered then I can translate the information to other calls. In the arima 'R' code there is a reference to .Call(R_TSconv, a, b) If from the console I type: .Call(R_TSConv, c(1,-1), c(1,-1)) I get: Error: object R_TSConv not found If I do getNativeSymbolInfo(R_TSConv) I get: Error in FUN(R_TSConv[[1L]], ...) : no such symbol R_TSConv What am I missing? Thank you. Kevin __ 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] color and fontfamily in lattice
Try this:
dados - data.frame(varsep = factor(rep(1:2,10)),
i = runif(20))
library(lattice)
font.settings - list(
font = 2,
cex = 2,
fontfamily = serif)
my.theme - list(
box.umbrella = list(col = red),
box.rectangle = list(col = purple),
box.dot = list(col = blue, pch = 15),
par.xlab.text = font.settings,
par.ylab.text = font.settings,
axis.text = font.settings)
bwplot(varsep ~ i, dados,
xlab = names(dados)[1],
ylab = names(dados)[2],
panel = function(...) {
panel.grid(v = -1, h = 0)
panel.bwplot(...)
},
par.settings = my.theme)
Type trellis.par.get() at the R command line to see other parameters
you can change. To change the settings on all plots, you can remove
the par.settings from the call to bwplot and simply use:
trellis.par.set(theme = my.theme)
HTH,
--sundar
2009/2/3 Leandro Marino lean...@cesgranrio.org.br:
Hi,
I am having some problems using bwplot(lattice) in my data. I want change
some parameters:
1) Fontfamily to serif
2) The size of the font
3) Put it in a bold face
4) Change de color of the lines
How can I do that?! Now, I am using this to plot my boxplot.
dados - data.frame(varsep=as.factor(rep(1:2,10)),i=runif(20))
bwplot(dados[,'varsep']~dados[,'i'],xlab=names(dados)[2],ylab=names(dados)[1],panel
=function(...){panel.grid(v = -1, h =
0);panel.bwplot(...)},font=2,fontfamily='serif')
Thanks for any help on advance and sorry about my English.
Atenciosamente,
Leandro Lins Marino
Centro de Avaliação
Fundação CESGRANRIO
Rua Santa Alexandrina, 1011 - 2º andar
Rio de Janeiro, RJ - CEP: 20261-903
R (21) 2103-9600 R.:236
0 (21) 8777-7907
( lean...@cesgranrio.org.br
Aquele que suporta o peso da sociedade
é precisamente aquele que obtém
as menores vantagens. (SMITH, Adam)
Antes de imprimir pense em sua responsabilidade e compromisso com o MEIO
AMBIENTE
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Re: [R] xyplot with lowess curves
You'll need a custom panel function. It would also help if you
provided a reproducible example:
xyplot (
SnowLineElevation ~ Year | Model,
data = data,
panel = function(x, y, col, ...) {
col - ifelse(panel.number() == 1, red, green)
panel.xyplot(x, y, col = blue, ...)
panel.loess(x, y, col = col)
},
ylim = c(0,1800),
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
Alternatively, you can use the group argument in conjunction with the panels:
xyplot(SnowLineElevation ~ Year | Model, data, groups = Model, type =
c(p, smooth))
if you want the points and the lines to be the same color.
--sundar
On Mon, Feb 2, 2009 at 10:20 AM, Hutchinson,David [PYR]
david.hutchin...@ec.gc.ca wrote:
I am trying to change the attributes of the lowess lines fit to an
xyplot command, but have been unsuccessful in my search of the online
help. Right now, both the points and lowess line come out in the same
color (blue). I am unsure how I can change the properties of the lowess
line separately.
xyplot (
SnowLineElevation ~ Year | Model,
data = data,
ylim = c(0,1800),
type = c('p','smooth'),
col = 'blue',
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
Any help would be much appreciated,
Dave
[[alternative HTML version deleted]]
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