[R] pie chart
Hi All, I have txt file like : $ cat data.txt US 10 UK 12 Ind 4 Germany 14 France 8 rawdata - read.table(file='data.txt',sep='\t' , header=FALSE) rawdata V1 V2 1 US 10 2 UK 12 3 Ind 4 4 Germany 14 5 France 8 I want to draw pie chart for the above data. How to split rawdata into : con - c(US,UK,Ind,Germany,France); total - (10,12,4,14,8) Any help will be appreciate. Thanks Rg Mohan L __ 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 to add a new column filled with value 1
Dear All, I have a data frame with 5 column and 201 row data. I want to add one more column between column 1 and 2 with value of 1. So the new column has to be the second column filled with 1. Any help will be appreciated. Thanks for your time. Thanks Rg Mohan L __ 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 to merge two data frame if name matches
Dear All, I have two data like this : $cat main.csv name,id,memory,storage mohan,1,100.20,1.10 ram,1,200,100 kumar,1,400,50 xxx,1,100,40 aaa,1,800,45 mount,1,200,80 main - read.csv(file='main.csv',sep=',' , header=TRUE) main nameidmemory storage 1 mohan 1100.2 10 2 ram 1 200.0 100 3 kumar1 400.0 50 4 xxx 1 100.0 40 5 aaa 1 800.0 45 6 mount1 200.0 80 $cat other.csv name,ip,bsent,breceived mohan,1,12.00,0.01 xxx,1,00.00,1.110 kumat,1,1.00,1.00 mmm,1,10.00,8.08 own,1,20.13,12.08 per,1,1.89,0.89 other - read.csv(file='other.csv',sep=',' , header=TRUE) other nameip bsent breceived 1 mohan 1 12.00 0.01 2 xxx 1 0.00 1.11 3 kumat1 1.00 1.00 4 mmm 1 10.00 8.08 5 own 1 20.13 12.08 6 per 1 1.89 0.89 I want to merge ip,bsent,breceived column to main , If the name in the the main data frame is there in the other data frame . some thinng like this: nameid memory storage ip bsent breceived mohan1100 20 1 12.00 0.01 ram1200 1000 00.000.00 kumar1400 50 1 1.00 1.00 xxx1100 40 1 00.001.110 aaa 1800 45 0 00.0000.00 mount1200 80 0 00.0000.00 If in case the name in the main data frame does not there in the other data frame, simple I want to add zero to ip,bsent,breceived value. I hope this can be done with R. Any help will appricated. Thanks for you time. Thanks Rg Mohan L __ 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] using variable in rmysql query
Dear All, I am using this query it returns id : id - dbGetQuery(con1,SELECT id FROM tenants WHERE name LIKE '%consim%') But In my case the string consim is there in another variable(it is coming from configuration file); str - consim I am trying to replace the string some this like, but it not working: id - dbGetQuery(con1,SELECT id FROM tenants WHERE name LIKE '%str%') I need help to replace the value of str . Any help will really appreciated . Thanks for your time. Mohan L __ 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 to sum up data frame
Dear All, I have a data frame like this: name ip Bsent Breceived a 10.00 0.00 a 2 1.43 19.83 a 1 0.00 0.00 a 21.00 1.00 b 10.00 2.00 b 3 0.00 2.00 b 2 2.00 0.00 b 2 2.00 0.00 b 1 24.40 22.72 c 1 1.00 1.00 c 1 2.00 1.00 c 1 2.00 1.00 c 1 90.97 15.70 d 0 0.00 0.00 d 1 30.00 17.14 I want to sum up the similar name into one row, like : name ip Bsent Breceived a62.43 20.83 b9 28.40 26.72 c d I need help to sum up. Thanks for your time. Thanks Rg Mohan L __ 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 required to remove \\N
Dear All, I have .csv file it looks like this : rawdata - read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t' , header=FALSE) head(rawdata,n=5) TenantDomain Owner Current State 1\\N ROOTadmin Running 2\\N ROOTadmin Stopped 3\\N ROOTadmin Running 4\\N ROOTadmin Running 5\\N ROOTadmin Running 20 DEMO ROOTadmin Stopped 21 DEMOROOT admin Stopped 22 Demo ROOT admin Stopped The first column contain the \\n up to 19 row. I need to replace the \\N value to Blankspace . Any help will really appreciated. Thanks for your time. Thanks Rg Mohan L __ 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 the current working directory to the location of the source file
On Fri, Jun 11, 2010 at 6:03 AM, Marcin Gomulka mrgo...@gmail.com wrote:
AFAIK a script run through source() does not have any legit way to learn
about it's own location.
I need this to make sure that the script will find its datafiles after I
move the whole directory. (The datafiles are in the same directory.)
Here is a hack I invented to work around it:
print(getwd())
source_pathname = get(ofile,envir = parent.frame())
source_dirname = dirname(source_pathname )
setwd(source_dirname)
print(getwd())
Question: Is there a better, cleaner way?
Hi,
I am using some thing like this to setup working directory:
##set working directory
pwd - getwd()
if (!is.null(pwd)) {
setwd(pwd)
} else {
print(Directory not found)
}
Thanks Rg
Mohan L
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and provide commented, minimal, self-contained, reproducible code.
[R] Re :help to aggregate data
Dear All, I have the data some thing like this, I am showing here three days data only: dummy.data - read.table(file='dummy.txt',sep='', header=TRUE) dummy.data StDate Domaindesc Logins 1 05/01/10xxx 10 2 05/01/10xxx 45 3 05/01/10xxx 2 4 05/01/10yyy 45 5 05/01/10yyy 20 6 05/01/10yyy 22 7 05/01/10zzz 34 8 05/01/10zzz 54 9 05/01/10zzz 1 10 05/01/10zzz 0 11 05/02/10yyy 32 12 05/02/10xxx 40 13 05/02/10zzz 23 14 05/02/10yyy 5 15 05/02/10zzz 12 16 05/02/10xxx 19 17 05/02/10xxx 23 18 05/02/10xxx 11 19 05/02/10yyy 9 20 05/02/10zzz 0 21 05/03/10xxx 2 22 05/03/10xxx 21 23 05/03/10xxx 6 24 05/03/10yyy 45 25 05/03/10yyy 43 26 05/03/10yyy 34 27 05/03/10yyy 41 28 05/03/10zzz 31 29 05/03/10zzz 19 30 05/03/10zzz 27 I trying to aggregate(sum) the Logins based on Domaindesc,StDate. I need like this : Domaindesc05/01/10 05/02/10 05/03/10 xxx 5793 29 yyy 8746 122 zzz Any help will be greatly appreciated. Thanks for your time. Mohan L [[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] Re : help to replace variable value
Dear All, I have a data frame data with Jan , Feb as column. I using like this data$Jan to take Jan column. I have the Jan in another variable var1 - Jan data$var1 I need to replace the variable value there . Any help will be greatly appreciated. Thanks Rg Mohan L [[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] Factor to Numeric
Dear All, I have a data frame with State and 12 Months as column. I want to convert all the 12 month column from factor to numeric. any help will be greatly appreciated. str(data) 'data.frame':33 obs. of 9 variables: $ State: Factor w/ 33 levels Andaman and Nicobar Islands,..: 1 2 3 4 5 6 7 8 9 10 ... $ April: Factor w/ 21 levels 0,1,10,116,..: 2 16 1 1 9 20 1 1 2 21 ... $ August : Factor w/ 18 levels 0,1,10,125,..: 1 14 1 1 9 3 1 1 2 5 ... $ February : Factor w/ 19 levels 0,1,103,..: 1 15 1 1 2 6 2 1 2 14 ... $ January : Factor w/ 19 levels 0,1,10,114,..: 1 15 1 1 11 12 2 1 1 4 ... $ July : Factor w/ 16 levels 0,1,122,..: 1 13 1 1 1 12 8 2 1 4 ... $ June : Factor w/ 18 levels 0,1,10,119,..: 1 14 1 1 9 10 9 2 1 4 ... $ May : Factor w/ 20 levels 0,1,10,111,..: 2 19 1 1 10 15 1 1 1 5 ... $ September: Factor w/ 19 levels 0,1,11,122,..: 1 18 1 1 14 3 2 1 2 9 ... Thanks Rg Mohan L [[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] Factor to Numeric
On Wed, May 26, 2010 at 1:13 PM, Ivan Calandra ivan.calan...@uni-hamburg.de wrote: Hi, What about: as.numeric(as.character(data$State)) ? What I what is, I want to convert all the column excluding State to factor to numeric in the data frame. So that I will send this data frame to another function to total . Any help. Thanks Rg Mohan L [[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 required to melt a data frame
I have the raw data with 9 column and 197977 row: dummy State Months Bedroom 1xxxJan 1 2xxxJan 2 3xxxJan 1 4yyyJan 1 5yyyJan 2 6yyyJan 1 7zzzJan 3 8zzzJan 1 9zzzJan 2 10 xxxFeb 3 11 xxxFeb 4 12 xxxFeb 2 13 yyyFeb 1 14 yyyFeb 2 15 yyyFeb 1 16 zzzFeb 2 17 zzzFeb 3 18 zzzFeb 3 dim(rawdata) [1] 197977 9 I want to do some calculation on the State Months Bedroom. So I am picking only three column using subset. subsets.dummy - subset(dummy, select=c(State,Months,Bedroom)) subsets.dummy State Months Bedroom 1xxxJan 1 2xxxJan 2 3xxxJan 1 4yyyJan 1 5yyyJan 2 6yyyJan 1 7zzzJan 3 8zzzJan 1 9zzzJan 2 10 xxxFeb 3 11 xxxFeb 4 12 xxxFeb 2 13 yyyFeb 1 14 yyyFeb 2 15 yyyFeb 1 16 zzzFeb 0 17 zzzFeb 3 18 zzzFeb 3 From the above data. I want below table for 1,2,3,4 Bedroom. Ex : for 1 Bedroom, I need how may(count of 1 Bedroom) 1 Bedroom is there in a particular state for month wise. State Jan Feb xxx 20 yyy 22 zzz 10 for 2 Bedroom : State Jan Feb xxx 11 yyy 11 zzz 10 I want to melt subsets.dummy data for the above requirement. To achive this I am doing like this: melt.data.frame(subsets.dummy,id.vars=c(State,Bedroom),measure.vars=Months) State Bedroom variable value 1xxx 1 Months Jan 2xxx 2 Months Jan 3xxx 1 Months Jan 4yyy 1 Months Jan 5yyy 2 Months Jan 6yyy 1 Months Jan 7zzz 3 Months Jan 8zzz 1 Months Jan 9zzz 2 Months Jan 10 xxx 3 Months Feb 11 xxx 4 Months Feb 12 xxx 2 Months Feb 13 yyy 1 Months Feb 14 yyy 2 Months Feb 15 yyy 1 Months Feb 16 zzz 2 Months Feb 17 zzz 3 Months Feb 18 zzz 3 Months Feb I trying to get a new data frame for 1 bedroom using cast. But I am not able to get the below data for 1 Bedroom using cost. State Jan Feb xxx 20 yyy 22 zzz 10 I not able to understand how to cost it. any help will be greatly appreciated. Thanks Rg Mohan L [[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] help required to melt a data frame
On Tue, May 25, 2010 at 6:59 PM, Hadley Wickham had...@rice.edu wrote: I trying to get a new data frame for 1 bedroom using cast. But I am not able to get the below data for 1 Bedroom using cost. State Jan Feb xxx 20 yyy 22 zzz 10 What do those numbers represent? Hadley The raw data is real estate property portal data. the numbers represent how many 1 Bedroom house has registered in xxx state on particular months. Then we will normalize the data and plot the graph for trend analysis. Thanks Rg Mohan L [[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] help required to melt a data frame
In that case, you probably want: subsets.melt - melt(subsets.dummy,id.vars=c(State,Bedroom, Months)) cast(subsets.melt, State ~ Months, fill = 0, fun = length) Hadley Dear Hadley, subsets.melt - melt(subsets.dummy,id.vars=c(State,Bedroom, Months)) subsets.melt State Bedroom Months 1xxx 1Jan 2xxx 2Jan 3xxx 1Jan 4yyy 1Jan 5yyy 2Jan 6yyy 1Jan 7zzz 3Jan 8zzz 1Jan 9zzz 2Jan 10 xxx 3Feb 11 xxx 4Feb 12 xxx 2Feb 13 yyy 1Feb 14 yyy 2Feb 15 yyy 1Feb 16 zzz 2Feb 17 zzz 3Feb 18 zzz 3Feb cast(subsets.melt, State ~ Months, fill = 0, fun = length) It show some thing like this : Using Months as value column. Use the value argument to cast to override this choice Error in `[.data.frame`(data, , variables, drop = FALSE) : undefined columns selected Thanks Mohan L [[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] help required to melt a data frame
On Tue, May 25, 2010 at 9:35 PM, Wu Gong gho...@gmail.com wrote: Do you mean count frequency of One Bedroom? table(dummy[dummy$Bedroom==1,][,1:2]) Dear learner, Thanks for you time. Yes, that is what I am trying to archive using melt and cost. any way it works for me . Thanks for your help. Thanks Rg Mohan L [[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 required state wise count
I have the data like this: dummy State Months No 1xxxJan 1 2xxxJan 2 3xxxJan 1 4yyyJan 1 5yyyJan 2 6yyyJan 1 7zzzJan 3 8zzzJan 1 9zzzJan 2 10 xxxFeb 3 11 xxxFeb 4 12 xxxFeb 2 13 yyyFeb 1 14 yyyFeb 2 15 yyyFeb 1 16 zzzFeb 2 17 zzzFeb 3 18 zzzFeb 3 I want for a month Jan how many count(row) is there for xxx and yyy.some thing like this: State Jan Feb Mar xxx 3 3 yyy 3 3 zzz 3 3 Any clue will be greatly appreciated Thanks Rg Mohan L [[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] help required state wise count
Hi Mohan, Try this: table(dummy$State,dummy$Months) Jim Thanks for your time. table(dummy$State,dummy$Months) in this case the state column becomes the row index. I want the state name as the first column. There may be a way to do it. Thanks Rg Mohan L [[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] rounding up to nearest integer
Dear All, I have a data frame data and the below is the str of data : $ Feb : int 1 1195 0 11 28 152 24 2 1 1470 ... $ Mar : int 0 1212 0 17 27 184 15 1 1 1311 ... $ Apr : int 2 1244 1 15 23 135 11 0 1 991 ... $ May : int 2 1158 2 10 23 111 16 1 1 1237 ... $ Jun : int 0 845 1 9 16 86 16 2 1 1129 ... $ Jul : int 0 832 0 7 16 68 9 1 0 994 ... $ Aug : int 0 1107 1 4 25 144 7 0 3 1260 ... $ Sep : int 2 1278 1 8 53 212 14 0 3 1375 ... $ Oct : int 3 1329 0 8 39 201 13 0 0 1340 ... $ Nov : int 1 1179 0 5 7 135 2 0 0 1153 ... $ Dec : int 0 1271 0 7 34 168 5 1 2 1792 ... $ Jan.10: int 0 1405 1 10 55 245 26 2 4 2806 ... $ Feb.10: int 0 1330 1 9 29 360 27 3 6 3492 ... $ Mar.10: int 0 1727 0 8 7 341 8 2 4 4578 ... $ Apr.10: int 0 1530 1 8 12 144 7 2 2 5453 ... I am doing some this like this : x - (data[,-(1:2)] - data[,2]) * prop.table(data[,2]) + data[,2] x Feb Mar Apr May Jun Jul 1 1.000.5971.4031.4030.597 0.597 2 1292.610851 1293.5003225 1295.1746211 1290.6749436 1274.2982103 1273.6180264 Now the str of x is : str(x) 'data.frame':32 obs. of 15 variables: $ Feb : num 1 1293 0 5 18 ... $ Mar : num 1 1294 0 5 18 ... $ Apr : num 1 1295 0 5 18 ... $ May : num 1 1291 0 5 18 ... $ Jun : num 1 1274 0 5 18 ... $ Jul : num 1 1274 0 5 18 ... $ Aug : num 1 1288 0 5 18 ... $ Sep : num 1 1297 0 5 18 ... $ Oct : num 1 1300 0 5 18 ... $ Nov : num 1 1292 0 5 18 ... $ Dec : num 1 1297 0 5 18 ... $ Jan.10: num 1 1303.6 0 5 18 ... $ Feb.10: num 1 1300 0 5 18 ... $ Mar.10: num 1 1320 0 5 18 ... $ Apr.10: num 1 1310 0 5 18 ... I need to round up the data frame some thing like this : Feb Mar Apr May Jun Jul 1 1 11 1 1 1 2 1293 1294 1295 1291 1274 1274 there may be a way to round up the nearest integer. any help will be greatly appropriated. Thanks Rg Mohan L [[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] Re : Indexing array to 1000
Dear All, I have an array some thing like this: avglog January February March April May June July August September 60102 83397 56774 48785 49010 40572 38175 47037 51402 The class of avglog array. class(avglog) [1] array str(avglog) num [1:9(1d)] 60102 83397 56774 48785 49010 ... - attr(*, dimnames)=List of 1 ..$ : chr [1:9] January February March April ... I have to normalize this avglog array to 1000. I mean, I need to devide 1000/avglog[1] and have to multiply this to all the elements in the array and need to plot graph Month Vs Index. To achive this I am doing the below code. I am feeling there may be a simple way to do this. value - matrix (avglog) value [,1] [1,] 60102 [2,] 83397 [3,] 56774 [4,] 48785 [5,] 49010 [6,] 40572 [7,] 38175 [8,] 47037 [9,] 51402 day1Avg - value[1] day1Avg [1] 60102 ID - (1000/day1Avg) ID [1] 0.01663838 index - value*ID index [,1] [1,] 1000. [2,] 1387.5911 [3,] 944.6275 [4,] 811.7034 [5,] 815.4471 [6,] 675.0524 [7,] 635.1702 [8,] 782.6195 [9,] 855.2461 monthcount - length(avglog) Month - c(1:monthcount) trend - cbind(Month,c(index)) colnames(trend) - c(Month,Index) trend Month Index [1,]1 1000. [2,]2 1387.5911 [3,]3 944.6275 [4,]4 811.7034 [5,]5 815.4471 [6,]6 675.0524 [7,]7 635.1702 [8,]8 782.6195 [9,]9 855.2461 any help will be greatly appreciated. Thanks Rg Mohan L [[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] help to indexing data frame
On Fri, May 21, 2010 at 11:27 AM, Mohan L l.mohan...@gmail.com wrote: Dear All, I have data some thing like this: sample StateJan Feb Mar A 1 1 1 B1298 12931294 C00 0 D55 5 E 18 18 18 I need to multiply Jan column *1000 and divided by the same number, like this : data[,-(1:2)] * ((data[,2] *1000)/data[ ,2]) data[,-(1:2)] * 1000/data[ ,2] It will work as I expected. I have 0 in the C row. After the calculation it gives NaN zero place. But I need there zero instead of NaN. There may be some solution for this. Thanks Rg Mohan L [[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] Re : Manipulating Data Frames
Dear All, I have data some thisng like this : data - read.csv(file='ipsample.csv',sep=',' , header=TRUE) data State Jan Feb Mar Apr May Jun 1 AAA11022 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 00012 1 4 DDD5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111 86 from this data frame, I took Jan as base and calculating weightage like this : basemonth.sum - sum(data[[2]]) basemonth.sum [1] 1390 basemonth.data - data[[2]] basemonth.data [1]1 129805 18 68 weightage - basemonth.data / basemonth.sum weightage [1] 0.0007194245 0.9338129496 0.00 0.0035971223 0.0129496403 [6] 0.0489208633 The above is the weightage for base month Jan. Now I need to calculate weighted states data. What I need to do is : (((Feb[i]-Jan[1])*weightage)+Jan[1]) for all column. The Jan column is fixed. I need to do the calculation in all the column Feb, Mar etc... StateJan Feb Mar 1 AAA1(((Feb[1]-Jan[1])*weightage[1])+Jan[1]) (((Mar[1]-Jan[1])*weightage[1])+Jan[1]) 2 BBB 1298 (((Feb[2]-Jan[2])*weightage[2])+Jan[2]) (((Mar[2]-Jan[1])*weightage[2])+Ja[1]) 3 CCC 0 4 DDD5 5 EEE 18 6 FFF 68 I am struggling with this . I have framed a logic using for loop. But it seems me very bad logic. Any help will be greatly appreciated. Thanks Rg Mohan L [[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 to indexing data frame
Dear All, I have data some thing like this: sample StateJan Feb Mar A 1 1 1 B1298 12931294 C00 0 D55 5 E 18 18 18 I need to multiply Jan column *1000 and divided by the same number, like this : data[,-(1:2)] * ((data[,2] *1000)/data[ ,2]) I doing some thing wrong with the above to get below result. What I actually need is : StateJan FebMar A 1*1000/1 1*1000/1 1*1000/1 B1298*1000/1298 1293*1000/1298 1294 C00 0 D55 5 E 18 18 18 Here Jan base for me . Any help will be greatly appreciated. Thanks Rg Mohan L [[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] Re : Adding column sum to new row in data frame
Dear All, I have data some thing like this: State Jan Feb Mar Apr May Jun AAA 1 1 0 2 2 0 BBB 1298 1195 1212 1244 1158 845 CCC 0 0 0 1 2 1 DDD 5 11 17 15 10 9 EEE 18 28 27 23 23 16 FFF 68 152 184 135 111 86 I want to sum all the column(Jan, Feb, Mar ...) and have to merge the total at last row. like this: StateJanFebMarAprMayJunAAA110220BBB12981195121212441158845CCC 000121DDD51117 15109EEE182827232316FFF6815218413511186Total1390 1387 1440 1420 1306 957 I am doing some thing like this, but I don't know how to merge Total to data Or I don't know there may be a alternative way. data - read.csv(file='ipsample.csv',sep=',' , header=TRUE) data State Jan Feb Mar Apr May Jun 1 AAA11022 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 00012 1 4 DDD5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111 86 attributes(data) $names [1] State Jan Feb Mar Apr May Jun $class [1] data.frame $row.names [1] 1 2 3 4 5 6 x - data[,2:ncol(data)] x Jan Feb Mar Apr May Jun 111022 0 2 1298 1195 1212 1244 1158 845 300012 1 45 11 17 15 10 9 5 18 28 27 23 23 16 6 68 152 184 135 111 86 Total - sapply(x,sum,na.rm=T) Total Jan Feb Mar Apr May Jun 1390 1387 1440 1420 1306 957 I hope there may be alternative way. Any help will be appreciated. Thanks Rg Mohan L [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re : Adding column sum to new row in data frame
Hi Joshua, rbind(data, c(Total,apply(data[,-1], 2, sum, na.rm=TRUE))) Yes. This what exactly I want. Thanks for your time. If your State column is a factor, it will return a warning that NAs were introduced (but the totals will still be at the bottom). Yes. is.factor(data$State) [1] TRUE State column is a factor, it will return warning message like this . Warning message: In `[-.factor`(`*tmp*`, ri, value = Total) : invalid factor level, NAs generated I know that this is common factor vs character problem in R. I think there may be a safe way to handle this: data - read.csv(file='ipsample.csv',sep=',' , header=TRUE) a - rbind(data, c(Total,apply(data[,-1], 2, sum, na.rm=TRUE))) Warning message: In `[-.factor`(`*tmp*`, ri, value = Total) : invalid factor level, NAs generated a State Jan Feb Mar Apr May Jun 1 AAA11022 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 00012 1 4 DDD5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111 86 7 NA 1390 1387 1440 1420 1306 957 How to safely avoid this warning massage? Now I have NA instead of Total in last row State column. How to I replace it as Total? Thanks Rg Mohan L [[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] Re : Adding column sum to new row in data frame
How to safely avoid this warning massage? Now I have NA instead of Total in last row State column. How to I replace it as Total? Dear All, The below link provides a very good explanation of Creating factor variables and way to avoid the warning message http://www.ats.ucla.edu/stat/R/modules/factor_variables.htm Now it works for me without warning message ( invalid factor level, NAs generated ) according to the document above , I did below to avoid the warning: data - read.csv(file='ipsample.csv',sep=',' , header=TRUE) data State Jan Feb Mar Apr May Jun 1 AAA11022 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 00012 1 4 DDD5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111 86 a - rbind(data, c(Total,apply(data[,-1], 2, sum, na.rm=TRUE))) Warning message: In `[-.factor`(`*tmp*`, ri, value = Total) : invalid factor level, NAs generated a State Jan Feb Mar Apr May Jun 1 AAA11022 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 00012 1 4 DDD5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111 86 7 NA 1390 1387 1440 1420 1306 957 We can see that instead of Total, the label was NA. To do this correctly, I have added the new level, Total, to the factor column data$State using the factor function with the levels argument. Then I can finally add an element to the factor variable from the new level. here is the steps levels(data$State) [1] AAA BBB CCC DDD EEE FFF data$State - factor(data$State,levels=c(levels(data$State),Total)) data$State [1] AAA BBB CCC DDD EEE FFF Levels: AAA BBB CCC DDD EEE FFF Total levels(data$State) [1] AAA BBB CCC DDD EEE FFF Total x - rbind(data, c(Total,apply(data[,-1], 2, sum, na.rm=TRUE))) Now the above works without warning. x State Jan Feb Mar Apr May Jun 1 AAA11022 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 00012 1 4 DDD5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111 86 7 Total 1390 1387 1440 1420 1306 957 I think I am doing right. If I miss understood anything. Please guide me I am beginer to R. Thanks Rg Mohan L [[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] Re : Adding column sum to new row in data frame
On Thu, May 20, 2010 at 10:31 AM, Joshua Wiley jwiley.ps...@gmail.comwrote: Dear Mohan, First, I would like to modify my code slightly to: data - rbind(data,data.frame(State=Total,t(apply(data[,-1], 2, sum, na.rm=TRUE This actually will add a 7th level to your factor automatically. The reason I wanted to change from using c() to data.frame() is that if one uses c(), all the columns are converted to character (this has to do with different methods for rbind, see ?rbind particularly the Details and Value section which describe the different methods for rbind and what its behavior will be if it is using the default method). This may not be an issue, but it would hamper any subsequent calculations you may wish to perform on your data. Thanks for your great help. Mohan L [[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] Re : append new line in existing graph
Dear All, I am executing my R script from PHP using the shell_exec. $return = shell_exec(/usr/bin/R --slave --args .$Domain.,.$Gender. /var/www/html/trends/newTrend.R); In newTrend.R, I am plotting graph and save in pdf file using the below code. It will create a pdf file with two line. pdf(file=/var/www/html/trends/bmtrend.pdf, height=7.5, width=10.5) g_range - range(0,index , ReqDataindex) plot(index, type=o, col=blue, ylim=g_range,axes=FALSE, ann=FALSE) axis(1, at=1:7, lab=c(1:daycount)) axis(2, las=1, at=25*0:g_range[2]) box() lines(ReqDataindex, type=o, pch=22, lty=2, col=red) Now what I want do is: I want to execute shell_exec() second time with different argument and I want to add one more line to existing (bmtrend.pdf) file. I mean if the bmtrend.pdf file does not exist, my script has to create the file. If the bmtrend.pdf file is already exist, I want to append new line in existing file. any help would be greatly appreciated. Thanks for you time. Mohan L [[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] extract required data from already read data
Hi all, I have data like this: sample - read.csv(file=sample.csv,sep=,,header=TRUE) sample stdate Domainsex age Login 1 01/11/09xxx FeMale 25 2 2 01/11/09xxx FeMale 35 4 3 01/11/09xxx Male 1830 4 01/11/09xxx Male 31 3 5 02/11/09xxx Male 3211 6 02/11/09xxx Male 31 1 7 02/11/09xxx FeMale 29 1 8 02/11/09xxx FeMale 23 5 9 03/11/09xxx FeMale 25 9 10 03/11/09xxx FeMale 35 6 11 03/11/09xxx Male 18 3 12 03/11/09xxx Male 31 0 13 04/11/09xxx Male 3225 14 04/11/09xxx Male 31 1 15 04/11/09xxx FeMale 29 0 16 01/11/09yyy FeMale 25 2 17 01/11/09yyy FeMale 35 4 18 01/11/09yyy Male 1830 19 01/11/09yyy Male 31 3 20 02/11/09yyy Male 3211 21 02/11/09yyy Male 31 1 22 02/11/09yyy FeMale 29 1 23 02/11/09yyy FeMale 23 5 24 03/11/09yyy FeMale 25 9 25 03/11/09yyy FeMale 35 6 I have done the graph for all the data. Now I want to get data from particular Domain and Sex from above data, then I need to do the calculation , eg : # if both are true if(Domain == xxx sex==FeMale) I need the data in another variable, like this 1 01/11/09xxx FeMale 25 2 2 01/11/09xxx FeMale 35 4 3 02/11/09xxx FeMale 29 1 I have failed to achieve it . any help would greatly appreciated. Thanks Rg Mohan L [[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] extract required data from already read data
On Fri, May 7, 2010 at 10:51 AM, Joshua Wiley jwiley.ps...@gmail.comwrote: Try something like: sample[which(sample$Domain==xxx sample$sex==FeMale), ] Hope that helps, Hi Josh, It works. Thanks for your time. I am using source() to run program, like source (sample.R) I want to know how to pass the argument to my sample.R, some this like source(sample.R, xxx,FeMale) Is it possible? any link will greatly appreciated. Thanks Rg Mohan L [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] make a column from the row names
Dear All, avglog 01/11/09 02/11/09 03/11/09 04/11/09 9.75 4.50 4.50 8.67 avglog1 - data.frame(avglog) avglog1 avglog 01/11/09 9.75 02/11/09 4.50 03/11/09 4.50 04/11/09 8.67 The first column isnt a column, It's the row names. I makeing a column from the row names by using the following value1$Day - rownames(value1) value1 avglog Day 01/11/09 9.75 01/11/09 02/11/09 4.50 02/11/09 03/11/09 4.50 03/11/09 04/11/09 8.67 04/11/09 But I want like this : Dayavglog Index 119.759.75*100 224.504.50*100 334.504.50*100 448.678.67*100 How to achieve it? Any help will be appreciated. Thanks Rg Mohan L [[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] Re :argument is not numeric or logical
Hi all, I have data size of : dim(sample) [1] 3594317 The first column is stdate - is date ( 01/11/2009 00:00:00,02/11/2009 00:00:00,02/11/2009 00:00:00 etc... ) Login is 13th column - is numbers (12,0,1 erc...) The below operation return the following error. sample1 - read.csv(file=sample1.csv,sep=,,header=TRUE) avglog - with(sample1, tapply(Login, stdate, mean)) Warning messages: 1: In mean.default(X[[1L]], ...) : argument is not numeric or logical: returning NA 2: In mean.default(X[[2L]], ...) : argument is not numeric or logical: returning NA 3: In mean.default(X[[3L]], ...) : argument is not numeric or logical: returning NA 4: In mean.default(X[[4L]], ...) : argument is not numeric or logical: returning NA 5: In mean.default(X[[5L]], ...) : argument is not numeric or logical: returning NA 6: In mean.default(X[[6L]], ...) : argument is not numeric or logical: returning NA 7: In mean.default(X[[7L]], ...) : argument is not numeric or logical: returning NA avglog 01/11/09 00:00 02/11/09 00:00 03/11/09 00:00 04/11/09 00:00 05/11/09 00:00 NA NA NA NA NA How to fix this issue? Thanks Rg Mohan L [[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] Re :argument is not numeric or logical
Your code has a different name for the sample object. And it would be more informative if you offered str on sample1. This is the result of str on sample1. str(sample1) 'data.frame':35943 obs. of 17 variables: $ stdate : Factor w/ 7 levels 01/11/09 00:00,..: 1 1 1 1 1 1 1 1 1 1 ... $ domain : Factor w/ 15 levels Assamese,Bengali,..: 1 1 1 1 1 1 1 1 1 1 ... $ gender : Factor w/ 2 levels FeMale,Male: 1 1 1 1 1 1 1 1 2 2 ... $ postedby : Factor w/ 10 levels Brother,Daughter,..: 2 6 6 7 7 7 7 9 1 5 ... $ caste : Factor w/ 424 levels ,Caste no bar,..: 1 171 189 130 182 214 394 32 394 373 ... $ age: Factor w/ 15 levels 20,20,21,..: 7 15 8 9 11 12 4 9 9 10 ... $ occupation : Factor w/ 54 levels ,Accounts/Finance Professional,..: 43 43 3 2 43 33 43 53 36 14 ... $ religion : Factor w/ 17 levels ,Buddhist,..: 11 7 7 7 7 7 13 7 14 10 ... $ city : Factor w/ 561 levels ,24 Parganas,..: 365 202 116 365 135 143 202 339 404 1 ... $ country: Factor w/ 115 levels Afghanistan,..: 47 47 47 47 47 47 47 47 47 11 ... $ education : Factor w/ 12 levels ,Bachelors - Arts/ Science/ Commerce/ Others,..: 2 5 8 2 2 8 2 2 9 9 ... $ profiletype: Factor w/ 1 level Free: 1 1 1 1 1 1 1 1 1 1 ... $ Login : Factor w/ 419 levels .00,1.00,10.00,..: 114 283 217 216 14 1 2 2 407 327 ... $ EISent : Factor w/ 250 levels .00,1.00,10.00,..: 1 132 23 132 2 1 1 1 1 2 ... $ EIReceived : num 12 5 43 0 8 2 11 12 4 0 ... $ PMSent : Factor w/ 298 levels .00,1.00,10.00,..: 1 1 1 1 1 1 1 1 1 1 ... $ PMReceived : num 8 4 61 1 12 8 0 9 1 0 ... What I am trying is: I have 7 days data in .csv format. I need to the average Login for first day(1/11/2009) , from this average I have to calculate index divisor(ID). Then I have to calculate the average Login for all 7 days and need to multiple each day Login average with index divisor(ID), like this stdate LoginAverage Index 1/11/2009 29.50 LoginAverage*ID This what I am trying? Any help will be appreciated Thanks for your time. Thanks Rg Mohan L [[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] Re :argument is not numeric or logical
Your code has a different name for the sample object. And it would be more informative if you offered str on sample1. This is the result of str on sample1. str(sample1) 'data.frame':35943 obs. of 17 variables: $ stdate : Factor w/ 7 levels 01/11/09 00:00,..: 1 1 1 1 1 1 1 1 1 1 ... So stdate is not a date variable but that is probably not your problem since it looks like the time portion of your not-dates are all 00:00. tapply can work with this information snipped $ Login : Factor w/ 419 levels .00,1.00,10.00,..: 114 283 217 216 14 1 2 2 407 327 ... So Login is not a numeric class variable. Read the FAQ about the proper way to convert that variable to numeric without loosing information. The Login column contain all numeric values . I don't understand why I need to convert that variable to numeric? login column contain data like this : sample$Login [35881] .00 1.00 .00 1.00 .00 .00 1.00 2.00 [35889] 1.00 3.00 .00 .00 .00 1.00 .00 32.00 [35897] 2.00 4.00 .00 17.003.00 1.00 12.00 .00 [35905] 8.00 .00 3.00 7.00 .00 17.00.00 4.00 [35913] 16.004.00 20.007.00 .00 22.0018.00 1.00 [35921] 3.00 3.00 .00 .00 37.003.00 12.00 3.00 [35929] 2.00 11.00.00 5.00 77.001.00 4.00 1.00 [35937] 9.00 18.003.00 3.00 4.00 14.002.00 The mode of Login is numeric only : mode(sample$Login) [1] numeric And I am also using as.numeric(as.character(sample$Login), still it has some warning and NA . I don't understand why this is happening. could you explain? avglog - with(sample1, tapply(as.numeric(as.character(sample$Login)), stdate, mean)) Warning message: In tapply(as.numeric(as.character(sample$Login)), stdate, mean) : NAs introduced by coercion avglog 01/11/09 00:00 02/11/09 00:00 03/11/09 00:00 04/11/09 00:00 05/11/09 00:00 22.04866 16.65358 NA 15.84970 16.58600 06/11/09 00:00 07/11/09 00:00 16.20743 18.52379 Thanks Rg Mohan L [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re :argument is not numeric or logical
I have no way to determining _why_ it is not numeric, but it simply is ... not. Your input method turned it into a factor. Perhaps there was one missing delimiter, or there was a stray character in one of the entries in the file. Who knows. Why waste time arguing? Follow the directions for fixing the problem. Hi David, Thanks for your time. I thinks it works now. str(assame$Login) Factor w/ 419 levels .00,1.00,10.00,..: 114 283 217 216 14 1 2 2 407 327 ... is.numeric(assame$Login) [1] FALSE as.numeric(assame$Login) //convert to numerics str(as.numeric(assame$Login)) num [1:35943] 114 283 217 216 14 1 2 2 407 327 ... the I did some thing like this : avglog - with(assame, tapply(as.numeric(Login), stdate, mean) ) avglog 01/11/09 00:00 02/11/09 00:00 03/11/09 00:00 04/11/09 00:00 05/11/09 00:00 145.0176 135.5207 133.0390 131.1457 132.3732 06/11/09 00:00 07/11/09 00:00 129.6357 133.4521 Am I right ? Thanks Rg Mohan L [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re :argument is not numeric or logical
as.numeric(assame$Login) //convert to numerics Nooo. You did not do what what I suggested earlier. Do not reach for your keyboard in the R console until you have read the FAQ section regarding converting factors to numeric. If you don't want to read the FAQ Hi David, Now I understand the trick : as.numeric(as.character(Login)) avglog - with(assame,tapply(as.numeric(as.character(Login)),stdate,mean)) avglog 01/11/09 00:00 02/11/09 00:00 04/11/09 00:00 05/11/09 00:00 06/11/09 00:00 22.04866 16.65358 15.84970 16.58600 16.20743 07/11/09 00:00 18.52379 But still it will not work : avglog - with(assame,tapply(as.numeric(levels(Login)[as.integer(Login)]),stdate,mean)) Error in tapply(as.numeric(levels(Login)[as.integer(Login)]), stdate, : arguments must have same length Thanks Rg Mohan L [[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] Re :reading large file
On Sat, May 1, 2010 at 11:55 AM, Tal Galili tal.gal...@gmail.com wrote: Hi Mohan, Check: dim(bmtrend) If the output is like the dimension of your data, then it would appear you succeeded in reading in the data :) Yes . You are right. It reads all the data. Thanks you. Mohan L [[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] Average Login based on date
Hi All, I have the data like this : sample - read.csv(file=sample.csv,sep=,,header=TRUE) sample stdate Domainsex age Login 1 01/11/09xxx FeMale 25 2 2 01/11/09xxx FeMale 35 4 3 01/11/09xxx Male 1830 4 01/11/09xxx Male 31 3 5 02/11/09xxx Male 3211 6 02/11/09xxx Male 31 1 7 02/11/09xxx FeMale 29 1 8 02/11/09xxx FeMale 23 5 9 03/11/09xxx FeMale 25 9 10 03/11/09xxx FeMale 35 6 11 03/11/09xxx Male 18 3 12 03/11/09xxx Male 31 0 13 04/11/09xxx Male 3225 14 04/11/09xxx Male 31 1 15 04/11/09xxx FeMale 29 0 16 01/11/09yyy FeMale 25 2 17 01/11/09yyy FeMale 35 4 18 01/11/09yyy Male 1830 19 01/11/09yyy Male 31 3 20 02/11/09yyy Male 3211 21 02/11/09yyy Male 31 1 22 02/11/09yyy FeMale 29 1 23 02/11/09yyy FeMale 23 5 24 03/11/09yyy FeMale 25 9 25 03/11/09yyy FeMale 35 6 26 03/11/09yyy Male 18 3 27 03/11/09yyy Male 31 0 28 04/11/09yyy Male 3225 29 04/11/09yyy Male 31 1 30 04/11/09yyy FeMale 29 0 I need to fine the average login on 01/11/09 and 02/11/09 etc... like below stdate AverageLogin 01/11/09 9.75 02/11/09 . 03/11/09 ... How do I find the average Login based on date? Thanks for your time. Mohan L [[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] Re :reading large file
On Sat, May 1, 2010 at 11:02 AM, Mohan L l.mohan...@gmail.com wrote: Hi All, I am new to R and the mailing list. I have a data file (.xls) format with little bit large (16 column,35000 rows ) data file. I am trying to read this file for calculation. I have converted the file into .csv format and read like that : bmtrend - read.csv(file=simple.csv,head=TRUE,sep=,) bmtrend will display something in the middle of display [ reached get Option(max.print) -- omitted 30061 rows ]] It will not show the full list of data. please suggest me best way to read large file? Thanks Rg Mohan L I send the mail without subject, I apologies for inconvenience. [[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] Re :
Hi All, I am new to R and the mailing list. I have a data file (.xls) format with little bit large (16 column,35000 rows ) data file. I am trying to read this file for calculation. I have converted the file into .csv format and read like that : bmtrend - read.csv(file=simple.csv,head=TRUE,sep=,) bmtrend will display something in the middle of display [ reached get Option(max.print) -- omitted 30061 rows ]] It will not show the full list of data. please suggest me best way to read large file? Thanks Rg Mohan L [[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.
