Writing a single output file

This should get you close:
# get file names
setwd('/temp')
fileNames <- list.files(pattern = "file.*.csv")
fileNames
[1] "file1.csv" "file2.csv" "file3.csv" "file4.csv"
input <- do.call(rbind, lapply(fileNames, function(.name){
+     .data <- read.table(.name, header = TRUE, as.is = TRUE)
+     # add file name to the data
+     .data$file <- .name
+     .data
+ }))
input
date yield_rate      file
1 12/23/2010       5.25 file1.csv
2 12/22/2010       5.19 file1.csv
3 12/23/2010       5.25 file2.csv
4 12/22/2010       5.19 file2.csv
5 12/23/2010       5.25 file3.csv
6 12/22/2010       5.19 file3.csv
7 12/23/2010       5.25 file4.csv
8 12/22/2010       5.19 file4.csv
require(reshape)
in.melt <- melt(input, measure = 'yield_rate')
cast(in.melt, date ~ file)
date file1.csv file2.csv file3.csv file4.csv
1 12/22/2010      5.19      5.19      5.19      5.19
2 12/23/2010      5.25      5.25      5.25      5.25

Dear R helpers!

Let me first wish all of you "Merry Christmas and Very Happy New year 2011"

"Christmas day is a day of Joy and Charity,
May God make you rich in both" - Phillips Brooks

## ----------------------------------------------------------------------------------------------------------------------------

I have a process which generates number of outputs. The R code for the same is as given below.

for(i in 1:n)
{
write.csv(output[i], file = paste("output", i, ".csv", sep = ""), row.names = FALSE)
}

Depending on value of 'n', I get different output files.

Suppose n = 3, that means I am having three output csv files viz. 'output1.csv', 'output2.csv' and 'output3.csv'

output1.csv
date?????????????? yield_rate
12/23/2010??????? 5.25
12/22/2010??????? 5.19
.................................
.................................

output2.csv

date?????????????? yield_rate

12/23/2010??????? 4.16

12/22/2010??????? 4.59

.................................

.................................

output3.csv

date?????????????? yield_rate

12/23/2010??????? 6.15

12/22/2010??????? 6.41

.................................

.................................

Thus all the output files have same column names viz. Date and yield_rate. Also, I do need these files individually too.

My further requirement is to have a single dataframe as given below.

Date???????????? yield_rate1?????????????? yield_rate2??????????????? yield_rate3
12/23/2010?????? 5.25????????????????????????? 4.16????????????????????????? 6.15
12/22/2010?????? 5.19 ? ? ? ? ? ? ? ? ? ? ? ?? 4.59 ? ? ? ? ? ? ? ? ? ? ? ?? 6.41
...............................................................................................
...............................................................................................

where yield_rate1 = output1$yield_rate and so on.

One way is to simply create a dataframe as

df = data.frame(Date = read.csv('output1.csv')$Date, yield_rate1 =? read.csv('output1.csv')$yield_rate,?? yield_rate2 = read.csv('output2.csv')$yield_rate,
yield_rate3 = read.csv('output3.csv')$yield_rate)

However, the problem arises when I am not aware how many output files are there as n can be 5 or even 100.

So is it possible to write some loop or some function which will enable me to read 'n' files individually and then keeping "Date" common, only pickup the yield_curve data from each output file.

Thanking in advance for any guidance.

Regards

Amy

? ? ? ?[[alternative HTML version deleted]]

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What is the problem that you are trying to solve?

Writing a single output file

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