Hi

Hi all,

I have solved my question

Output<- matrix(0, length(x), 1)

for (i in 1:length(x)){

files<- paste('KT', i, '.csv', sep = '')

da1<- read.csv(files)

Output[i,1]<-da1[1,2]
}
2 comments:

1: use seq:_along(x) rather than 1:length(x) to protect against length 0 
objects.

2: use list.files(pattern="^KT.*\\.csv$") to get all the files you were 
looking for from the current directory, hence no need to construct the 
filenames.

That way it is probably a one-liner (untested):

Output <- sapply(list.files(pattern="^KT.*\\.csv$"), function(x) 
read.csv(x)[1,2])

Uwe Ligges
2011/12/14 Trying To learn again<tryingtolearnagain at gmail.com>

Hi all,

I have 100 csv files always with this information (I Attach two example
excels)

KT80.csv contains:
      ,"x"
  1,127.188271604938

KT80.csv contains

      ,"x"
  1,1.06329287351545

I have three questions:

First

When I created this input files I used write.csv and it authomaticaly
creates the structure of the file as you can see.

Second. If I try to read csv it appears:

  read.csv(KT2.csv,header=T, sep='.')
Error en read.table(file = file, header = header, sep = sep, quote =
quote,  :
   objeto 'KT2.csv' no encontrado

But the file is in the right directy....

Third (supossing I can solve the format files)

My main question is:

I want to create a matrix (or data frame) for instance from file KT80.csv
to KT81.csv so that the rows are without header and position.

127.18
1.06

Wich function would help me?

I think I should use something like

ffor (i in 80:81){

files<- paste('KT', i, '.csv', sep = '')

myfun<- function(d) {

da1<- read.csv(d)

Output[i,1]=da1[1,2]

}

lout<- lapply(files, myfun)

}

But results that says Output doesn?t exists?

I must create before? How?

Many Thans???

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