How do I loop over the rows of a data frame without relying on locational references to Column in R

Question

I have figured out how to create a new column on my data frame that = TRUE if the character string in "Column 5" is contained within the longer string in "Column 6" - can I do this by referring to the names of my columns rather than using [r,c] locational references?

rows = NULL

for(i in 1:length(excptn1[,1]))
{
    rows[i] <- grepl(excptn1[i,5],excptn1[i,6], perl=TRUE)
}

As a programmer I'm nervous about referring to things as "Column 5 and Column 6"...I want to refer to the names of the variables captured in those columns so that I'm not reliant on my source file always having the columns in the identical order. Furthermore I might forget about that locational reference and add something earlier in the code that causes the locational reference to fail later...when you can think in terms of the names of the columns in general (rather than their particular ordering at a point in time) it's a lot easier to build robust production strength code.

I found a related question on this site and it uses the same kind of locational references I want to avoid...

How do I perform a function on each row of a data frame and have just one element of the output inserted as a new column in that row

While R does seem very flexible it seems to lack a lot of features that you'd want in scaleable, production strength code...but I'm hoping I'm wrong and can learn otherwise.

Thanks!

David Robinson · Answer 1 · 2014-03-06T00:47:39.873

2

You could refer to the columns by name rather than by index in two ways:

rows[i] <- grepl(excptn1[i,"colname"],excptn1[i,"othercolname"], perl=TRUE)

or

rows[i] <- grepl(excptn1$colname[i],excptn1$othercolname[i], perl=TRUE)

Finally, note that most R programmers would do this as:

rows = sapply(1:nrow(excptn), grepl(excptn1$colname[i],excptn1$othercolname[i], perl=TRUE))

One thing this avoids is the overhead of increasing the size of the vector in each iteration.

edited Mar 06 '14 at 00:47

answered Mar 06 '14 at 00:16

David Robinson

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Can you explain this further? The first argument in `grepl` should be a regular expression, as far as I know. – Rich Scriven Mar 06 '14 at 00:32
@RScriv: My apologies: I had forgotten that `grepl` vectorizes across the second argument but not the first. Edited to correct it. – David Robinson Mar 06 '14 at 00:47
Thanks! I was missing the quotes around my attempt to reference my columns. – user3381203 Mar 06 '14 at 01:03
Point taken on the apply, sapply, tapply fucntions but I'm trying to keep things more simple and literal for now since I'm new to R – user3381203 Mar 06 '14 at 01:03

score 0 · Answer 2 · edited Jan 11 '15 at 16:29

If you want to do this faster, use stri_match_first_regex function from stringi package.

Example:

require(stringi)

ramka <- data.frame(foo=letters[1:3],bar=c("ala","ma","koteczka"))

> ramka
  foo      bar
1   a      ala
2   b       ma
3   c koteczka

> stri_match_first_regex(str=ramka$bar, pattern=ramka$foo)
     [,1]
[1,] "a" 
[2,] NA  
[3,] "c"

How do I loop over the rows of a data frame without relying on locational references to Column in R

2 Answers2