R apply function on data frame column - r

Question

I need to , efficiently, parse one of my dataframe column (a url string)
and call a function (strsplit) to parse it, e.g.:
url <- c("www.google.com/nir1/nir2/nir3/index.asp")
unlist(strsplit(url,"/"))
My data frame : spark.data.url.clean looks like this:
classes url
[107,662,685,508,111,654,509] drudgereport.com/level1/level2/level3
This df has 100k rows and I don't want to loop/iterate over it, parse each url separately and write the results to a new data frame.
What I DO need/want is to create a new 5 column data frame:
df.result <- data.frame(fullurl = as.character(),baseurl=as.character(), firstlevel = as.character(), secondlevel=as.character(),thirdlevel=as.character(),classificaiton=as.character())
call one of the "apply" family function over spark.data.url.clean$url
and to write the results to the new data frame df.result such that the first column (fullurl) will be populated with the relevant spark.data.url.clean$url, the 2nd to 5th columns will be populated with the relevant results from applying
unlist(strsplit(url,"/"))
- taking the only the first, 2nd, 3rd and 4th elements from the resulted vector and putting it in the first,2nd, 3rd and 4th columns in df.result and finally putting the spark.data.url.clean$classes in the new data frame columns df.result$classificaiton
Sorry for the complication and let me know if anything need to be further cleared out.

Answer 1

There is no need for apply, as far as I see it.
Try this:
spark.data.url.clean <- data.frame(classes = c(107,662,685,508,111,654,509),
url = c("drudgereport.com/level1/level2/level3", "drudgeddddreport.com/levelfe1/lefvel2/leveel3",
"drudgeaasreport2.com/lefvel13/lffvel244/fel223", "otherurl.com/level1/second/level3",
"whateversite.com/level13/level244/level223", "esportsnow.com/first/level2/level3",
"reeport2.com/level13/level244/third"), stringsAsFactors = FALSE)
df.result <- spark.data.url.clean
names(df.result) <- c("classification", "fullurl")
df.result[c("baseurl", "firstlevel", "secondlevel", "thirdlevel")] <- do.call(rbind, strsplit(df.result$fullurl, "/"))

Answer 2

You could consider using the package splitstackshape to do this; we can use its cSplit-function. Setting drop to F ensures that the original column is preserved. Not that it returns a data.table, not a data.frame.
library(splitstackshape)
output <- cSplit(dat,2,sep="/", drop=F)
data used:
dat <- data.frame(classes="[107,662,685,508,111,654,509]",
url="drudgereport.com/level1/level2/level3")

Answer 3

Here's an option with data.table which should be pretty fast. If your data looks like this:
> df
# classes url
#1 [107,662,685,508,111,654,509] drudgereport.com/level1/level2/level3
You can do the following:
library(data.table)
setDT(df) # convert to data.table
cols <- c("baseurl", "firstlevel", "secondlevel", "thirdlevel") # define new column names
df[, (cols) := tstrsplit(url, "/", fixed = TRUE)[1:4]] # assign new columns
Now, the data looks like this:
> df
# classes url baseurl firstlevel secondlevel thirdlevel
#1: [107,662,685,508,111,654,509] drudgereport.com/level1/level2/level3 drudgereport.com level1 level2 level3

Answer 4

The simple solution is to use:
apply(row, 2, function(col) {})