Is there a better way to parse strings than this?

Hi:

Here's one approach:

strings <- c(
"A5.Brands.bought...Dulux",
"A5.Brands.bought...Haymes",
"A5.Brands.bought...Solver",
"A5.Brands.bought...Taubmans.or.Bristol",
"A5.Brands.bought...Wattyl",
"A5.Brands.bought...Other")

slist <- strsplit(strings, '\\.\\.\\.')

Hi Everyone,


I needed to parse some strings recently.

The code I've wound up using seems rather clunky, and I was wondering if
anyone had any suggestions on a better way?

Basically I do the following:

1) Use substr() to do the parsing
2) Use regexpr() to find the location of the string I want to parse on, I
then pass this onto substr()
3) Use nchar() as the stop input to substr() where necessary



I've got a simple example of the parsing code I used below. It takes
questionnaire variable names that includes the question and the brand it
was answered for and then parses it so the variable name and the brand are
in separate columns. I then use this to restructure the data from
unstacked to stacked, but that's another story.

# this is the data set
test

[1] "A5.Brands.bought...Dulux"
[2] "A5.Brands.bought...Haymes"
[3] "A5.Brands.bought...Solver"
[4] "A5.Brands.bought...Taubmans.or.Bristol"
[5] "A5.Brands.bought...Wattyl"
[6] "A5.Brands.bought...Other"

# Where do I want to parse?
break1 <- ?regexpr('...',test, fixed=TRUE)
break1

[1] 17 17 17 17 17 17
attr(,"match.length")
[1] 3 3 3 3 3 3

# Put Variable name in a variable
str1 <- substr(test,1,break1-1)
str1

[1] "A5.Brands.bought" "A5.Brands.bought" "A5.Brands.bought"
"A5.Brands.bought"
[5] "A5.Brands.bought" "A5.Brands.bought"

# Put Brand name in a variable
str2 <- substr(test,break1+3, nchar(test))
str2

[1] "Dulux" ? ? ? ? ? ? ? "Haymes" ? ? ? ? ? ? ?"Solver"
[4] "Taubmans.or.Bristol" "Wattyl" ? ? ? ? ? ? ?"Other"

x <- c("A5.Brands.bought...Dulux", "A5.Brands.bought...Haymes",

do.call(rbind, strsplit(x, "...", fixed = TRUE))

# or
xa <- sub("...", "\1", x, fixed = TRUE)
read.table(textConnection(xa), sep = "\1", as.is = TRUE)

Hi Everyone,


I needed to parse some strings recently.

The code I've wound up using seems rather clunky, and I was wondering if
anyone had any suggestions on a better way?

Basically I do the following:

1) Use substr() to do the parsing
2) Use regexpr() to find the location of the string I want to parse on,

then pass this onto substr()
3) Use nchar() as the stop input to substr() where necessary



I've got a simple example of the parsing code I used below. It takes
questionnaire variable names that includes the question and the brand it
was answered for and then parses it so the variable name and the brand

in separate columns. I then use this to restructure the data from
unstacked to stacked, but that's another story.

# this is the data set
test

[1] "A5.Brands.bought...Dulux"
[2] "A5.Brands.bought...Haymes"
[3] "A5.Brands.bought...Solver"
[4] "A5.Brands.bought...Taubmans.or.Bristol"
[5] "A5.Brands.bought...Wattyl"
[6] "A5.Brands.bought...Other"

# Where do I want to parse?
break1 <- ?regexpr('...',test, fixed=TRUE)
break1

[1] 17 17 17 17 17 17
attr(,"match.length")
[1] 3 3 3 3 3 3

# Put Variable name in a variable
str1 <- substr(test,1,break1-1)
str1

[1] "A5.Brands.bought" "A5.Brands.bought" "A5.Brands.bought"
"A5.Brands.bought"
[5] "A5.Brands.bought" "A5.Brands.bought"

# Put Brand name in a variable
str2 <- substr(test,break1+3, nchar(test))
str2

[1] "Dulux" ? ? ? ? ? ? ? "Haymes" ? ? ? ? ? ? ?"Solver"
[4] "Taubmans.or.Bristol" "Wattyl" ? ? ? ? ? ? ?"Other"

x <- c("A5.Brands.bought...Dulux", "A5.Brands.bought...Haymes",

do.call(rbind, strsplit(x, "...", fixed = TRUE))

# or
xa <- sub("...", "\1", x, fixed = TRUE)
read.table(textConnection(xa), sep = "\1", as.is = TRUE)

I was trying strsplit(string,"\.\.\.") as per the suggestion in Venables
and Ripleys book to "(use '\.' to match '.')", which is in the Regular
expressions section.

I noticed that in the suggestions sent to me people used:
strsplit(test,"\\.\\.\\.")


Could anyone please explain why I should have used "\\.\\.\\." rather than
"\.\.\."?

Thanks for the suggestions, they were all exactly what I was looking for.
(I knew that had to be a more elegant way then my brute force method)

One question though.

I was playing around with strsplit but couldn't get it to work, I realised
my problem was that I was using "." as the string.

I was trying strsplit(string,"\.\.\.") as per the suggestion in Venables
and Ripleys book to "(use '\.' to match '.')", which is in the Regular
expressions section.

I noticed that in the suggestions sent to me people used:
strsplit(test,"\\.\\.\\.")


Could anyone please explain why I should have used "\\.\\.\\." rather than
"\.\.\."?

cat("\\.\\.\\.\n")

Hi Everyone,


I needed to parse some strings recently.

The code I've wound up using seems rather clunky, and I was wondering if
anyone had any suggestions on a better way?

Basically I do the following:

1) Use substr() to do the parsing
2) Use regexpr() to find the location of the string I want to parse on, I
then pass this onto substr()
3) Use nchar() as the stop input to substr() where necessary



I've got a simple example of the parsing code I used below. It takes
questionnaire variable names that includes the question and the brand it
was answered for and then parses it so the variable name and the brand are
in separate columns. I then use this to restructure the data from
unstacked to stacked, but that's another story.

# this is the data set
test

[1] "A5.Brands.bought...Dulux"
[2] "A5.Brands.bought...Haymes"
[3] "A5.Brands.bought...Solver"
[4] "A5.Brands.bought...Taubmans.or.Bristol"
[5] "A5.Brands.bought...Wattyl"
[6] "A5.Brands.bought...Other"

# Where do I want to parse?
break1 <- ?regexpr('...',test, fixed=TRUE)
break1

[1] 17 17 17 17 17 17
attr(,"match.length")
[1] 3 3 3 3 3 3

# Put Variable name in a variable
str1 <- substr(test,1,break1-1)
str1

[1] "A5.Brands.bought" "A5.Brands.bought" "A5.Brands.bought"
"A5.Brands.bought"
[5] "A5.Brands.bought" "A5.Brands.bought"

# Put Brand name in a variable
str2 <- substr(test,break1+3, nchar(test))
str2

[1] "Dulux" ? ? ? ? ? ? ? "Haymes" ? ? ? ? ? ? ?"Solver"
[4] "Taubmans.or.Bristol" "Wattyl" ? ? ? ? ? ? ?"Other"



Thanks for any and all suggestions


Chris Howden
Founding Partner
Tricky Solutions
Tricky Solutions 4 Tricky Problems
Evidence Based Strategic Development, IP Commercialisation and Innovation,
Data Analysis, Modelling and Training
(mobile) 0410 689 945
(fax / office) (+618) 8952 7878
chris at trickysolutions.com.au

I was trying strsplit(string,"\.\.\.") as per the suggestion in Venables
and Ripleys book to "(use '\.' to match '.')", which is in the Regular
expressions section.

I noticed that in the suggestions sent to me people used:
strsplit(test,"\\.\\.\\.")


Could anyone please explain why I should have used "\\.\\.\\." rather

"\.\.\."?