regexpr - ignore all special characters and punctuation in a string
You can use the [:alnum:] regex class with gsub.
str1 <- "What a nice day today! - Story of happiness: Part 2."
str2 <- "What a nice day today: Story of happiness (Part 2)"
gsub("[^[:alnum:]]", "", str1) == gsub("[^[:alnum:]]", "", str2)
[1] TRUE
The same can be done with the stringr package if you really are partial to
it.
library(stringr)
On Mon, Apr 20, 2015 at 9:10 AM, Sven E. Templer <sven.templer at gmail.com>
wrote:
Hi Dimitri,
str_replace_all is not in the base libraries, you could use 'gsub' as well,
for example:
a = "What a nice day today! - Story of happiness: Part 2."
b = "What a nice day today: Story of happiness (Part 2)"
sa = gsub("[^A-Za-z0-9]", "", a)
sb = gsub("[^A-Za-z0-9]", "", b)
a==b
# [1] FALSE
sa==sb
# [1] TRUE
Take care of the extra space in a after the '-', so also replace spaces...
Best,
Sven.
On 20 April 2015 at 16:05, Dimitri Liakhovitski <
dimitri.liakhovitski at gmail.com> wrote:
I think I found a partial answer: str_replace_all(x, "[[:punct:]]", " ") On Mon, Apr 20, 2015 at 9:59 AM, Dimitri Liakhovitski <dimitri.liakhovitski at gmail.com> wrote:
Hello! Please point me in the right direction. I need to match 2 strings, but focusing ONLY on characters, ignoring all special characters and punctuation signs, including (), "", etc.. For example: I want the following to return: TRUE "What a nice day today! - Story of happiness: Part 2." == "What a nice day today: Story of happiness (Part 2)" -- Thank you! Dimitri Liakhovitski
-- Dimitri Liakhovitski
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