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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