Using regex to parse string patterns from a single string

Question

I have read in product pricing for some products. As you will see below, not every product pricing string is set up the same. What I am trying to do is to parse out the sub-strings I do not want.

Below is the code I have which works, but there has to be a more efficient way to do this.

tmp1 = p_pricing.replace("from ", "")
tmp1 = tmp1.replace("Options Available on Open Box", "")
tmp1 = tmp1.replace("Open Box Price: From ", "")
tmp1 = re.sub(r'\([^)]*\)', '', tmp1)
tmp1 = re.split("[$]", tmp1)

Below is a small sample of my pricing string:

$11.99($6.00 per item)$14.99
from $13.99$18.25
$9.89($4.94 per item)$14.99
from $9.83($3.28 per item) 
from $15.99$29.99
from $84.99$104.95
from $9.83($3.28 per item) 
$3.47
$94.99$129.99
from $14.34$19.90
from $25.01$65.00Options Available on Open Box

Wiktor Stribiżew · Accepted Answer · 2021-06-07 20:15:01Z

It seems you just want to get the numeric values of all prices in each string.

You can use

re.findall(r'\$(\d+(?:\.\d+)?)', text)

See the regex demo.

Details

\$ - a $ char
(\d+(?:\.\d+)?) - Capturing group 1: one or more digits, and then an optional occurrence of a . and one or more digits.

See the Python demo:

import re
pattern = r"\$(\d+(?:\.\d+)?)"
text = "$11.99($6.00 per item)$14.99\nfrom $13.99$18.25\n$9.89($4.94 per item)$14.99\nfrom $9.83($3.28 per item) \nfrom $15.99$29.99\nfrom $84.99$104.95\nfrom $9.83($3.28 per item) \n$3.47\n$94.99$129.99\nfrom $14.34$19.90\nfrom $25.01$65.00Options Available on Open Box"
print( re.findall(pattern, text) )

Output:

['11.99', '6.00', '14.99', '13.99', '18.25', '9.89', '4.94', '14.99', '9.83', '3.28', '15.99', '29.99', '84.99', '104.95', '9.83', '3.28', '3.47', '94.99', '129.99', '14.34', '19.90', '25.01', '65.00']

The fourth bird · Accepted Answer · 2021-06-08 06:36:19Z

As you are replacing from and opening till closing parenthesis in your code using $[^)]*$ with an empty string, you can get all the prices outside of the parenthesis by matching from an opening parenthesis till a closing parenthesis.

Then use an alternation | and capture what you want to keep.

The digits are in capture group 1.

\([^()]*\)|\$(\d+(?:\.\d+))

$[^()]*$ Match from an opening till closing parenthesis
| Or
\$ Match a dollar sign
(\d+(?:\.\d+)) Capture group 1 Match 1+ digits and an optional decimal part

See a regex demo or a Python demo

Example code

import re

pattern = r"\([^()]*\)|\$(\d+(?:\.\d+))"

s = "$11.99($6.00 per item)$14.99 from $13.99$18.25 $9.89($4.94 per item)$14.99 from $9.83($3.28 per item) from $15.99$29.99 from $84.99$104.95 from $9.83($3.28 per item) $3.47 $94.99$129.99 from $14.34$19.90 from $25.01$65.00Options Available on Open Box"

print([s for s in re.findall(pattern, s) if s])

Output

['11.99', '14.99', '13.99', '18.25', '9.89', '14.99', '9.83', '15.99', '29.99', '84.99', '104.95', '9.83', '3.47', '94.99', '129.99', '14.34', '19.90', '25.01', '65.00']

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