Need to find '$word;' pattern in string

Question

I have big text file and I have to find all words starts with '$' and ends with ';' like $word;.

import re

text = "$h;BREWERY$h_end;You've built yourself a brewery."
x = re.findall("$..;", text)
print(x)

I want my output like ['$h;', '$h_end;'] How can I do that?

`x = re.findall("\$[a-zA-Z_]+;", text)` – rdas Jun 01 '20 at 12:32 — rdas, Jun 01 '20 at 12:32

score 2 · Accepted Answer · answered Jun 01 '20 at 12:43

I have to find all words starts with '$' and ends with ';' like $word;.

I would do:

import re
text = "$h;BREWERY$h_end;You've built yourself a brewery."
result = re.findall('\$[^;]+;',text)
print(result)

Output:

['$h;', '$h_end;']

Note that $ needs to be escaped (\$) as it is one of special characters. Then I match 1 or more occurences of anything but ; and finally ;.

score 1 · Answer 2 · answered Jun 01 '20 at 12:32

You may use

\$\w+;

See the regex demo. Details:

import re

text = "$h;BREWERY$h_end;You've built yourself a brewery."
x = re.findall(r"\$\w+;", text)
print(x) # => ['$h;', '$h_end;']

2 Answers2