What is the regex to replace a single space between a closing brace and a letter?
Question:
I want to replace all occurrences of
- case 1: a single space between 2 letters
- case 2: a single space between a closing brace
} and a letter
with times such that
'a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}'
becomes
'a^{3} + 3 \times a^{2} \times b + 3 \times a \times b^{2} + b^{3}'
I can successfully do the first case with bsb but I don’t know how to do for the second one.
My attempt below just produces
'a^{3} + 3 \times a^{2} b + 3 \times a \times b^{2} + b^{3}'
where I fail to put \times between a^{2} and b.
import re
import sympy
from IPython.display import display
a,b = sympy.symbols('a b')
f = (a+b)**3
expr = sympy.latex(f.expand())
display(expr)
expr = re.sub(r'(bsb)|(IDONTKNOW)', r' \times ', expr)
display(expr)
Answers:
Like this?
re.sub(r'(?<=[}w])s(?=w)', r' times ', expr)
The lookarounds are assertions which do not include the text in the match. (?<=...) requires the text before the main match to be either } or an alphanumeric, and (?=...) requires the text after the main match to be an alphanumeric.
(w is not exactly "an alphanumeric"; it matches numbers and underscores in addition to alpbabetics. Probably replace it with e.g. [A-Za-z] if you strictly need to match only English alphabetics.)
I want to replace all occurrences of
- case 1: a single space between 2 letters
- case 2: a single space between a closing brace
}and a letter
with times such that
'a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}'
becomes
'a^{3} + 3 \times a^{2} \times b + 3 \times a \times b^{2} + b^{3}'
I can successfully do the first case with bsb but I don’t know how to do for the second one.
My attempt below just produces
'a^{3} + 3 \times a^{2} b + 3 \times a \times b^{2} + b^{3}'
where I fail to put \times between a^{2} and b.
import re
import sympy
from IPython.display import display
a,b = sympy.symbols('a b')
f = (a+b)**3
expr = sympy.latex(f.expand())
display(expr)
expr = re.sub(r'(bsb)|(IDONTKNOW)', r' \times ', expr)
display(expr)
Like this?
re.sub(r'(?<=[}w])s(?=w)', r' times ', expr)
The lookarounds are assertions which do not include the text in the match. (?<=...) requires the text before the main match to be either } or an alphanumeric, and (?=...) requires the text after the main match to be an alphanumeric.
(w is not exactly "an alphanumeric"; it matches numbers and underscores in addition to alpbabetics. Probably replace it with e.g. [A-Za-z] if you strictly need to match only English alphabetics.)