Regex matching a set of characters on a given index

Question

I have a string and a position in this string. I want to find out if the last non space character before this position is one of the characters in a given set. Can I do this using regex? I couldn't figure it out on my own.

Example with a set of characters (?, |, :):

foo('blah? test', pos=6) is True

foo('blah? test', pos=7) is False

Answer 1

With some help from Regex:

In [93]: def is_matched(text, pos, chars='?|!'): 
    ...:     text = text[:pos] 
    ...:     matched = re.search(r'.*(\S)(?=\s*$)', text) 
    ...:     return matched.group(1) in chars if matched else False 
    ...:                                                                                                                                                                                                    

In [94]: is_matched('blah? test', pos=6)                                                                                                                                                                    
Out[94]: True

In [95]: is_matched('blah? test', pos=7)                                                                                                                                                                    
Out[95]: False

.*(\S)(?=\s*$) :

• .* matches any characters upto last non-space character

• (\S) matches the last non-space character and put it in a captured group

• The zero-width postive lookahead (?=\s*$) makes sure the pattern is followed by zero-more spaces only till end

Answer 2

Assuming that you want 0 indexed strings

def foo(text, pos):
    return text[pos] in ['?','|','!']

foo('blah? test', pos=4) // True
foo('blah? test', pos=5) // False

Answer 3

You don't really need regex for this. You can very easily use any and list comprehension .

s = 'blah? test'
print(any(v in s[4] for v in '?|!'))

Returns True .

Changing s[4] to s[5] results in False

Answer 4

You don't need to use a regex:

def foo(s, pos, chars='?|!'):
    for i in range(pos - 1, -1, -1):
        if s[i] == ' ':
            continue
        return s[i] in chars
    return False

print(foo('blah? test', pos=6))

If you had to use a regex:

def foo(s, pos, chars='?|!'):
    l = re.findall(r'[^ ]', s[:pos]) # find all non-blank characters in first pos - 1 characters
    if not l:
        return False
    return l[-1] in chars

Answer 5

You don't need regex here. Remove spaces at the end of a slice, if they are, and compare the last char

def is_matched(text, pos, chars='?|!'): 
   return text[:pos].rstrip()[-1] in chars

is_matched('blah? test', pos=6) #True
is_matched('blah? test', pos=7) #False