New regex module fuzzy function error value. Python

Question

im trying out the fuzzy function of the new regex module. in this case, i want there to find a match for all strings with <= 1 errors, but i'm having trouble with it

import regex

statement = 'eol the dark elf'
test_1 = 'the dark'
test_2 = 'the darc' 
test_3 = 'the black'

print regex.search('{}'.format(test_1),statement).group(0) #works

>>> 'the dark' 

print regex.search('{}'.format(test_1){e<=1},statement).group(0)

>>> print regex.search('{}'.format(test_1){e<=1},statement).group(0) #doesn't work 
                                          ^
SyntaxError: invalid syntax

i have also tried

print regex.search('(?:drk){e<=1}',statement).group(0) #works
>>> 'dark'

but this . . .

print regex.search(('(?:{}){e<=1}'.format(test_1)),statement).group(0) #doesn't work
>>> SyntaxError: invalid syntax

after serach you need only one (. It should be so: print regex.search('(?:{}){e<=1}'.format(test_1)).group(0) — TobiMarg
– TobiMarg, Commented Jul 3, 2013 at 19:08

Martin Ender · Accepted Answer · 2013-07-03 20:42:02Z

1

In your first snippet, you forgot to put the {e<=1} in a string. In your final snippet, I think the problem is, that format tries to deal with the {e<=1} itself. So either you use concatenation:

print regex.search(test_1 + '{e<=1}', statement).group(0)

or you escape the literal braces, by doubling them:

print regex.search('{}{{e<=1}}'.format(test_1), statement).group(0)

This can then easily be extended to

print regex.search('{}{{e<={}}}'.format(test_1, num_of_errors), statement).group(0)

edited Jul 3, 2013 at 20:42

answered Jul 3, 2013 at 19:28

Martin Ender

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2 Comments

O.rka Over a year ago

i'm a little confused on how to make the error number a variable. lets say error = 2 ... it doesn't work when i try '{e<={}}'.format(error)

Martin Ender Over a year ago

@draconisthe0ry I just realized, literal braces can be escaped by doubling them. I'll edit the answer

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