Regex to find variants of “Google”
$begingroup$
From the HackerRank question definition:
The word google can be spelled in many different ways.
E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...
Because
g = G
o = O = 0 = () = = <>
l = L = I
e = E = 3
That's the problem here to solve.
And the match has to be only this single word, nothing more, nothing
less.
E.g.
"g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
False, "hey google" = False
What I'm looking in review: How can I make my code more pythonic?
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import re
g = ["g", "G"]
o = ["o", "O", "0", "()", "", "<>"]
l = ["l", "L", "I"]
e = ["e", "E", "3"]
regex = (g, o, o, g, l, e)
regex = ((re.escape(y) for y in x) for x in regex)
regex = ("(?:{})".format("|".join(x)) for x in regex)
regex = "^{}$".format("".join(regex))
print(bool(re.match(regex, input())))
python python-3.x programming-challenge regex
edited 8 hours ago
200_success
129k15153416
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
$endgroup$
add a comment |
$begingroup$
From the HackerRank question definition:
The word google can be spelled in many different ways.
E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...
Because
g = G
o = O = 0 = () = = <>
l = L = I
e = E = 3
That's the problem here to solve.
And the match has to be only this single word, nothing more, nothing
less.
E.g.
"g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
False, "hey google" = False
What I'm looking in review: How can I make my code more pythonic?
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import re
g = ["g", "G"]
o = ["o", "O", "0", "()", "", "<>"]
l = ["l", "L", "I"]
e = ["e", "E", "3"]
regex = (g, o, o, g, l, e)
regex = ((re.escape(y) for y in x) for x in regex)
regex = ("(?:{})".format("|".join(x)) for x in regex)
regex = "^{}$".format("".join(regex))
print(bool(re.match(regex, input())))
python python-3.x programming-challenge regex
edited 8 hours ago
200_success
129k15153416
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
$endgroup$
add a comment |
$begingroup$
From the HackerRank question definition:
The word google can be spelled in many different ways.
E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...
Because
g = G
o = O = 0 = () = = <>
l = L = I
e = E = 3
That's the problem here to solve.
And the match has to be only this single word, nothing more, nothing
less.
E.g.
"g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
False, "hey google" = False
What I'm looking in review: How can I make my code more pythonic?
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import re
g = ["g", "G"]
o = ["o", "O", "0", "()", "", "<>"]
l = ["l", "L", "I"]
e = ["e", "E", "3"]
regex = (g, o, o, g, l, e)
regex = ((re.escape(y) for y in x) for x in regex)
regex = ("(?:{})".format("|".join(x)) for x in regex)
regex = "^{}$".format("".join(regex))
print(bool(re.match(regex, input())))
python python-3.x programming-challenge regex
edited 8 hours ago
200_success
129k15153416
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
$endgroup$
From the HackerRank question definition:
The word google can be spelled in many different ways.
E.g. google, g00gle, g0oGle, g<>0gl3, googl3, GooGIe etc...
Because
g = G
o = O = 0 = () = = <>
l = L = I
e = E = 3
That's the problem here to solve.
And the match has to be only this single word, nothing more, nothing
less.
E.g.
"g00gle" = True, "g00gle " = False, "g google" = False, "GGOOGLE" =
False, "hey google" = False
What I'm looking in review: How can I make my code more pythonic?
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import re
g = ["g", "G"]
o = ["o", "O", "0", "()", "", "<>"]
l = ["l", "L", "I"]
e = ["e", "E", "3"]
regex = (g, o, o, g, l, e)
regex = ((re.escape(y) for y in x) for x in regex)
regex = ("(?:{})".format("|".join(x)) for x in regex)
regex = "^{}$".format("".join(regex))
print(bool(re.match(regex, input())))
python python-3.x programming-challenge regex
python python-3.x programming-challenge regex
edited 8 hours ago
200_success
129k15153416
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
edited 8 hours ago
200_success
129k15153416
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
edited 8 hours ago
200_success
129k15153416
edited 8 hours ago
200_success
129k15153416
edited 8 hours ago
200_success
129k15153416
129k15153416
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
asked 8 hours ago
422_unprocessable_entity422_unprocessable_entity
1,88231750
1,88231750
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.
The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.
Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).
import re
def regex_options(options):
options = (re.escape(o) for o in options)
return "(?:{})".format("|".join(options))
g = regex_options(["g", "G"])
o = regex_options(["o", "O", "0", "()", "", "<>"])
l = regex_options(["l", "L", "I"])
e = regex_options(["e", "E", "3"])
regex = "^{}$".format("".join((g, o, o, g, l, e)))
print(bool(re.match(regex, input())))
If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
$endgroup$
2
$begingroup$
I like this solution better, because it doesn't repeatedly redefineregexand change its type.
$endgroup$
– 200_success
7 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.
The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.
Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).
import re
def regex_options(options):
options = (re.escape(o) for o in options)
return "(?:{})".format("|".join(options))
g = regex_options(["g", "G"])
o = regex_options(["o", "O", "0", "()", "", "<>"])
l = regex_options(["l", "L", "I"])
e = regex_options(["e", "E", "3"])
regex = "^{}$".format("".join((g, o, o, g, l, e)))
print(bool(re.match(regex, input())))
If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
$endgroup$
2
$begingroup$
I like this solution better, because it doesn't repeatedly redefineregexand change its type.
$endgroup$
– 200_success
7 hours ago
add a comment |
$begingroup$
Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.
The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.
Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).
import re
def regex_options(options):
options = (re.escape(o) for o in options)
return "(?:{})".format("|".join(options))
g = regex_options(["g", "G"])
o = regex_options(["o", "O", "0", "()", "", "<>"])
l = regex_options(["l", "L", "I"])
e = regex_options(["e", "E", "3"])
regex = "^{}$".format("".join((g, o, o, g, l, e)))
print(bool(re.match(regex, input())))
If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
$endgroup$
2
$begingroup$
I like this solution better, because it doesn't repeatedly redefineregexand change its type.
$endgroup$
– 200_success
7 hours ago
add a comment |
$begingroup$
Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.
The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.
Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).
import re
def regex_options(options):
options = (re.escape(o) for o in options)
return "(?:{})".format("|".join(options))
g = regex_options(["g", "G"])
o = regex_options(["o", "O", "0", "()", "", "<>"])
l = regex_options(["l", "L", "I"])
e = regex_options(["e", "E", "3"])
regex = "^{}$".format("".join((g, o, o, g, l, e)))
print(bool(re.match(regex, input())))
If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
$endgroup$
Honestly there's not much code to comment on, it's also a programming challenge, so there's not much incentive to make it any better than it is. I too would have done it your way.
The only objective criticism I have is I'd just add a function regex_options to build the non-capturing group.
Other than that I'd apply this to the creation of g, o, l and e. As I think it's a little cleaner. But you may want to perform it before "".join, and after regex = (g, o, o, g, l, e).
import re
def regex_options(options):
options = (re.escape(o) for o in options)
return "(?:{})".format("|".join(options))
g = regex_options(["g", "G"])
o = regex_options(["o", "O", "0", "()", "", "<>"])
l = regex_options(["l", "L", "I"])
e = regex_options(["e", "E", "3"])
regex = "^{}$".format("".join((g, o, o, g, l, e)))
print(bool(re.match(regex, input())))
If this were professional code, I'd suggest using an if __name__ == '__main__': block, and possibly a main function.
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
answered 7 hours ago
PeilonrayzPeilonrayz
25.5k337107
25.5k337107
2
$begingroup$
I like this solution better, because it doesn't repeatedly redefineregexand change its type.
$endgroup$
– 200_success
7 hours ago
add a comment |
2
$begingroup$
I like this solution better, because it doesn't repeatedly redefineregexand change its type.
$endgroup$
– 200_success
7 hours ago
2
2
$begingroup$
I like this solution better, because it doesn't repeatedly redefine
regex and change its type.$endgroup$
– 200_success
7 hours ago
$begingroup$
I like this solution better, because it doesn't repeatedly redefine
regex and change its type.$endgroup$
– 200_success
7 hours ago
add a comment |
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