python regex: match a string with only one instance of a character

Question

Suppose there are two strings:

$1 off delicious ham.
$1 off delicious $5 ham.

In Python, can I have a regex that matches when there is only one $ in the string? I.e., I want the RE to match on the first phrase, but not on the second. I tried something like:

re.search(r"\$[0-9]+.*!(\$)","$1 off delicious $5 ham.")

..saying "Match where you see a $ followed by anything EXCEPT for another $." There was no match on the $$ example, but there was also no match on the $ example.

Thanks in advance!

Simple test method for checking:

def test(r):
  s = ("$1 off $5 delicious ham","$1 off any delicious ham")    
  for x in s:
    print x
    print re.search(r,x,re.I)
    print ""

I'm shoe-horning an RE into a tool that doesn't let me add any additional logic. So I can't do "if count(str,"$") > 1: pass". I need the re to just not match on such a string. — Chris, Jul 02 '10 at 14:52

MattH · Accepted Answer · 2010-07-02T14:49:42.263

12

>>> import re
>>> onedollar = re.compile(r'^[^\$]*\$[^\$]*$')
>>> onedollar.match('$1 off delicious ham.')
<_sre.SRE_Match object at 0x7fe253c9c4a8>
>>> onedollar.match('$1 off delicious $5 ham.')
>>>

Breakdown of regexp:
^ Anchor at start of string
[^\$]* Zero or more characters that are not $
\$ Match a dollar sign
[^\$]* Zero or more characters that are not $
$ Anchor at end of string

edited Jul 02 '10 at 14:49

answered Jul 02 '10 at 14:38

MattH

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Bonus points for the show-n-tell. Do you have a couple minutes to describe to me what's going on, here? – Chris Jul 02 '10 at 14:43
0 or more non dollars sign followed by one dollar followed by 0 or more non dollar sign – Xavier Combelle Jul 02 '10 at 14:52
1

You don't need to escape the `$` inside a character class. `r'^[^$]*\$[^$]*$'` works just as well. – Tim Pietzcker Jul 02 '10 at 18:35

score 7 · Answer 2 · answered Jul 02 '10 at 14:44

7

>>> '$1 off delicious $5 ham.'.count('$')
2
>>> '$1 off delicious ham.'.count('$')
1

answered Jul 02 '10 at 14:44

SilentGhost

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I agree, a regular expression would be overkill here. – Duncan Jul 02 '10 at 14:55

score 2 · Answer 3 · answered Jul 02 '10 at 14:42

2

You want to use the complement of a character class [^] to match any character other than $:

re.match(r"\$[0-9]+[^\$]*$","$1 off delicious $5 ham.")

The changes from your original are as follows:

.* replaced with [^\$]*. The new term [^\$] means any character other than $
$ appended to string. Forces the match to extend to the end of the string.
re.search replaced with re.match. Matches the whole string, rather than any subset of it.

answered Jul 02 '10 at 14:42

fmark

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Didn't pass! >>> test(r"\$[0-9]+[^\$]*$") $1 off $5 delicious ham <_sre.SRE_Match object at 0x00ED65D0> $1 off any delicious ham <_sre.SRE_Match object at 0x00ED65D0> – Chris Jul 02 '10 at 14:56

score 1 · Answer 4 · answered Jul 02 '10 at 14:39

1

re.search("^[^$]*\$[^$]*$",test_string)

answered Jul 02 '10 at 14:39

John La Rooy

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score 1 · Answer 5 · answered Jul 02 '10 at 14:40

1

^.*?\$[^$]*$

this should make the trick

answered Jul 02 '10 at 14:40

dweeves

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nice use of a non-greedy match – a'r Jul 02 '10 at 14:42

Hmm.. Didn't pass the test, though, unless there are some special flags I need:

    >>> test(r"^.*?\$[^$]*$")
    $1 off $5 delicious ham
    <_sre.SRE_Match object at 0x00ED6288>
    
    $1 off any delicious ham
    <_sre.SRE_Match object at 0x00ED6288>

– Chris Jul 02 '10 at 14:51

python regex: match a string with only one instance of a character

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