Re: Unpacking sequences and keywords in one function call
ram wrote:
> Stupid question #983098403:
>
> I can't seem to pass an unpacked sequence and keyword arguments to a
> function at the same time. What am I doing wrong?
>
> def f(*args, **kw):
> for a in args:
> print 'arg:', a
> for (k,v) in kw.iteritems():
> print k, '=', v
>
f(1,2)
> arg: 1
> arg: 2
>
f(*[1,2])
> arg: 1
> arg: 2
>
f(1,2, a=1)
> arg: 1
> arg: 2
> a = 1
>
f(*[1,2], a=1)
> File "", line 1
> f(*[1,2], a=1)
> ^
> SyntaxError: invalid syntax
>
> Thanks,
> Rick
>
I don't know if it's because there's some potential ambiguity (that I'm
not seeing), but yeah, you just can't do that. This should work though...
>>> f(*[1, 2], **{'a':1})
noah
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Re: re.match -- not greedy?
Carl Banks wrote:
> EXI-Andrews, Jack wrote:
>> the '*' and '+' don't seem to be greedy.. they will consume less in
>> order to match a string:
>>
> import re;re.match('(a+)(ab)','aaab').groups()
>> ('aa', 'ab')
>>
>> this is the sort of behaviour i'd expect from
>>'(a+?)(ab)'
>>
>> a+ should greedily consume a's at the expense of the string not matching
>
> It's called backtracking. If a match fails, the regexp engine
> recursively unwinds some of the greedy matching and tries again.
>
>> in real life, i am trying to match #defines:
>>
> re.match( '#define ([A-Za-z0-9_]+)([^(])', '#define
>> abc(d)').groups()
>> ('ab', 'c')
>>
>> i want this example to fail because the first character after a string
>> of letters is a '('
>> i want to match only #defines without parameters.
>
> Probably the easiest thing to do is to attempt to match zero or one
> opening parentheses, and bail out if it did end up matching the
> parenthesis.
>
> m = re.match(r"#define ([A-Za-z0-9_]+)(\(?)","#define abc(d)")
> if m and m.groups(2) != "(":
> ...
>
> (BTW, it's good practice to use a raw strings for the regular
> expressions as I did.)
>
Another way that seems clearer to me is to use negative lookahead
assertions.
>>>defPatt = re.compile(r'#define\s+(\w+)\b(?!\()')
>>> defPatt.match("#define abc(x)")
>>> defPatt.match("#define foo_BarBaz")
<_sre.SRE_Match object at 0xb7dd5820>
>>> defPatt.match("#define foo_BarBaz").groups()
('foo_BarBaz',)
In general '\w' is the same as [A-Za-z0-9_]
There are other considerations too... I don't know if
#define abc (x)
But the main thing here is the use of '\b' to require there to be a word
boundary at the end your match and a (?!\() to require that the match is
not followed by a '('
see http://docs.python.org/lib/re-syntax.html
noah
>> so what's the definition of greedy?
>
> It means match as much you can *while getting a match*.
>
> Just remember regexp parsing is not a one way street: if it hits a dead
> end, it'll go back and try another path. In fact, greediness usually
> doesn't affect *whether* a regexp matches; it only affects the
> groupings. I'm suddenly curious if there are any cases at all where
> greediness changes whether it finds a match.
>
>> (pls copy me on responses)
>
> Ah, too bad you won't see it then
>
>
> Carl Banks
>
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Re: Is time.time() < time.time() always true?
Ben Finney wrote: > Really? Where does Python guarantee that the left side *must* be > evaluated before the right side of a comparison? (If the right side > were to be evaluated first, the left might end up with a greater > value.) > http://docs.python.org/ref/evalorder.html -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to align words?
Robert R. wrote:
> Hello,
>
> i would like to write a piece of code to help me to align some sequence
> of words and suggest me the ordered common subwords of them
>
> s0 = "this is an example of a thing i would like to have".split()
> s1 = "another example of something else i would like to have".split()
> s2 = 'and this is another " example " but of something ; now i would
> still like to have'.split()
> ...
> alist = (s0, s1, s2)
>
> result should be : ('example', 'of', 'i', 'would', 'like', 'to', 'have'
>
> but i do not know how should i start, may be have you a helpful
> suggestion?
> a trouble i have if when having many different strings my results tend
> to be nothing while i still would like to have one of the, or maybe,
> all the best matches.
>
> best.
>
Your requirements are a little vague... how are these three strings handled?
s1 = "hello there dudes"
s2 = "dudes hello there"
s3 = "there dudes hello"
they all share the 3 words, but what order do you want them back?
here is a simplistic approach using sets that results in a list of words
that are in all strings ordered arbitrarily by their order in the first
string ( it also doesn't worry about matches (or lack of) due to
punctuation and case and crap like that)
>>> strList = []
>>> strList.append('this is an example of a thing i would like to have')
>>> strList.append('another example of something else i would like to
have')
>>> strList.append('and this is another " example " but of something ;
now i would still like to have')
>>> [word for word in strList[0].split() if word in reduce(lambda x, y:
x.intersection(y), [set(str.split()) for str in strList])]
['example', 'of', 'i', 'would', 'like', 'to', 'have']
but you still have issues with mutiple matches and how they are handled
etc...
noah
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Re: best way to align words?
Noah Rawlins wrote:
>
> >>> strList = []
> >>> strList.append('this is an example of a thing i would like to have')
> >>> strList.append('another example of something else i would like to
> have')
> >>> strList.append('and this is another " example " but of something ;
> now i would still like to have')
> >>> [word for word in strList[0].split() if word in reduce(lambda x, y:
> x.intersection(y), [set(str.split()) for str in strList])]
> ['example', 'of', 'i', 'would', 'like', 'to', 'have']
>
I think that ends up doing the set reduction over and over for every
word in the first string, so you probably want to move that outside the
list comprehension
noah
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