On Sat, 05 Feb 2011 14:38:29 +0100, Daniel Urban wrote:
> On Sat, Feb 5, 2011 at 14:08, Lisa Fritz Barry Griffin
> <lisaochba...@gmail.com> wrote:
>> Hi there,
>>
>> How can I do this in a one liner:
>>
>> maxCountPerPhraseWordLength = {}
>> for i in range(1,MAX_PHRASES_LENGTH+1):
>> maxCountPerPhraseWordLength[i] = 0
>
> maxCountPerPhraseWordLength = {0 for i in range(1,MAX_PHRASES_LENGTH+1)}
Unfortunately not. In versions before Python 2.7, it will give a
SyntaxError. In 2.7 or better it gives a set instead of a dict:
>>> MAX_PHRASES_LENGTH = 5
>>> maxCountPerPhraseWordLength={0 for i in range(1,MAX_PHRASES_LENGTH+1)}
>>> maxCountPerPhraseWordLength
set([0])
This will do the job most simply:
>>> dict(zip(range(1, MAX_PHRASES_LENGTH+1), [0]*MAX_PHRASES_LENGTH))
{1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
Or this:
>>> dict((i, 0) for i in range(1, MAX_PHRASES_LENGTH+1))
{1: 0, 2: 0, 3: 0, 4: 0, 5: 0}
But perhaps the most simple solution is to just leave the dict empty, and
instead of fetching items with dict[i], use dict.get(i, 0) or
dict.setdefault(i, 0).
--
Steven
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