simultaneous assignment

[John Salerno]

Re: simultaneous assignment

bruno at modulix wrote:[color=blue]
> John Salerno wrote:[color=green]
>> Bruno Desthuilliers wrote:
>>[color=darkred]
>>> But my question (sorry, it may not have been clear) was more along the
>>> line of : "why do you worry about identity in the given snippet ?".[/color]
>>
>> Actually, I kind of thought that maybe it *didn't* matter in this
>> particular example anyway, so my question was meant to be more general
>> than I might have written it. It seems like the identity issue can be a
>> problem in certain cases, although I can't think of a case right now! :)[/color]
>
> something
> a = b = []
> a.append(1)
> print b
>
>[/color]

Ah ha!

[Edward Elliott]

Re: simultaneous assignment

Steve R. Hastings wrote:[color=blue]
> You could also use a function that counts all different values in a list,
> reducing the list to a dictionary whose keys are the unique values from
> the list.[/color]

Wouldn't reducing to a set instead of a dict make more sense if all you want
to do is count uniq elements?

[Steve R. Hastings]

Re: simultaneous assignment

On Wed, 03 May 2006 17:51:03 +0000, Edward Elliott wrote:
[color=blue]
> Steve R. Hastings wrote:[color=green]
>> You could also use a function that counts all different values in a list,
>> reducing the list to a dictionary whose keys are the unique values from
>> the list.[/color]
>
> Wouldn't reducing to a set instead of a dict make more sense if all you want
> to do is count uniq elements?[/color]

My apologies for not explaining tally() better.

The dict has one key for each unique element, and the value associated
with each key is a count of how many times that element appeared in the
original list.

lst = ['a', 'b', 'b', 'c', 'c', 'c']
d = iterwrap.tally( lst)
print d # prints something like: {'a': 1, 'c': 3, 'b': 2}


If you didn't care how many times the values appeared in the original
list, and just just wanted the unique values, then a set would be perfect.

If you happen to have tally(), it is an easy way to solve the original
problem: figure out whether a list has exactly one true value in it.

d = tally(bool(x) for x in lst)
if d[True] == 1:
print "and there was much rejoicing"

--
Steve R. Hastings "Vita est"
steve@hastings. org http://www.blarg.net/~steveha

[Dave Hansen]

Re: simultaneous assignment

On Tue, 02 May 2006 18:52:48 GMT in comp.lang.pytho n, John Salerno
<johnjsal@NOSPA Mgmail.com> wrote:

[...][color=blue]
>
>Yeah, after trying some crazy things, I just wrote it this way:
>
>def truth_test(seq) :
> truth = 0
> for item in seq:
> if item:
> truth += 1
> if truth == 1:
> return True
> else:
> return False[/color]

You could replace those last four lines with

return truth == 1
[color=blue]
>
>Not sure I like having to keep a counter though, but the other stuff I[/color]

Well, if you want something minimalist, you could try

def truth_test(seq) :
return sum(1 for item in seq if item) == 1

Though I'm not sure it's really any clearer...
[color=blue]
>did was really convoluted, like checking to see if the first item was
>True, and if it was, popping it from the list and iterating over the
>rest of the items (needless to say, the in-place change wasn't helpful).[/color]

Perhaps something like

def truth_test(seq) :
found = False
for item in seq:
if item:
if found:
return False
found = True
return found

Gets you an early exit, anyway...

All code untested. Regards,
-=Dave

--
Change is inevitable, progress is not.

[Paul Rubin]

Re: simultaneous assignment

Dave Hansen <iddw@hotmail.c om> writes:[color=blue]
> Well, if you want something minimalist, you could try
>
> def truth_test(seq) :
> return sum(1 for item in seq if item) == 1[/color]

def truth_test(seq) :
return sum(map(bool, seq)) == 1

[Boris Borcic]

Re: simultaneous assignment

Steve R. Hastings wrote:
[color=blue]
> You could also use a function that counts all different values in a list,
> reducing the list to a dictionary whose keys are the unique values from
> the list. I got the idea from a discussion here on comp.lang.pytho n; I
> called my version of it tally().
>
> d = tally(bool(x) for x in seq)
> print d[True] # prints how many true values in seq
> print d[False] # prints how many false values in seq
>
>
> tally() is in my iterwrap.py module, which you can get here:
>
> http://home.blarg.net/~steveha/iterwrap.tar.gz
>[/color]
[color=blue][color=green][color=darkred]
>>> from itertools import groupby
>>> tally = lambda it : dict((x,sum(1 for _ in y)) for x,y in groupby(sorted( it)))
>>> tally('abbcabbc ca')[/color][/color][/color]
{'a': 3, 'c': 3, 'b': 4}