MC <[EMAIL PROTECTED]> writes:
> Hi!
>
> Brutal, not exact answer, but:
>
> a = range(5)
> b = range(3)
> print zip(a+[None]*(len(b)-len(a)),b+[None]*(len(a)-len(b)))
You reinvented map(None,a,b).
'as
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[EMAIL PROTECTED] writes:
C> hi
> suppose i have 2 lists, a, b then have different number of elements,
> say len(a) = 5, len(b) = 3
> >>> a = range(5)
> >>> b = range(3)
> >>> zip(b,a)
> [(0, 0), (1, 1), (2, 2)]
> >>> zip(a,b)
> [(0, 0), (1, 1), (2, 2)]
>
> I want the results to be
> [(0, 0), (1,
[EMAIL PROTECTED] wrote:
> suppose i have 2 lists, a, b then have different number of elements,
> say len(a) = 5, len(b) = 3
a = range(5)
b = range(3)
zip(b,a)
> [(0, 0), (1, 1), (2, 2)]
zip(a,b)
> [(0, 0), (1, 1), (2, 2)]
>
> I want the results to be
> [(0, 0), (1, 1), (2, 2)
Hi!
Brutal, not exact answer, but:
a = range(5)
b = range(3)
print zip(a+[None]*(len(b)-len(a)),b+[None]*(len(a)-len(b)))
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@-salutations
Michel Claveau
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On Apr 4, 4:53 pm, [EMAIL PROTECTED] wrote:
> elements, say len(a) = 5, len(b) = 3
> >>> a = range(5)
> >>> b = range(3)
...
> I want the results to be
> [(0, 0), (1, 1), (2, 2) , (3) , (4) ]
> can it be done?
A bit cumbersome, but at least shows it's possible:
>>> def superZip( a, b ):
c
hi
suppose i have 2 lists, a, b then have different number of elements,
say len(a) = 5, len(b) = 3
>>> a = range(5)
>>> b = range(3)
>>> zip(b,a)
[(0, 0), (1, 1), (2, 2)]
>>> zip(a,b)
[(0, 0), (1, 1), (2, 2)]
I want the results to be
[(0, 0), (1, 1), (2, 2) , (3) , (4) ]
can it be done?
thanks
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