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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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 ):
common = min(
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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[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) , (3) , (4) ]
can it
[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, 1), (2, 2) , (3) , (4)
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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