zip list with different length

[s99999999s2003]

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

[ginstrom]

On Apr 4, 4:53 pm, [e-mail] 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( len(a), len(b) )
results = zip( a[:common], b[:common] )
if len( a ) < len( b ):
a = b
return results + [ (x,) for x in a[common:] ]

>>superZip( range( 5 ), range( 3 ) )

[(0, 0), (1, 1), (2, 2), (3,), (4,)]

>>superZip( range( 3 ), range( 5 ) )

[(0, 0), (1, 1), (2, 2), (3,), (4,)]

>>superZip( range( 0 ), range( 5 ) )

[(0,), (1,), (2,), (3,), (4,)]

>>superZip( range( 3 ), range( 3 ) )

[(0, 0), (1, 1), (2, 2)]

Regards,
Ryan Ginstrom

[MC]

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)))

--
@-salutations

Michel Claveau

[Peter Otten]

s9************@yahoo.com 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 be done?
thanks

from itertools import izip, chain, repeat, takewhile, starmap

def zip_longest(*seqs):
padded = [chain(izip(s), repeat(())) for s in seqs]
return takewhile(bool, starmap(sum, izip(izip(*padded), repeat(()))))

Just to bring itertools to your attention :-)

Peter

[Alexander Schmolck]

s9************@yahoo.com writes:

Chi

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?

map(lambda *t: filter(lambda x: x is not None,t),a,b)

'as

[Alexander Schmolck]

MC <XX*****@XX.XmclaveauX.comwrites:

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