Clever Geek Handbook

Multiply Adjacent Elements

I have a set of integers like (1, 2, 3, 4, 5)

and I want to create a tuple (1*2, 2*3, 3*4, 4*5)

by multiplying adjacent elements. Can you do this with a single layer?

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5 answers


Short and sweet. Remember that zip

only works up to the shortest input.



print tuple(x*y for x,y in zip(t,t[1:]))
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>>> t = (1, 2, 3, 4, 5)
>>> print tuple(t[i]*t[i+1] for i in range(len(t)-1))
(2, 6, 12, 20)


Not the most pythonic solutions though.

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If t is your tuple:

>>> tuple(t[x]*t[x+1] for x in range(len(t)-1))
(2, 6, 12, 20)

And another solution with a lovely map:

>>> tuple(map(lambda x,y:x*y, t[1:], t[:-1]))
(2, 6, 12, 20)

Edit: And if you're worried about extra chunk memory, you can use islice from itertools, which will iterate over your tuple (thanks @eyquem):

>>> tuple(map(lambda x,y:x*y, islice(t, 1, None), islice(t, 0, len(t)-1)))
(2, 6, 12, 20)
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tu = (1, 2, 3, 4, 5)

it = iter(tu).next
it()
print tuple(a*it() for a in tu)

I have timed various codes:

from random import choice
from time import clock
from itertools import izip

tu = tuple(choice(range(0,87)) for i in xrange(2000))

A,B,C,D = [],[],[],[]

for n in xrange(50):

    rentime = 100

    te = clock()
    for ren in xrange(rentime): # indexing
        tuple(tu[x]*tu[x+1] for x in range(len(tu)-1))
    A.append(clock()-te)

    te = clock()
    for ren in xrange(rentime): # zip
        tuple(x*y for x,y in zip(tu,tu[1:]))
    B.append(clock()-te)

    te = clock()
    for ren in xrange(rentime): #i ter
        it = iter(tu).next
        it()
        tuple(a*it() for a in tu)
    C.append(clock()-te)

    te = clock()
    for ren in xrange(rentime): # izip
        tuple(x*y for x,y in izip(tu,tu[1:]))
    D.append(clock()-te)


print 'indexing ',min(A)
print 'zip      ',min(B)
print 'iter     ',min(C)
print 'izip     ',min(D)

result

indexing  0.135054036197
zip       0.134594201218
iter      0.100380634969
izip      0.0923947037962

izip is better than zip: - 31%

My solution is not that bad (I don't think so by the way): -25% relative to zip, 10% longer than izip champion

I'm surprised indexing is not faster than zip: nneonneo is right, zip is acceptable

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I like recipes fromitertools

:

from itertools import izip, tee

def pairwise(iterable):
    xs, ys = tee(iterable)
    next(ys)
    return izip(xs, ys)

print [a * b for a, b in pairwise(range(10))]

Result:

[0, 2, 6, 12, 20, 30, 42, 56, 72]
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