Can this be written as a python reduce function?

Can you make this more pytho开发者_如何学Cnic by using the map and/or reduce functions? it just sums the products of every consecutive pair of numbers.

topo = (14,10,6,7,23,6)
result = 0
for i in range(len(topo)-1):
    result += topo[i]*topo[i+1]

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This is the nicest way I can think of:

import operator
sum(map(operator.mul, topo[:-1], topo[1:]))

Edit: I've just found out there's a better way to do this:

import operator
import itertools

def pairwise(iterable):
    a, b = itertools.tee(iterable)
    next(b, None)
    return a, b

def sum_products(l):
    return sum(itertools.imap(operator.mul, *pairwise(l)))

Credit for pairwise function goes to the itertools documentation.

This is faster and uses less memory. Of course, it's less concise.

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You can use map and reduce like this, but I'm not convinced it's more pythonic:

reduce( lambda x, y: x + y, map( lambda x, y: x * y, topo[:-1], topo[1:]) )

Probably simpler is this sum + generator expression:

sum(topo[x] * topo[x+1] for x in xrange(len(topo)-1))

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This works:

mult = lambda (x, y): x * y
pairs = zip(list(topo), list(topo)[1:])
result = sum(map(mult, pairs))

but is probably harder to understand.

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Instead of map using a list comprehension should work:

>>> topo = (14,10,6,7,23,6)
>>> sum((x*y for x,y in zip(topo[:-1],topo[1:])))
541
>>> 

or

>>> sum((topo[i]*topo[i+1] for i in range(len(topo)-1)))
541

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I wouldn't call this pythonic, though it looks cooler, reduce doesn't fit in here:

def func(first, *rest):
    return reduce(lambda (x,y),z:(x+y*z,z), rest, (0,first))[0]

Note the usage of (x,y),z is 2.x only.

---

With reduce and Python < 3.x:

from itertools import tee, izip

#recipe from http://docs.python.org/library/itertools.html#recipes
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

reduce(lambda s, (x, y):s + x * y, pairwise(topo), 0)

with map:

from operator import mul
from itertools import tee

a, b = tee(topo)
next(b, None)

sum(map(mul, a, b))

---

this can also get your answer

a= [14,10,6,7,23,6]
reduce(lambda a,b: a+b,  map(lambda (x,y): x*y, map(lambda i:(a[i],a[i+1]), range(len(a)-1))  ) )