Python 3.3 functions on pairs in a list - list

Question

I am trying to create a program that will find the difference between all pairs in a list. For example
[2,4,6]
Would then make a list containing the difference
[2,2]
Is there a way to do this

Answer 1

Itertools Recipes: pairwise
from itertools import tee
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return zip(a, b)
def diffs(iterable):
return [b - a for a, b in pairwise(iterable)]
print(diffs([2,4,6]))

Answer 2

[L[i+1] - L[i] for i in range(len(L)-1)] will do it.
Some other ways also using a list comprehension:
[L[i+1] - L[i] for i in range(len(L[:-1]))]
[L[i] - L[i-1] for i in range(1, len(L[1:]))]
Using map:
list(map(lambda i: L[i+1]-L[i], range(len(L[:-1]))))
list(map(lambda i: L[i]-L[i-1], range(1, len(L[1:]))))
Using map and the operator module:
list(map(operator.sub, L[1:], L[:-1]))
Using zip (this one is probably the nicest way, imo):
[x - y for x, y in zip(L[1:], L[:-1])]
A more verbose approach if you aren't familiar with list comprehensions or with map (GET FAMILIAR!):
def differences(L1,L2):
L = []
for V1,V2 in zip(L1,L2):
L.append(V2-V1)
return L
diffs = differences(L[:-1],L[1:])
And a similar, but much better way to do it using a generator:
def differences(L1,L2):
for V1,V2 in zip(L1,L2):
yield V2-V1
diffs = list(differences(L[:-1],L[1:]))
And here is the generator comprehension equivalent of the above generator(notice it's almost exactly the same as the last list comprehension above, except it uses the list function instead of brackets):
list(V2-V1 for V1,V2 in zip(L[:-1],L[1:]))
Study all of these ways of doing it very closely and you will learn a lot of Python.