CODEWITHSUNDEEP
pythonpython-2.7list-comprehensionpython-itertools
CoMartel
CoMartel

Reputation: 3591

Product with descreasing values

I have a 3 lists :

a = [10, 9, 8, 7, 6]
b = [8, 7, 6, 5, 4, 3]
c = [6, 5, 4, 3, 2]

I need to get all the permutations obtained with itertools.product(), BUT only if the values are decreasing: 

[10, 8, 6] # is good
[6, 8, 4]  # is not good, since 8 > 6

Is there a simple way to do it or should I go with list comprehension and conditions ?

Upvotes: 1

Views: 67

Answers (4)

Shagun Bhatt
Shagun Bhatt

Reputation: 96

You can refer following code which does not have list comprehensions:

from itertools import product
a = [10, 9, 8, 7, 6]
b = [8, 7, 6, 5, 4, 3]
c = [6, 5, 4, 3, 2]
for result in product(a,b,c):
    if sorted(result, reverse = True) == list(result):
            print result

Upvotes: 2

Chris_Rands
Chris_Rands

Reputation: 41218

You could do this with a list comprehension by looping over theitertools.product iterator and extracting only those returned items that are sorted in reverse:

[item for item in product(a,b,c) if sorted(item, reverse = True) == list(item)]

Example:

from itertools import product
a = [10,9,8,7,6]
b = [8, 7, 6, 5, 4, 3]
c = [6, 5, 4, 3, 2]
[item for item in product(a,b,c) if sorted(item, reverse = True) == list(item)]
# [(10, 8, 6), (10, 8, 5), (10, 8, 4), (10, 8, 3), (10, 8, 2) ...continues

Upvotes: 3

Alex Fargus
Alex Fargus

Reputation: 122

If you don't want to use list comprehensions for some reason:

def decreasing(l):
    return all(a >= b for a, b in zip(l[:-1], l[1:]))


filter(decreasing, product(a, b, c))

Upvotes: 0

Tom Fuller
Tom Fuller

Reputation: 5349

This is a simple one line solution

>>> mylist = [10, 9, 8, 7, 6]
>>> all(earlier >= later for earlier, later in zip(mylist, mylist[1:]))
True
>>> mylist = [10, 9, 7, 8, 6]
>>> all(earlier >= later for earlier, later in zip(mylist, mylist[1:]))
False

I found this here:

Determine if a list is in descending order

Upvotes: 0

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