I have a function named checker(nums) that has an argument that will later receive a list. What i want to do with that list is to check if each other element is greater or equal to the previous one.
Example:
I have a list [1, 1, 2, 2, 3] and i have to check if it fulfills the condition.
Since it does, the function should return True
My code:
def checker(nums):
for x in range(len(nums)):
if x+1<len(nums):
if nums[x] <= nums[x+1] and nums[-1] >= nums[-2]:
return True
This will only run once and return True if the first condition is true. I've seen a statement if all and am unsure of how to use it.
Your function can be reduced to this:
def checker(nums):
return all(i <= j for i, j in zip(nums, nums[1:]))
Note the following:
zip loops through its arguments in parallel, i.e. nums[0] & nums[1] are retrieved, then nums[1] & nums[2] etc.i <= j performs the actual comparison.() ensures that each value of the condition, i.e. True or False is extracted one at a time. This is called lazy evaluation.all simply checks all the values are True. Again, this is lazy. If one of the values extracted lazily from the generator expression is False, it short-circuits and returns False.To avoid the expense of building a list for the second argument of zip, you can use itertools.islice. This option is particularly useful when your input is an iterator, i.e. it cannot be sliced like a list.
from itertools import islice
def checker(nums):
return all(i <= j for i, j in zip(nums, islice(nums, 1, None)))
Another iterator-friendly option is to use the itertools pairwise recipe, also available via 3rd party more_itertools.pairwise:
# from more_itertools import pairwise # 3rd party library alternative
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 checker(nums):
return all(i <= j for i, j in pairwise(nums))
Another alternative is to use a functional approach instead of a comprehension:
from operator import le
def checker_functional(nums):
return all(map(le, nums, nums[1:]))
Your code can in fact be reduced to checking if nums is sorted, e.g.
def checker(nums):
return sorted(nums) == nums
This does exactly what you expect, e.g.
>>> checker([1, 1, 2, 2, 3])
True
>>> checker([1, 1, 2, 2, 1])
False
Similar solution to @jp_data_analysis using more_itertools.windowed
>>> from more_itertools import windowed
>>> nums = [1, 1, 2, 2, 3]
>>> all(i <= j for i, j in windowed(nums, 2))
True
And for scientific purposes (not recommended code), here is a more functional approach
>>> from operator import le
>>> from itertools import starmap
>>> all(starmap(le, windowed(nums, 2)))
True