Python: pairwise comparison between two lists: list a >= list b?

Question

If I want to like to check if all the elements in a list

a: [1 2 3 6]

are greater than or equal to their corresponding elements in another list

b: [0 2 3 5]

If a[ i ] > b[ i ] for all i 's, then return true, otherwise false. Is there a logical function for this? such as a>b? Thanks

Answer 1

You could just do

all(x >= y for x,y in zip(a,b))

which has the advantage of short-circuit evaluation: if it finds any x < y it returns False immediately.

Answer 2

If you actually want to compare every element in a against b you actually just need to check against the max of b so it will be an 0(n) solution short circuiting if we find any element less than the max of b:

mx = max(b)
print(all(x >= mx for x in a))

For pairwise you can use enumerate:

print(all(x >= b[ind] for ind,x in enumerate(a)))

Or using hughbothwell's zip idea use itertools.zip:

from itertools import izip
print(all(x >= y for x,y  in izip(a,b)))

Or numpy:

print(np.greater_equal(a,b).all())

print(np.all(a >= b))

Answer 3

You can write out an explicit for loop, or you can do an inline double loop and an all as shown below with the interpreter.

>>> a = {1,2,3,4}
>>> b = {2,3,4,5}
>>> lst = [x>y for x in a for y in b]
>>> lst
[False, False, False, False, False, False, False, False, True, False, False, False, True, True, False,False]
>>> all(bool==True for bool in lst)
False

Answer 4

If you are willing to use numpy, you can do this with a logic function (and it's much faster than pure python list comparisons)

>>> from numpy import array
>>> a=array([1,2,3,4,5])
>>> b=array([3,0,3,1,2])
>>> a > b
array([False,  True, False,  True,  True], dtype=bool)
>>> 

Answer 5

How about this? Do I win the code-golf?

a=[1,3,8,6]
b=[0,2,3,5]

print all(map(cmp,a,b))

Answer 6

a = [1, 2, 3, 6]
b = [0, 2, 3, 5]

def check_lists(a, b):
    result = []

    for i in range(len(a)):
        result.append(a[i] >= b[i])

    return result

print check_lists(a, b)