comparing sub-list elements in list using python [duplicate]

Question

i have a list a=[[1,2],[3,4],[5,6]] and i need to check if all elements in sublist are in ascending order (e.g [1,2] is less than [3,4] and [5,6], and [3,4] is less than [5,6] and so on). i use the following function:

def FirstRuleLink (L):
    for i in range(0,len(L)):
        for j in range(0,len(L[i])):
            if L[i][0]<L[i+1][0] and L[i][1]<L[i+1][1]:
                return True
            else:
                return False

but the python gives me error message that index is out of range. so how could i change this code to get the correct output.

how do you think comparing [1, 2] to [3, 4] should work? — Ma0
– Ma0, Commented May 30, 2018 at 11:11
@p.deman right, it may or may not work depending on what OP wants. — hilberts_drinking_problem
– hilberts_drinking_problem, Commented May 30, 2018 at 11:13

Vikas Periyadath · Accepted Answer · 2018-05-30 11:37:23Z

0

Here a try :

a = [[1,2],[3,4],[5,6]]

def FirstRuleLink (a):
    for i,nxt_elmnt in zip(a, a[1:]):
        if i[0] <= nxt_elmnt[0] and i[1] <= nxt_elmnt[1]:
            pass
        else:
            return False

edited May 30, 2018 at 11:37

answered May 30, 2018 at 11:14

Vikas Periyadath

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2 Comments

Vikas Periyadath Over a year ago

@YakymPirozhenko thanks for finding mistakes, please take a look and let me know how I can improve it.

hilberts_drinking_problem Over a year ago

I would go with all(list(t) == sorted(t) for t in zip(*a)). Both this and your solution differs from OPs intention of using strict inequalities. Also, you probably mean to return True after the for loop.

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