比较2个不同的列表

Question

for example there are the following 2 list of lists

A = [[0, 1, 2, 1, 9], [1, 0, 0, 6, 0], [2, 0, 0, 15, 2], [1, 6, 15, 0, 7], [9, 0, 2, 7, 0]]

B = [[0, 19, 1, 0, 12], [19, 0, 2, 0, 0], [1, 2, 0, 0, 2], [0, 0, 0, 0, 3], [12, 0, 2, 3, 0]]

so what i want to do is, i want to compare each value in the 2 different list of lists like. i want to compare A[0][0] with B[0][0] but not A[0][1] with B[0][0]. so basically i want to compare like the first value in the first list in A with the respectively value in B and so on. how can i do this? thank you very much :) EDIT: sorry. my lengths of the lists were different. this is the updated version .

Answer 1

If you need to know if they are identical just run this:

list1 = [['first',1], ['second',2], ['third',3]]
list2 = [['first',1], ['second',2], ['third',3]]
print(sorted(list1) == sorted(list2))

Result: true

Answer 2

You could use a double list comprehension to always select the largest value and create a new, flat list:

>>> A = [[0, 1, 2, 1, 9], [1, 0, 0, 6, 0], [2, 0, 0, 15, 2], [1, 6, 15, 0, 7], [9, 0, 2, 7, 0]]
>>> B = [[0, 19, 1, 0, 12], [19, 0, 2, 0, 0], [1, 2, 0, 0, 2], [0, 0, 0, 0, 3], [12, 0, 2, 3, 0]]
>>> [max(a,b) for la,lb in zip(A,B) for a,b in zip(la,lb)]
[0, 19, 2, 1, 12, 19, 0, 2, 6, 0, 2, 2, 0, 15, 2, 1, 6, 15, 0, 7, 12, 0, 2, 7, 0]

If you want to keep the 2-D structure, you can use a nested list comprehension:

>>> [[max(a,b) for a,b in zip(la,lb)] for la,lb in zip(A,B)]
[[0, 19, 2, 1, 12], [19, 0, 2, 6, 0], [2, 2, 0, 15, 2], [1, 6, 15, 0, 7], [12, 0, 2, 7, 0]]

Answer 3

To check if two lists are equal you simply do equality check like this:

A == B

But if you are interested in knowing which elements of the sublists differ or alternatively, which positions or index "coordinates" for which the lists differ then you might want to do this:

A = [[0, 1, 2, 1, 9], [1, 0, 0, 6, 0], [2, 0, 0, 15, 2], [1, 6, 15, 0, 7], [9, 0, 2, 7, 0]]
B = [[0, 19, 1, 0, 12], [19, 0, 2, 0, 0], [1, 2, 0, 0, 2], [0, 0, 0, 0, 3], [12, 0, 2, 3, 0]]

differences = [
    (outer_idx, inner_idx)
    for outer_idx, (a, b) in enumerate(zip(A, B))
    for inner_idx, (a_element, b_element) in enumerate(zip(a, b))
    if a_element != b_element
]

print(differences)

# output
[(0, 1),
 (0, 2),
 (0, 3),
 (0, 4),
 (1, 0),
 (1, 2),
 (1, 3),
 (2, 0),
 (2, 1),
 (2, 3),
 (3, 0),
 (3, 1),
 (3, 2),
 (3, 4),
 (4, 0),
 (4, 3)]

All this means that for (0, 1) the lists are different. ie: A[0][1] != B[0][1] and so on.

Answer 4

#returns True if f is larger than s
def compare(f, s):
    print("comparing {} and {}".format(f,s))
    return f > s

def compare_lists(A,B):
    for sub_lists in zip(A,B):
        for first, second in zip(sub_lists[0], sub_lists[1]):
            compare(first, second)

You haven't specified what kind of compare you want. The other answers assume you want to check for equality. This is a more general approach. Put whatever compare logic you want into compare(f,s)

Edit: Your sublists contain different amounts of elements. This approach only compares according to the shortest sublist. So for [1,2] and [1] , only 1 and 1 is compared, since there is no matching entry for 2 from the first list.

Edit2: You just edited your post. I thought it was intentional that the lenghts were different...

Answer 5

Assuming you want to know if they are identical:

def compare(a,b):
    if (len(a) != len(b) ) return false;
    for i in range(0, len(a)):
       if ( len(a[i]) != len(b[i]) ) return false;
       for j in range(0, len(a[i])):
           if (a[i][j]!=b[i][j]) return false;
    return true;