Check if a list exists in given list of lists

[Daniel94]

I want to check if a list of tuples contains a specific tuple. If

numbers = [('a','a'), ('c','a')]
look = ('a','c')

"look" should not be contained in "numbers". But that's how my code thinks right now.

  def find_edge(self, src, dst):
        """
        Returns True if there's an edge from node `src` to node `dst`.
        """
        # Check if nodes exist
        if self.find_node(src) is True and self.find_node(dst) is True:
            src_node = self.return_head(src)

            src_edges = src_node.list_edges()
            src_edges_weightless = [x[:-1] for x in src_edges]

            possible_edge = (src, dst)

            for elem in src_edges_weightless:
                if collections.Counter(elem) == collections.Counter(possible_edge):
                    return True

        return False

So when looking if ("a","c") is contained in the list, return should be FALSE. However right now it returns TRUE. Any ideas why?

[Daniel94]

Solved it by adding changing to this:

            for elem in src_edges_weightless:
                if collections.Counter(elem) == collections.Counter(possible_edge):
                    if elem[0] == possible_edge[0] and elem[1] == possible_edge[1]:
                        return True

[deanhystad]

This is why your original program did not work:

import collection
x = collections.Counter(['a', 'b'])
y = collections.Counter(['b', 'a'])
print(x)
print(y)
print(x==y)

Output:
Counter({'a': 1, 'b': 1})
Counter({'b': 1, 'a': 1})
True

Counter does not care about order of entries in the source collection.

This is what your function does with all the extra stuff removed:

numbers = [('a','a'), ('c','a')]

def find_edge(nodes, x, y):
    for node in nodes:
        if x == node[0] and y == node[1]:
            return True
    return False

print(find_edge(numbers, 'a', 'c'))
print(find_edge(numbers, 'c', 'a'))

Output:
False
True