Python - Check if value is unique in dictionary of lists

Deyan 0
Viewed 38

Is there a way to check if a value is unique for a specific list, within a dictionary of lists? If yes, I would like to know in which key/list pair/s the value is present.

For example: {1: [3, 4, 5], 2: [4], 3: [5], 6: [3, 5, 9], 8: [3 ,8]}

I want to check if 3 is present in any of the other lists within the dictionary, besides 1: [3, 4, 5(compare it with all others). Since it is, then I must receive as a result 6 and 8, because these are the keys, which corresponding lists have this value.

dictionary list python
4 Answers

You can enter the value you are looking for and then loop over all the dictionaries to see if its in them.

Take this as pseudo code:

int val = 3

for(keys in mydict):
    if(mydict.key.contains(val())
         print(mydict.key)

You could also save the dictionaries in a list, so that you have a list of dicts with your corresponding value.

DennisM0

Not entirely sure if this is what you meant but:

def unique(list1):

    # initialize a null list:
    unique_list = []

    # traverse for all elements:
    for x in list1:
        # check if exists in unique_list or not
        if x not in unique_list:
            unique_list.append(x)
    # printing the list:
    for x in unique_list:
        print(x, end=' ')


list1 = [10, 20, 10, 30, 40, 40]
print("the unique values from 1st list is")
unique(list1)


list2 = [1, 2, 1, 1, 3, 4, 3, 3, 5]
print("\nthe unique values from 2nd list is")
unique(list2)

This code will loops through all the values and ensure that they are not repeated.

joeca0

You could write some simple helper functions like so:

my_dict = {1: [3, 4, 5], 2: [4], 3: [5], 6: [3, 5, 9], 8: [3 ,8]}

def keys_containing_element(dict_of_lists: dict, element: int) -> list:
    """Return all keys in dict_of_list whose list contains element"""

    return [key for key in dict_of_lists.keys() if element in dict_of_lists[key]]

def is_element_unique(dict_of_lists: dict, element: int) -> list:
    """Return true if element is in exactly one of the lists in dict_of_lists"""

    return True if len(keys_containing_element(dict_of_lists, element)) == 1 else False

Now,

print(keys_containing_element(my_dict, 3))
print(is_element_unique(my_dict, 3))
print(keys_containing_element(my_dict, 9))
print(is_element_unique(my_dict, 9))

outputs

[1, 6, 8]
False
[6]
True
ljdyer0

Try this:

def find_n(nums, target, exception):
    res = []
    for n, nums_list in nums.items():
        exception_key = [i for i in exception][0]
        if n != exception_key and nums_list != exception[exception_key]:
            if target in nums_list:
                res.append(n)
    return res


if __name__ == '__main__':
    nums = {1: [3, 4, 5], 2: [4], 3: [5], 6: [3, 5, 9], 8: [3, 8]}
    target = 3
    exception = {1: [3, 4, 5]}
    print(find_n(nums, target, exception))

Reuslt:

[6, 8]
Inozem0
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