unique (single) value in dict (or list)

[3Pinter]

Given any random dictionary (or a list doesnt matter), how to ask the dictionary if there is an unique value (so which only occurs once).

{
"1": "a",
"2": "b",
"1": "c",
"1": "c",
"1": "33",
"1": "a"
}

--> No unique value is found ... or technically 2 (b and 33) are found, but I only want 1.
so not desired result ... therefor errorhandling or so.

{
"1": "a",
"2": "b",
"1": "c",
"1": "c",
"1": "c",
"1": "a"
}

--> Unique value is found (b) ... do something.

What would your approach be?
Set returns all unique values but I'm not after that.

[ndc85430]

What have you tried?

[3Pinter]

Still unsure about the approach ndc85430, but thinking about something like this.

mylist = [ [1, "a"], [2, "b"], [1, "c"], [1, "c"],  [1, "c"], [1, "a"] ]

test = [a[1] for a in mylist]
# ['a', 'b', 'c', 'c', '33', 'a']

from collections import Counter
collection_counter = Counter(test)
for m in mylist:
    print "element {} num of times: {}".format(m[0], collection_counter.get(m[1]))

num_occurance = 0
newlist = ""
for key in collection_counter: 
    if collection_counter[key] == 1: 
        num_occurance = num_occurance + 1
        newlist = key
print num_occurance

if num_occurance == 1:
    print "unique value found: {}".format(newlist)
    print [x for x in mylist if x[1] == newlist]
else:
    print "bummer, no uniques found"
        
    

mylist = [ [1, "a"], [2, "b"], [1, "c"], [1, "c"],  [1, "c"], [1, "a"] ]
#[[2, 'b']]
#(annoying nested list ..grrr)

mylist = [ [1, "a"], [2, "b"], [1, "c"], [1, "c"],  [1, "33"], [1, "a"] ]
# bummer, no uniques found 
#(annoying nested list ..grrr)

[deanhystad]

Like this?

items1 = {1:1, 2:2, 3:3}
items2 = {1:1, 2:1, 3:2, 4:2, 5:3}
items3 = {1:1, 2:1, 3:3, 4:5, 5:4}

def find_unique(d):
    value_list = list(d.values())
    value_set = set(value_list)
    if len(value_set) == len(value_list):
        return None # All values are unique

    unique = None
    for value in value_set:
        if value_list.count(value) == 1:
            if unique is None:
                unique = value
            else:
                return None # Already found a unique value
    return unique

print(items1, '  Unique =', find_unique(items1))
print(items2, '  Unique =', find_unique(items2))
print(items3, '  Unique =', find_unique(items3))

Output:
{1: 1, 2: 2, 3: 3}   Unique = None
{1: 1, 2: 1, 3: 2, 4: 2, 5: 3}   Unique = 3
{1: 1, 2: 1, 3: 3, 4: 5, 5: 4}   Unique = None

[3Pinter]

yes. that looks like what I want to achieve.

your approach looks more logical than mine, thanks!

[3Pinter]

on second thoughts: how to return the dictionary item (key+val)?

I think that dictionary isn't the logical approach for this?