Question: The code I'm writing currently requires me to check if any item (at least one) in one dictionary is present in another dictionary.

Answer: 
How about this solution:

```
a = {"a":2, "b":4, "c":4, "d":4}
b = {"a":1, "e":1, "f":5}
print(any(a.items() & b.items()))

```

will yield output:
`False`

because there is no common items in `a` and `b`, while:

```
a = {"a":2, "b":4, "c":4, "d":4}
b = {"a":1, "b":4, "f":5}
print(any(a.items() & b.items()))

```

will yield output:
`True`

because there is one common item for `a` and `b`

It does not iterates over dictionaries directly, however as [juanpa-arrivillaga](https://stackoverflow.com/users/5014455/juanpa-arrivillaga) pointed out in comment, this solution technically uses iteration, as under the hood `any()` does iteration.
You can use `set` to check if at least one of the first `dict`'s keys and/or values exists in the second. You can do something like this:

```
a = {1:"a", 2:"b", 3:"c"}
b = {"foo":"hello", "bar":"hi", 2:"hoo"}
c = {"hello":"hoo", 1:"hii"}

def keys_exists(first:"dict", second:"dict") -> bool:
    # Or:
    # return not (set(first) - set(second)) == first.keys()
    return bool(set(first) & set(second))

def values_exists(first:"dict", second:"dict") -> bool:
    return bool(set(first.values()) & set(second.values()))

print("At least one of a keys exists in b keys: {0}".format(keys_exists(a,b)))
print("At least one of a keys exists in c keys: {0}".format(keys_exists(a,c)))
print("At least one of b keys exists in c keys: {0}".format(keys_exists(b,c)))

print("At least one of a values exists in b values: {0}".format(values_exists(a,b)))
print("At least one of a values exists in c values: {0}".format(values_exists(a,c)))
print("At least one of b values exists in c values: {0}".format(values_exists(b,c)))

```

Output:

```
At least one of a keys exists in b keys: True
At least one of a keys exists in c keys: True
At least one of b keys exists in c keys: False

At least one of a values exists in b values: False
At least one of a values exists in c values: False
At least one of b values exists in c values: True

```