Python dictionnaries have keys and values accessed using those keys.

You can access the keys as follows, your dict key will be stored in the key variable:

my_dict = {"a" : 1, "b" : 2}
for key in my_dict:
    print(key)

This will print:

a
b

You can then do any comparisons you want:

my_dict = {"a" : 1, "b" : 2}
for key in my_dict:
    if key == "a":
        return True
    else:
        return False

which can be improved to:

my_dict = {"a" : 1, "b" : 2}
print("a" in my_dict.keys())

You can then access the values for each key in your dict as follows:

my_dict = {"a" : 1, "b" : 2}
for key in my_dict:
    print(my_dict[key])

This will print:

1
2

I suggest you read more about dictionaries from the official Python documentation: https://docs.python.org/3.6/tutorial/datastructures.html#dictionaries

Looks like you're thinking about this strangely. All you need to do is check if the letter is in your score dict, and if it is, to add that number to your total.

def compare(word):
    res = 0
    for letter in word:
        if letter in score:
            res += score[letter]
    return res

However there's an easier way to do this. Since you're just using res as an accumulator, you can add score[letter] if it exists or 0 if it doesn't. This is easy using the dict.get method.

def compare(word):
    res = 0
    for letter in word:
        res += score.get(letter, 0)
        # dict.get(key, defaultvalue)
    return res

In fact you can even make it into an ugly lambda.

compare = lambda word: sum([scores.get(letter,0) for letter in word])

I think your problem is that a is a variable instead bing the string 'a'

this work

o_list = [{'h': 4, 'name': 'a'},{'h': 6, 'name': 'b'},{'h': 1, 'name': 'c'}]

min_node = 'a'

def f():
  for i, elem in enumerate(o_list):
    if min_node == o_list[i]['name']:
        return min_node
    else:
        return 'error in retrieving min F'

print f()

however it would be more elegant to do the following

o_list = [{'h': 4, 'name': 'a'},{'h': 6, 'name': 'b'},{'h': 1, 'name': 'c'}]

min_node = 'a'

def f():
  for elem in o_list:
    if min_node == elem["name"]:
        return min_node
    else:
        return 'error in retrieving min F'

print f()

If you want to know how many values match in both the dictionaries, you should have said that :)

Maybe something like this:

shared_items = {k: x[k] for k in x if k in y and x[k] == y[k]}
print(len(shared_items))

You need to iterate over the keys in crucial and compare each one against the dishes keys. So a directly modified version of your code.

for key in crucial.keys():
    if key in dishes.keys():
        print(dishes[key])

It could be better stated as (no need to specify .keys):

for key in crucial:
    if key in dishes:
        print(dishes[key])

Try:

mydict =  {"value for money": ["rescheduled", "cost", "low", "high", "simplicity", "booking", "price-performance", "satisfied", "satisfaction", "pricing", "prices"]}
mystring = "I love simplicity"
if any((word in mystring) for word in mydict["value for money"]):
    print("Found one.")