Check if value already exists within list of dictionaries?

Question

Here's one way to do it:

if not any(d['main_color'] == 'red' for d in a):
    # does not exist

The part in parentheses is a generator expression that returns True for each dictionary that has the key-value pair you are looking for, otherwise False.

---

If the key could also be missing the above code can give you a KeyError. You can fix this by using get and providing a default value. If you don't provide a default value, None is returned.

if not any(d.get('main_color', default_value) == 'red' for d in a):
    # does not exist

Maybe this helps:

a = [{ 'main_color': 'red', 'second_color':'blue'},
     { 'main_color': 'yellow', 'second_color':'green'},
     { 'main_color': 'yellow', 'second_color':'blue'}]

def in_dictlist(key, value, my_dictlist):
    for entry in my_dictlist:
        if entry[key] == value:
            return entry
    return {}

print in_dictlist('main_color','red', a)
print in_dictlist('main_color','pink', a)

Perhaps a function along these lines is what you're after:

 def add_unique_to_dict_list(dict_list, key, value):
  for d in dict_list:
     if key in d:
        return d[key]

  dict_list.append({ key: value })
  return value

Based on @Mark Byers great answer, and following @Florent question, just to indicate that it will also work with 2 conditions on list of dics with more than 2 keys:

names = []
names.append({'first': 'Nil', 'last': 'Elliot', 'suffix': 'III'})
names.append({'first': 'Max', 'last': 'Sam', 'suffix': 'IX'})
names.append({'first': 'Anthony', 'last': 'Mark', 'suffix': 'IX'})

if not any(d['first'] == 'Anthony' and d['last'] == 'Mark' for d in names):

    print('Not exists!')
else:
    print('Exists!')

Result:

Exists!

Just another way to do what the OP asked:

 if not filter(lambda d: d['main_color'] == 'red', a):
     print('Item does not exist')

filter would filter down the list to the item that OP is testing for. The if condition then asks the question, "If this item is not there" then execute this block.