Develop Reference

Parse a json/dictionary with same key values

I currently have a list variable that looks like this:
list_of_dicts = [{"Away_Team":"KC", "Home_Team":"NYY"},
{"Away_Team":"TB", "Home_Team":"MIA"},
{"Away_Team":"TOR", "Home_Team":"BOS"},
]
As you can see, there are multiple keys with the same names, pertaining to the game matchups.
When I try to use:
print(json.dumps(list_of_dicts[0], indent=4, sort_keys=True))
...it only prints out the first matchup due to the same keys:
{
"Away_Team": "KC",
"Home_Team": "NYY"
}
How can I convert this list_of_dicts variable into something like the following output so I can use it like a valid dictionary or json object?
{
"Away_Team_1":"KC", "Home_Team_1":"NYY",
"Away_Team_2":"TB", "Home_Team_2":"MIA",
"Away_Team_3":"TOR", "Home_Team_3":"BOS",
}
This output doesn't need to be exactly that if a better solution is available, this is just to give you an idea of how I'd like to be able to parse the data.
The list_of_dicts variable can be of varying sizes, I've shown 3 here, but it could contain 1 or 10 matchups, it varies, so the solution needs to be dynamic to that.

You can add suffixes to the keys with enumerate:
list_of_dicts2 = [{f"{k}_{i}":v for k,v in d.items()} for i,d in enumerate(list_of_dicts, start=1)]

One option is to use pandas:
pd.DataFrame(list_of_dicts).to_csv('filename.csv', index=False)
gives
Away_Team,Home_Team
KC,NYY
TB,MIA
TOR,BOS
Now the index is implied by the row, and if you load it back in you'll have those indices. Pandas also supports to_json if you are hard set on using json though. You can even recover your original list from a dataframe using .to_dict(orient='records')

Data structure is important. You really don't need a dictionary for this. Simply reducing to a list of tuples the first slot always the away team, and the second the home team.
list_of_dicts = [{"Away_Team":"KC", "Home_Team":"NYY"},
{"Away_Team":"TB", "Home_Team":"MIA"},
{"Away_Team":"TOR", "Home_Team":"BOS"},
]
l = [tuple(l.values()) for l in list_of_dicts]
output:
[('KC', 'NYY'), ('TB', 'MIA'), ('TOR', 'BOS')]
The problem with your proposed solution is iterating through dicts where you don't know the key name is cumbersome, this solution makes the data structure easy to decipher, transform, or manipulate.

How to sort a dictionary of dictionaries by their values python

I have a user dictionary that looks something like this:
{
user1: {
userLevel: 1,
userScore: 2
},
user2: {
userLevel: 5,
userScore: 16
},
user3: {
userLevel: 3,
userScore: 14
}
}
and I need to sort it by values, so that the user with the maximum level is 1st, and if some users have the same level then the one with a better score goes 1st. If some 2 users have the same level and score, any one of them can go first, it doesn't matter. I tried creating a double loop, but I think there might be a cleaner way to do that, I just can't figure it out.

There are already multiple answer on how to order a dictionnary by values. The only thing that's changing here is that your items are also dictionnaries.
The fonction sorted allows you to sort a dictionnary on a specific key.
What you're trying to do is to order the dictionnary on a specific key which is in this case the userScore.
Therefore you need to find a way to specify that this is the key on which your sortingis based.
You also need a way to itterate through your dictionnary. This is done with the .items() fonction.
If your big dictionnary is called a, then a.items() will return a list of tuples where each tuple has a key : user_x and a value : the dictionnary.
a.items() = [('user1', {'userLevel': 1, 'userScore': 2}), ('user2', {'userLevel': 5, 'userScore': 16}), ('user3', {'userLevel': 3, 'userScore': 14})]
Now you can itterate through these items and you need to specify that for each item that you parse when you iterate, you need to evaluate whether its key userScore is bigger or lower than the others.
So for each item x ( a tuple in our case ) they key userScore is retrieved with x[1]['userScore']
The sorting line of code should look something like this :
a = sorted(a.items(), key = lambda tuple : tuple[1]['userScore']

Check for string in list items using list as reference

I want to replace items in a list based on another list as reference.
Take this example lists stored inside a dictionary:
dict1 = {
"artist1": ["dance pop","pop","funky pop"],
"artist2": ["chill house","electro house"],
"artist3": ["dark techno","electro techno"]
}
Then, I have this list as reference:
wish_list = ["house","pop","techno"]
My result should look like this:
dict1 = {
"artist1": ["pop"],
"artist2": ["house"],
"artist3": ["techno"]
}
I want to check if any of the list items inside "wishlist" is inside one of the values of the dict1. I tried around with regex, any.
This was an approach with just 1 list instead of a dictionary of multiple lists:
check = any(item in artist for item in wish_list)
if check == True:
artist_genres.clear()
artist_genres.append()
I am just beginning with Python on my own and am playing around with the SpotifyAPI to clean up my favorite songs into playlists. Thank you very much for your help!

The idea is like this,
dict1 = { "artist1" : ["dance pop","pop","funky pop"],
"artist2" : ["house","electro house"],
"artist3" : ["techno","electro techno"] }
wish_list = ["house","pop","techno"]
dict2={}
for key,value in dict1.items():
for i in wish_list:
if i in value:
dict2[key]=i
break
print(dict2)

A regex is not needed, you can get away by simply iterating over the list:
wish_list = ["house","pop","techno"]
dict1 = {
"artist1": ["dance pop","pop","funky pop"],
"artist2": ["chill house","electro house"],
"artist3": ["dark techno","electro techno"]
}
dict1 = {
# The key is reused as-is, no need to change it.
# The new value is the wishlist, filtered based on its presence in the current value
key: [genre for genre in wish_list if any(genre in item for item in value)]
for key, value in dict1.items() # this method returns a tuple (key, value) for each entry in the dictionary
}
This implementation relies a lot on list comprehensions (and also dictionary comprehensions), you might want to check it if it's new to you.

Python `dict` indexed by tuple: Getting a slice of the pie

Let's say I have
my_dict = {
("airport", "London"): "Heathrow",
("airport", "Tokyo"): "Narita",
("hipsters", "London"): "Soho"
}
What is an efficient (no scanning of all keys), yet elegant way to get all airports out of this dictionary, i.e. expected output ["Heathrow", "Narita"]. In databases that can index by tuples, it's usually possible to do something like
airports = my_dict.get(("airport",*))
(but usually only with the 'stars' sitting at the rightmost places in the tuple since the index usually is only stored in one order).
Since I imagine Python to index dictionary with tuple keys in a similar way (using the keys's inherent order), I imagine there might be a method I could use to slice the index this way?
Edit1: Added expected output
Edit2: Removed last phrase. Added '(no scanning of all keys)' to the conditions to make it clearer.

The way your data is currently organized doesn't allow efficient lookup - essentially you have to scan all the keys.
Dictionaries are hash tables behind the scenes, and the only way to access a value is to get the hash of the key - and for that, you need the whole key.
Use a nested hierarchy like this, so you can do a direct O(1) lookup:
my_dict = {
"airport": {
"London": "Heathrow",
"Tokyo": "Narita",
},
"hipsters": {
"London": "Soho"
}
}

Check "airport" is present in the every key in the dictionary.
Demo:
>>> [value for key, value in my_dict.items() if "airport" in key]
['Narita', 'Heathrow']
>>>
Yes, Nested dictionary will be better option.
>>> my_dict = {
... "airport": {
... "London": "Heathrow",
... "Tokyo": "Narita",
... },
... "hipsters": {
... "London": "Soho"
... }
... }
>>>
>>> if "airport" in my_dict:
... result = my_dict["airport"].values()
... else:
... result = []
...
>>> print result
['Heathrow', 'Narita']
>>>

What I'd like to avoid, if possible, is to go through all dictionary keys and filter them down.
Why? Why do you think Python is doing the equivalent of a DB full table scan? Filtering a dictionary does not mean sequential scanning it.
Python:
[value for key, value in my_dict.items() if key[0] == "airport"]
Output:
['Narita', 'Heathrow']

Retrieve a key-value pair from a dict as another dict

I have a dictionary that looks like:
{u'message': u'Approved', u'reference': u'A71E7A739E24', u'success': True}
I would like to retrieve the key-value pair for reference, i.e. { 'reference' : 'A71E7A739E24' }.
I'm trying to do this using iteritems which does return k, v pairs, and then I'm adding them to a new dictionary. But then, the resulting value is unicode rather than str for some reason and I'm not sure if this is the most straightforward way to do it:
ref = {}
for k, v in charge.iteritems():
if k == 'reference':
ref['reference'] = v
print ref
{'reference': u'A71E7A739E24'}
Is there a built-in way to do this more easily? Or, at least, to avoid using iteritems and simply return:
{ 'reference' : 'A71E7A739E24' }

The trouble with using iteritems is that you increase lookup time to O(n) where n is dictionary size, because you are no longer using a hash table

If you only need to get one key-value pair, it's as simple as
ref = { key: d[key] }
If there may be multiple pairs that are selected by some condition,
either use dict from iterable constructor (the 2nd version is better if your condition depends on values, too):
ref = dict(k,d[k] for k in charge if <condition>)
ref = dict(k,v for k,v in charge.iteritems() if <condition>)
or (since 2.7) a dict comprehension (which is syntactic sugar for the above):
ref = {k,d[k] for k in charge if <condition>}
<same as above>

I dont understand the question:
is this what you are trying to do:
ref={'reference',charge["reference"]}