CODEWITHSUNDEEP
pythondictionary
RnRoger
RnRoger

Reputation: 690

Update dictionary key based on value

I have the following dictionary:

players = {'Roger': ['player1', 'dA'], 'Luka': ['player2', 'sK']}.

I want to update the key that contains 'player2', but I can't update players[Luka] because 'player2' is not always Luka. How can I select keys linked to 'player2'?

(for those wondering, dA = Ace of Diamonds, sK = King of Spades. These values will also be different every time).

Edit: Here's the part of my code: (It won't run because I left out a lot of clutter)

qPlayers = 2  #Amount of players
def game(qPlayers):
    players[nr]["value"].append(new_card)
    deal_to =[]
    for player in players:
        deal_to.append(player)
    deal(qPlayers,deck,players,deal_to)

def setup(qPlayers):
    playerhands = []
    for x in range(1,qPlayers+1):
        player = {}
        player["name"] = input("Enter your name, player {}\n>>>".format(x))
        playerhands.append(player)
    return playerhands      


def deal(qPlayers,deck,players,deal_to):
    nr = -1
    for player in deal_to:
        nr +=1
        new_card = getCard(deck)   #getCard produces a random card like sA, k9, cQ

Upvotes: 0

Views: 102

Answers (3)

ImportanceOfBeingErnest
ImportanceOfBeingErnest

Reputation: 339200

You can iterate the dictionary to find the key to update. dict.iteritems() will do that job in python 2.7.

players = {'Roger': ['player1', 'dA'], 'Luka': ['player2', 'sK']}

for key, value in players.iteritems():
    if value[0] == "player2":
        players[key][1] = "sQ"

print players

Upvotes: 1

nbrooks
nbrooks

Reputation: 18233

You have two options here:

  1. Search through the entire directory, checking for the value which contains 'player2'. This ends up being inefficient, and makes the dictionary structure somewhat useless.

  2. Use a list instead of a dictionary, and model your data accordingly, similar to the example below.

An advantage of this data structure is that you don't need the redundant player1/player2 identifiers, since the list index provides that implicitly. To reference player2 you'd take the second element from the list (players[1] since indexing starts at 0).

players = [
    {'Name' : 'Roger', 'Value' : 'dA'},
    {'Name' : 'Luka', 'Value' : 'sK'}
]

Upvotes: 1

A. Sokol
A. Sokol

Reputation: 336

You can keep track of which key in the dict contains your player2 value, or else you will have to iterate through the dict until you find a key that does. If you frequently need to search based on something internal to the value, you may want to reconsider the data structure you are using to store this data.

def findKeyWithValue(value):
    for playerName, playerData in players.items():
        if value in playerData:
            # your code here

Upvotes: 2

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