Randomly increment a key in a python dictionary until all of the keys have the same value?-CodePudding

I have 5 groups - A,B,C,D,E. I want to pick a random letter within the group letters, but I need exactly 4 instances of each letter

So in the end I want four As, four Bs, four Cs four Ds, and four Es, but I want to pick them randomly.

Using a dictionary was the best way I thought I could do this. I can keep track of how many letters I have this way, but I'm not certain how to write out code such that each letter only appears four times.

import random
random.seed(1)

groups = {
          'A' : 0,
          'B' : 0,
          'C' : 0,
          'D' : 0,
          'E' : 0,
          }

#return a random letter between A and E, the keys of the group dictionary
def random_letter():
    return random.choice(list(groups.keys()))    

while (groups['A'] != 4) and (groups['B'] != 4): #What can I put here so that it makes all of the groups A to E have only 4. using 'and' here isnt working im assuming because im using it incorrectly
    groups[random_letter()]  = 1 
    
print(list(groups.keys())) #should return ['A', 'B', 'C', 'D', 'E']
print(list(groups.values())) #should return [4,4,4,4,4]

CodePudding user response：

I think this is the simplest way I could do it.

str_options = "abcde" * 4
options_as_list = list(str_options)
random.shuffle(options_as_list)
print(options_as_list)

CodePudding user response：

Why don't you use random.sample:

import random

res = random.sample('ABCDE', counts=[4]*5, k=20)

If all the letters always have the same number of occurrences:

res = random.sample('ABCDE'*4, 20)

CodePudding user response：

Copy the desired choices into a list and shuffle it only once:

import random
random.seed(1)

groups = {
    'A': 0,
    'B': 0,
    'C': 0,
    'D': 0,
    'E': 0
}

full_list = [*groups]*4
random.shuffle(full_list)
print(full_list)

CodePudding user response：

from random import choice


groups = dict.fromkeys("ABCDE", 0)

while (key_pool := [key for key in groups if groups[key] != 4]):
    key = choice(key_pool)
    print(f"Incrementing {key}")
    groups[key]  = 1

Explanation: dict.fromkeys("ABCDE", 0) is a convenient short-hand for creating a dictionary, in this case, with the keys A through E, all mapping to the value 0.

while (key_pool := [key for key in groups if groups[key] != 4]): is pretty dense. We keep iterating as long as there exist keys which do not map to a count of 4. We generate that list of keys with the list comprehension, which we assign to key_pool via the so-called "walrus operator".

key = choice(key_pool) picks a random key from the collection of possible keys (keys that don't map to 4.)

groups[key] = 1 increments the value of the chosen key.

That being said, it's not immediately obvious from your question what you actually intend to do with this dictionary. Re-reading your question makes me wonder if what you really want is just a string with the letters A through E, each appearing four times in some random order?

If that's the case, maybe something like this?:

from random import sample

chars = "ABCDE"

print("".join(sample(chars*4, k=len(chars)*4)))

chars*4 will yield the string 'ABCDEABCDEABCDEABCDE' random.sample will pick k characters from that string (when an element is picked, that element will not be considered in successive samples). "".join(...) joins a collections of strings into a single string.