Dragonfly_Qiaoqiao Tan

[Shirley425]

No description provided.

[mikellewade]

Great job on Adagrams, Qiaoqiao!

[mikellewade]

Great job on the loop here! I love that you used .items and .extend!

[mikellewade]

Recall that .pop not only removes the element at the specified position i but also returns it. With that we could rewrite what you have to the following:

    while len(letter_list) < 10:
        i = randint(0, len(letter_pool_list)-1)
        letter_list.append(letter_pool_list.pop(i))
    return letter_list  

[mikellewade]

✨

[mikellewade]

We should use multiple lines (see here in the style guide about using multilines and indentation to create data structures).

[mikellewade]

Though this is functionally sound, I don't feel as if this communicates the intention behind what's going on here as readily as something like if len(word) >= 7 . This is how you will conventionally see the length of things checked in Python.

[mikellewade]

Another approach you could take to this is use a cumulative weight. This would cut the list of 98 letters from our code. We would essentially make another dictionary cumulating the weights:

LETTER_POOL = {'A': 9, 'B': 2, 'C': 2, 'D': 4, 'E': 12}
cumulative_weights = {'A': 9, 'B': 11, 'C': 13, 'D': 17, 'E': 29} # Total Weight: 29

Letter	Count	Cumulative Weight
A	9	9 (1-9 → 'A')
B	2	11 (10-11 → 'B')
C	2	13 (12-13 → 'C')
D	4	17 (14-17 → 'D')
E	12	29 (18-29 → 'E')

We could then randomly select a number between 1 - total weight. After that we would loop through the cumulative_weights dictionary and if the random number we got is less than the cumulative weight we are looking at on the current iteration we will select that that letter.

Note: The chances of choosing any number between 1 - 9 from a pile of numbers 1 - 98 is the same as choosing one of the 9 A's from the pile of 98 letters.

This code could look something like this:

Click to view the draw_letters function

import random

LETTER_POOL = {
    'A': 9, 'B': 2, 'C': 2, 'D': 4, 'E': 12, 'F': 2, 'G': 3, 'H': 2, 'I': 9, 
    'J': 1, 'K': 1, 'L': 4, 'M': 2, 'N': 6, 'O': 8, 'P': 2, 'Q': 1, 'R': 6, 
    'S': 4, 'T': 6, 'U': 4, 'V': 2, 'W': 2, 'X': 1, 'Y': 2, 'Z': 1
}

def compute_cumulative_weights(letter_pool):
    cumulative_weights = {}
    total_weight = 0
  
    for letter, weight in letter_pool.items():
        total_weight += weight
        cumulative_weights[letter] = total_weight
      
    return cumulative_weights, total_weight

def draw_letters():
    cumulative, total_weight = compute_cumulative_weights(LETTER_POOL)
    selected_letters = []

    for _ in range(10):
        rand_value = random.randint(1, total_weight)  # Pick a random number
      
        for letter, cumulative_weight in cumulative.items():
            if rand_value <= cumulative_weight:
                selected_letters.append(letter)
                break

    return selected_letters

[mikellewade]

Great job on breaking these up! It makes the flow a lot easier when you create essentially two sub-problems. One being to find the tied words for highest score and then the other being to implement the tie breaking logic. Perfect way to approach a problem! ⭐️