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Tigers- Selam Chaka #60

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70aea46
action plan added
thaocodes Sep 26, 2022
39dc166
passed wave-1 first 2 tests
s-chaka Sep 26, 2022
7731788
completed wave-1
s-chaka Sep 27, 2022
89d5e07
wave-02 3 tests passed and 2 failed
s-chaka Sep 27, 2022
44957ea
Completed wave-2
s-chaka Sep 27, 2022
3f3ed4a
wave 3 progress
thaocodes Sep 27, 2022
829880b
wave 3 complete
thaocodes Sep 27, 2022
c97227d
Wave 2 tests passed
s-chaka Sep 27, 2022
fb8d2e4
merged wave 2 tests
s-chaka Sep 27, 2022
16412e3
wave 4, 5/7 tests passed
thaocodes Sep 28, 2022
0ae543f
Merge branch 'master' of https://github.com/selamchaka/adagrams-py
s-chaka Sep 28, 2022
a8df5ca
All tests passed yeyygit status
s-chaka Sep 28, 2022
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2 changes: 2 additions & 0 deletions README.md
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Original file line number Diff line number Diff line change
@@ -1,4 +1,6 @@
# AdaGrams
## Plan of Action
process each part of a wave on our own and try to work through it and code what we can. Check in on eachother for progress or questions. Get together to review code. Attend extra help sessions if needed.

## Skills Assessed

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148 changes: 143 additions & 5 deletions adagrams/game.py
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Original file line number Diff line number Diff line change
@@ -1,11 +1,149 @@
def draw_letters():
pass
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
}

SCORE_CHART = {
'A': 1,
'B': 3,
'C': 3,
'D': 2,
'E': 1,
'F': 4,
'G': 2,
'H': 4,
'I': 1,
'J': 8,
'K': 5,
'L': 1,
'M': 3,
'N': 1,
'O': 1,
'P': 3,
'Q': 10,
'R': 1,
'S': 1,
'T': 1,
'U': 1,
'V': 4,
'W': 4,
'X': 8,
'Y': 4,
'Z': 10
}

def draw_letters():
letter_freq = {}
random_letters=[]
selected_letters =[]
while len(random_letters) < 26:
r = random.choice(list(LETTER_POOL.keys()))
random_letters.append(r)
for letter in random_letters:
if letter in letter_freq:
letter_freq[letter] += 1
else:
letter_freq[letter] = 1
if letter_freq[letter] <= LETTER_POOL[letter]:
selected_letters.append(letter)
return (selected_letters[0:10])

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This looks good, overall but remember that it is not accounting for the different probabilities of each letter! We could take that into account by creating a collection of some sort that holds every possible value (12 E's, 9 A's, etc) and then making our selections from that collection and removing them as they are picked!

def uses_available_letters(word, letter_bank):
pass
word_str = ""
temp_word = ""
letter_bank_dic = {}
word_dic ={}

for letter in word:
if str.upper(letter) in letter_bank:
temp_word += letter
word_str += str.upper(letter)
for letter in letter_bank:
if letter in letter_bank_dic:
letter_bank_dic[str.upper(letter)] += 1
else:
letter_bank_dic[str.upper(letter)] = 1
for letter in word_str:
if letter in word_dic:
word_dic[str.upper(letter)] += 1
else:
word_dic[str.upper(letter)] = 1
if word_dic[letter] > letter_bank_dic[letter]:
return False
elif temp_word != word:
return False
elif temp_word == word:
return True
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I see exactly what y'all were doing here. The problem is that you are making a variety of extra dictionaries and strings that just aren't necessary. In cases like this, it would be much better to make a deep copy of the letter_bank. From there, you can simply manipulate your copy.

The process would be as follows:

  1. Make a deep copy of the letter bank.
  2. Go through each letter in the word and see if it is in the letter bank copy.
    a) If it is, remove it from the letter bank copy.
    b) If it isn't return False
  3. If you make it through the word without returning False, return True.

This will take into account multiples of the same letter!

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I see exactly what y'all were doing here. The problem is that you are making a variety of extra dictionaries and strings that just aren't necessary. In cases like this, it would be much better to make a deep copy of the letter_bank. From there, you can simply manipulate your copy.

The process would be as follows:

  1. Make a deep copy of the letter bank.
  2. Go through each letter in the word and see if it is in the letter bank copy.
    a) If it is, remove it from the letter bank copy.
    b) If it isn't return False
  3. If you make it through the word without returning False, return True.

This will take into account multiples of the same letter!


def score_word(word):
pass
if word == "":
return 0
score_word = 0
upper_word = str.upper(word)

for letter in upper_word:
for key, value in SCORE_CHART.items():
if letter == key:
score_word += value

if len(word) in range(7, 11):
score_word += 8

return score_word

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This looks great! It's a simple and effective implementation of the score method!


def get_highest_word_score(word_list):
pass
highest_word = []
highest_value = 0
tie_word = []
for word in word_list:
word_score = score_word(word)
if word_score > highest_value:
highest_word = word
highest_value = word_score
tie_word = [highest_word]
elif word_score == highest_value: # tie case
tie_word.append(word)

if len(tie_word) > 1 :
if len(tie_word[0]) > len(tie_word[1]) and len(tie_word[0])==10 :
return (tie_word[0], highest_value)
elif len(tie_word[0]) < len(tie_word[1]) and len(tie_word[1])!=10 :
return (tie_word[0], highest_value)
elif len(tie_word[0]) == len(tie_word[1]):
return(tie_word[0], highest_value)
else:
return(tie_word[1], highest_value)

return (highest_word, highest_value)

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Once again, you are creating several extra lists that we don't necessarily need. This works and is a great first pass, but we could definitely pare it down a bit! One implementation would have been to have a tuple that just holds the highest scoring word and the highest score. From there, we can do the following:

  1. Make a tuple to hold the first word and its score.
  2. For each word in the list, score it and compare it to the score in your highest word tuple.
  3. If the score is higher, replace the tuple with the new word.
  4. If it is a tie, run the if/else statements you have there toward the bottom of the function!

This really just avoids having to write a tie_word list. We can kind of figure out what the highest word in place.






1 change: 1 addition & 0 deletions tests/test_wave_01.py
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Original file line number Diff line number Diff line change
Expand Up @@ -65,3 +65,4 @@ def test_letter_not_selected_too_many_times():
# Assert
for letter in letters:
assert letter_freq[letter] <= LETTER_POOL[letter]
pass