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Python Function "Not Defined" in VSCODE [closed]

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class Hand:
def __init__(self, hand):
self.hand = hand
def rank_to_index_converter(rank):
if rank == 2:
return 0
elif rank == 3:
return 1
elif rank == 4:
return 2
elif rank == 5:
return 3
elif rank == 6:
return 4
elif rank == 7:
return 5
elif rank == 8:
return 6
elif rank == 9:
return 7
elif rank == 10:
return 8
elif rank == 'J':
return 9
elif rank == 'Q':
return 10
elif rank == 'K':
return 11
elif rank == 'A':
return 1
else:
raise ValueError
def suit_to_index_converter(suit):
if suit == 'C':
return 0
elif suit == 'D':
return 1
elif suit == 'H':
return 2
elif suit == 'S':
return 3
else:
raise ValueError
def is_consecutive(hand):
### BEGIN YOUR SOLUTION
first = 0
string = ""
count = 0
for x in hand:
#print(x[0])
if x[0] == ("J"):
if first != 10 and (count != 0):
return False
string = x[0]
elif x[0] == ("Q"):
if string != "J" and (count != 0):
return False
string = x[0]
elif x[0] == ("K"):
if string != "Q" and (count != 0):
return False
string = x[0]
elif x[0] == ("A"):
if string != "K" and (count != 0):
return False
string = x[0]
elif x[0] != ("J" or "Q" or "K" or "A"):
if (x[0] != (first + 1)) and (count != 0):
if (hand[0][0] != "A"):
return False
first = x[0]
count = count + 1
return True
def is_suit(hand):
suit = hand[0][1]
for x in hand:
if x[1] != suit:
return False
return True
def is_flush(hand):
return is_suit(hand) and not is_consecutive(hand)
def is_four_of_a_kind(hand):
count = 0
pair_num = 0
lst = [0 for i in range(13)]
for x in hand:
index = rank_to_index_converter(x[0])
lst[index] += 1
for x in lst:
if x == 4:
return True
return False
def is_full_house(hand):
count = 0
pair_num = 0
lst = [0 for i in range(13)]
for x in hand:
index = rank_to_index_converter(x[0])
lst[index] += 1
three_pair = False
pair = False
for x in lst:
if x == 3:
three_pair = True
if x == 2:
pair = True
if three_pair and pair:
return True
return False
def is_flush(hand):
count = 0
while count < (len(hand) - 2):
if hand[count][1] != hand[count + 1][1]:
return False
count += 1
return True
def is_straight(hand):
if is_consecutive(hand) and not is_suit(hand):
return True
return False
def is_three_of_a_kind(hand):
lst = [0 for i in range(13)]
for x in hand:
index = rank_to_index_converter(x[0])
lst[index] += 1
three_pair = False
for x in lst:
if x == 3:
three_pair = True
if three_pair and not is_full_house(hand) and not is_suit(hand):
return True
return False
def is_two_pair(hand):
lst = [0 for i in range(13)]
for x in hand:
index = rank_to_index_converter(x[0])
lst[index] += 1
two_pair = False
counter = 0
switch = 0
while counter != len(lst):
if lst[counter] == 2:
switch = 1
if lst[counter] == 2 & switch == 1:
two_pair = True
if two_pair and not is_full_house(hand) and not is_suit(hand) and not is_four_of_a_kind(hand) and not is_three_of_a_kind(hand):
return True
return False
def is_pair(hand):
lst = [0 for i in range(13)]
for x in hand:
index = rank_to_index_converter(x[0])
lst[index] += 1
pair = False
for x in lst:
if x == 2:
pair = True
if pair and not is_full_house(hand) and not is_suit(hand):
return True
return False
def maximum_value(hand):
sum = 0
for x in hand:
sum += rank_to_index_converter(x[0])
return sum
The is_full_house function uses the rank_to_index_converter function. The rank_to_index_converter function is defined above the is_full_house function. Why am I receiving an error that the rank_to_index_converter function is not defined?
I am lost on where to proceed; therefore, I haven't tried anything, besides copying and pasting the rank_to_index_converter function into the is_full_house function; however, it is more convenient if the rank_to_index_converter is a separate function. That is why I want to solve this issue.

How to sort dictionary values from Highest to lowest?

My question would be how I withdraw winner based on highest score?
since u cant sort in dictionaries
i tried this with lists, but then name wouldnt appear, only the score...
a = {'name_a':0}
b = {'name_b':0}
c = {'name_c':0}
d = {'name_d':0}
e = {'name_e':0}
print("Each time someone scores a point, the letter of his name is typed in lowercase. If someone loses a point, the letter of his name is typed in uppercase")
score = input('Enter series of charachters indicating who scored a poitn: ')
for i in score:
if i == 'a':
a['name_a'] += 1
if i == 'A':
a['name_a'] -= 1
if i == 'b':
b['name_b'] += 1
if i == 'B':
b['name_b'] -= 1
if i == 'c':
c['name_c'] += 1
if i == 'C':
c['name_c'] -= 1
if i == 'd':
d['name_d'] += 1
if i == 'D':
d['name_d'] -= 1
if i == 'e':
e['name_e'] += 1
if i == 'E':
e['name_e'] -= 1
print(a,b,c,d,e)
print('Winner is: ', )

This will work:
max((i, name) for d in (a,b,c,d,e) for name, i in d.items())[1]

you probably want to use a single dict, instead of one for each, like such:
scores = {
'a': 0,
'b': 0,
'c': 0,
'd': 0,
'e': 0,
}
And then you can keep track of the highest scoring player as points are calculated:
point_scored = input('Enter series of charachters indicating who scored a point: ')
for i in point_scored:
if not scores.get(i) is None:
scores[i] += 1
elif not scores.get(i.lower()) is None:
scores[i.lower()] -= 1
else:
print(str(i) + ' is not a valid player...')
winner = max(scores, key=scores.get)
print(scores)
print('Winner is ' + winner)

max_key = ""
max_val = 0
for key, value in d.items():
if (value > max_val):
max_val = value
max_key = key
Is this what you mean?

i found the answer
winner = (sorted(d.items(), key = lambda x: int(x[1]), reverse = True))

Iterating through list (Python)

I have a list that extends on like this and I would like to sort them based on the first 2 digits of the second part of each. I'm really rushed right now so some help would be nice.
collection = ["81851 19AJA01", "68158 17ARB03", "104837 20AAH02",
I tried this and it didn't work. I'm not doing this for a class I'd really appropriate some help
for x in collection:
counter = 0
i=0
for y in len(str(x)):
if (x[i] == '1'):
counter == 1
elif (x[i] == '2'):
counter == 2
elif x[i] == '0' and counter == 2:
counter = 2
elif x[i] == '9' and counter == 1:
counter = 3
elif x[i] == '8' and counter == 1:
counter = 4
elif x[i] == '7' and counter == 1:
counter = 5
i = i + 1
if (counter==2):
freshmen.append(x)
elif (counter==3):
sophomores.append(x)
elif (counter==4):
juniors.append(x)
elif (counter==5):
seniors.append(x)

Use the key function to define a custom sorting rule:
In [1]: collection = ["81851 19AJA01", "68158 17ARB03", "104837 20AAH02"]
In [2]: sorted(collection, key=lambda x: int(x.split()[1][:2]))
Out[2]: ['68158 17ARB03', '81851 19AJA01', '104837 20AAH02']

Python Recursion Double letters

So what the task is, is that your supposed to write a recursion function that counts the amount of "double" letters in a string, So for example the string "hmmm" would return 1 and the string "hmmmm" would return 2 and that a string "abb" would return 1. My code is here:
def num_double_letters(astr):
if astr == "" or len(astr) == 1:
return 0
elif len(astr) == 2:
if astr[0] == astr[1]:
return 1 + num_double_letters(astr[1:])
else:
return 0 + num_double_letters(astr[1:])
elif astr[0] != astr[1]:
return 0 + num_double_letters(astr[1:])
elif astr[0] == astr[1] != astr[2]:
return 1 + num_double_letters(astr[1:])
elif astr[0] == astr[1] == astr[2]:
return 0 + num_double_letters(astr[1:])
My problem is that a string with 4 same letters = 1 when its supposed to = 2. And also is there a cleaner way to do this?

I think you've made it a bit complicated for yourself... there's no need to go deeper into the recursion once the length of your string is 2, and you want to advance by 2, not 1 when you find a double to count the way I think you do. Try this:
def num_double_letters(astr):
if astr == "" or len(astr) == 1:
return 0
elif len(astr) == 2:
if astr[0] == astr[1]:
return 1
else:
return 0
elif astr[0] != astr[1]:
return 0 + num_double_letters(astr[1:])
elif astr[0] == astr[1]:
return 1 + num_double_letters(astr[2:])
print(num_double_letters('hmm'))
print(num_double_letters('hmmm'))
print(num_double_letters('hmmmm'))
Output:
1
1
2
You might consider the following more Pythonic and concise:
def num_double_letters(astr):
if len(astr) < 2:
return 0
if astr[0] == astr[1]:
return 1 + num_double_letters(astr[2:])
return num_double_letters(astr[1:])

Given a String, What is the Length of the One of the Longest WFF in Polish Notation?

I'm trying to write a version of always popular Count-A-WFF section of the WFF 'N Proof game (no copyright infringement intended) in Python. Alright, not so popular.
I think I have everything up and running up as desired for up to the case of a 4 letter string.
def maximum_string(s):
if cs(s) == True:
return len(s)
elif len(s) == 2:
l1 = [cs(s[0]), cs(s[1])]
if True in l1:
return len(s) - 1
else:
return 0
elif len(s) == 3:
first = s[0] + s[1]
second = s[0] + s[2]
third = s[1] + s[2]
l1 = [cs(first), cs(second), cs(third)]
if True in l1:
return len(s) - 1
l2 = [cs(s[0]), cs(s[1]), cs(s[2])]
if True in l2:
return len(s) - 2
else:
return 0
elif len(s) == 4:
first = s[0]+s[1]+s[2]
second = s[0]+s[1]+s[3]
third = s[1]+s[2]+s[3]
fourth = s[0]+s[2]+s[3]
l1 = [cs(first), cs(second), cs(third), cs(fourth)]
if True in l1:
return 3
first = s[0] + s[1]
second = s[0] + s[2]
third = s[0] + s[3]
fourth = s[1] + s[2]
fifth = s[1] + s[3]
sixth = s[2] + s[3]
l2 = [cs(first), cs(second), cs(third), cs(fourth), cs(fifth), cs(sixth)]
if True in l2:
return 2
first = s[0]
second = s[1]
third = s[2]
fourth = s[3]
l3 = [cs(first), cs(second), cs(third), cs(fourth)]
if True in l3:
return 1
else:
return 0
def cs(string):
global length_counter, counter, letter
counter = 1
length_counter = 0
letters_left = len(string)
while letters_left != 0 and length_counter < len(string):
letter = string[length_counter]
if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
counter += 1
elif letter == 'N':
counter += 0
else:
counter -= 1
length_counter += 1
letters_left -= 1
if counter == 0 and len(string) == length_counter:
return True
else:
return False
The maximum_string helper function is intended to, given any string S, find the length of one of the longest possible wffs that you can make from just the letters of S. Of course, I can continue the pattern I currently have for the maximum_string helper function up to a length of 13. But, combinatorial explosion is evident. Thus, is there a more elegant way to finish off the maximum string helper function?

In effect one of the functions I had earlier would return a distance of how far away a string is from having a permutation in Polish notation. Thus this was surprisingly simpler to fix than I expected. Here's what I was looking for:
def maximum_string(string):
global length_counter, counter, letter
counter = 1
length_counter = 0
letters_left = len(string)
while letters_left != 0 and length_counter < len(string):
letter = string[length_counter]
if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
counter += 1
elif letter == 'N':
counter += 0
else:
counter -= 1
length_counter += 1
letters_left -= 1
if ('p' in string) or ('q' in string) or ('r' in string) or ('s' in string) or ('t' in string) or ('u' in string):
return len(string) - abs(counter)
else:
return 0