Question

in problem set 6, part 4, does anyone know a slick and easy way to get all the subsets for the elements in your hand?

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

Here's what I did. may not be slick and easy, but it works. I use two other functions: bubbleSort and dict_to_string (included at the bottom)

def pick_best_word_faster(hand, rearrange_dict):


"""
Return the highest scoring word from points_dict that can be made with the
given hand.

Return '.' if no words can be made with the given hand.

hand: dictionary: (string -> int)
"""

max_score = 0
best_word = '.'
# dict_to_string is a function I created to
# convert a dictionary of letters to a string
my_hand = dict_to_string(hand)
for i in range(1, len(my_hand) + 1):
S = list(itertools.combinations(my_hand,i))
for item in S:
ordered_string = string.join(bubbleSort(list(item)), '')
if ordered_string in rearrange_dict:
temp_word = rearrange_dict[ordered_string][0]
if points_dict[temp_word] > max_score:
max_score = points_dict[temp_word]
best_word = temp_word
return best_word

def bubbleSort(L):
"""returns a sorted list when passed the list, L"""
swapped = True
while swapped:
swapped = False
#print L
for i in range(len(L) - 1):
if L[i] > L[i+1]:
temp = L[i]
L[i] = L[i+1]
L[i+1] = temp
swapped = True
return L

def dict_to_string(dictionary):
"""
Returns a string consisting of all letters in dictionary
"""
list_of_letters = ()
list_pointer = 0
for letter in dictionary:
if dictionary[letter] != 1:
for i in range(0, dictionary[letter]):
list_of_letters += (letter,)
list_pointer += 1
else:
list_of_letters += (letter,)
list_pointer += 1
return string.join(list_of_letters, '')

Answer 3

Thanks! itertools.combinations is exactly what I was missing... :)