Your first comment is misleading (it's the itertools module not the permutations module) and useless anyway. Remove it.
There's absolutely no reason to rename permutations to prm. Don't do it.
num_of_cases would be better called num_cases, simply because it's more conventional.
Your take input comment is misplaced. It should be at the place it refers to.
counter is too vague: what are you counting? I would call it num_matched_anagrams.
You don't need to convert scrambled_word to a list. permutations accepts any iterable. You don't need to call str on the output of raw_input; it already returns a string.
The prior improvements make the comment take input really obvious, so you can remove that.
The later comments are often misaligned. Have them in line with the text.
prm_list can be generated with just list(permutations(scrambled_word)). This removes two comments.
prm_list should just be called word_orders or similar; don't bother naming things by their type.
check is badly named; it should be called something like find_anagrams.
check should not use file, which is deprecated, but open. It should also use with to handle the file.
check should save the dictionary between runs. In fact, this shouldn't really be check's job at all.
The words should have their newlines trimmed, not the other way around.
check should only return anagrams, not deal with appending.
check should have sensibly named parameters.
check should just call permutations itself.
You do
possible_words = list(set(possible_words))
but why bother having possible_words as a list at all?
Instead of if len(possible_words) == 0, do if not possible_words. I would also switch the if around.
num_matched_anagrams+=1 should have spaces: num_matched_anagrams += 1. The comment is obvious, so remove it.
Now to deal with speed.
words should be a set (since it's a set of words) and you loop over the permutations first. This allows a check of:
def find_anagrams(scrambled_word, words):
for word_order in permutations(scrambled_word):
potential_word = ''.join(word_order)
if potential_word in words:
yield potential_word
which is much faster.
We can improve this further by doing all the work upfront by making a cannonical, sorted version. This generates a mapping:
sorted word → [list of words that sort to this word]
This can be generated like so:
from collections import defaultdict
word_anagrams = defaultdict(set)
with open('dictionary.txt') as dictionary:
for word in dictionary:
word = word.rstrip()
word_anagrams[sorted(word)].add(word)
This then makes find_anagrams just word_anagrams[sorted(scrambled_word)]. This has high initial overhead (0.5 seconds for me with 170k words) but makes subsequent lookups insanely fast.
This gives the code:
from collections import defaultdict
word_anagrams = defaultdict(set)
with open('dictionary.txt') as dictionary:
for word in dictionary:
word = word.rstrip()
word_anagrams[''.join(sorted(word))].add(word)
num_cases = int(raw_input("Input number of words needed to be unscrambled: "))
num_matched_anagrams = 1
while num_matched_anagrams <= num_cases:
scrambled_word = raw_input("Input scrambled word: ")
sorted_word = ''.join(sorted(scrambled_word))
if word_anagrams[sorted_word]:
for word in word_anagrams[sorted_word]:
print "Possible Word for Scrambled Word #" + str(num_matched_anagrams) + ": " + word
else:
print "No matches found"
num_matched_anagrams += 1
The loop can be replaced with for num_matched_anagrams in range(1, num_cases+1).
The whole thing should be put into a main function.
The dictionary parsing could be extracted to another function.
The print should use formatting and I'd add brackets for Python 3 compatibility:
print("Possible Word for Scrambled Word #{}: {}".format(num_matched_anagrams, word))
To complete Python 3 compatibility, add this to the top:
try:
input = raw_input
except NameError:
pass
This gives:
from collections import defaultdict
try:
input = raw_input
except NameError:
pass
def parse_dictionary(filename):
word_anagrams = defaultdict(set)
with open(filename) as dictionary:
for word in dictionary:
word = word.rstrip()
word_anagrams[''.join(sorted(word))].add(word)
return word_anagrams
def main():
word_anagrams = parse_dictionary("dictionary.txt")
num_cases = int(input("Input number of words needed to be unscrambled: "))
for num_matched_anagrams in range(1, num_cases+1):
scrambled_word = input("Input scrambled word: ")
sorted_word = ''.join(sorted(scrambled_word))
if word_anagrams[sorted_word]:
for word in word_anagrams[sorted_word]:
print("Possible Word for Scrambled Word #{}: {}".format(num_matched_anagrams, word))
else:
print("No matches found")
if __name__ == "__main__":
main()
Note that this is the same technique as Prashant gave.
