比较元组列表，根据条件确认子集？

Question

Given a word and a list of words, I have to find the list elements/words that can be built using letters (count of letter matters) of the given word. I have tried to use Counter object from collections and a definition of python 2.7's cmp() function (I'm using 3.6.5).

I have since come to realize this approach seems to be bad practice for such a problem (Earlier, I was trying to use counter object-dictionaries to compare). The reason my program doesn't work is because the compare_fn relies on '>','<' operations between lists, which give result based on lexicographical order ( referred from here ). So even though 'raven' can be made from 'ravenous', the program below will fail because of the order of char in a sorted list.

from collections import Counter    
word = 'ravenous'
candidates = ["raven", "mealwheel", "rasputin"]

def count_fn(mystr):
    return sorted(list(Counter(mystr).items()))

def compare_fn (c1,c2):
    return ((c1>c2) - (c1<c2))

list_word =  count_fn(word)
list_candidates = list(map(count_fn,candidates))
cmp_list = [compare_fn(list_word,i) for i in list_candidates]
cmp_list
#[-1, -1, -1]    #should be [1,-1,-1]

So, for below two lists, how can I confirm that list_candidates[0] is a subset of list_word . Please note that the comparison ('a',1) in list_word against ('a',1) in list_candidates[i] could also be ('a',5) in list_word against ('a',1) in list_candidates[i] ; both cases are true.

print(list_word)
#[('a', 1), ('e', 1), ('n', 1), ('o', 1), ('r', 1), ('s', 1), ('u', 1), ('v', 1)]
print(list_candidates[0])
#[('a', 1), ('e', 1), ('n', 1), ('r', 1), ('v', 1)]

Answer 1

I think using counters is a good choice. Don't turn them into lists. I purposely returned [True, False, False] instead of [1, -1, -1], but you can change that easily.

Moreover: I used a list comprehension instead of map, beacause it is more current in python, but the semantic is the same.

from collections import Counter
word = 'ravenous'
candidates = ["raven", "mealwheel", "rasputin"]

def count_fn(mystr):
    return Counter(mystr)

def compare_fn (c1,c2):
    return all(c1[char] >= count for char, count in c2.items())

counter_word =  count_fn(word)
list_candidates = [count_fn(candidate) for candidate in candidates]
cmp_list = [compare_fn(counter_word, i) for i in list_candidates]
print(cmp_list)