Improve the time efficiency of the anagram string function - python-3.x

Question

How can I improve the time efficiency of my algorithm to solve the following problem?
Problem:
Write a function that takes in a dictionary and query, two string arrays.
It should return an array of integers where each element i contains the number
of anagrams of the query[i] that exists in the dictionary. An anagram of a string is
another string with the same characters in the same frequency, in any order.
Example: "bca", "abc", "cba", "cab" are all anagram of "abc".
Current Solution
from collections import defaultdict
def IsAnagram(word):
# a list of 26 elements, each represents the number of occurrence of every english letter in the word
lst = [0] * 26
for c in word:
lst[ord(c) - 97] += 1
return lst
def anagram_string(query, dictionary):
d = defaultdict(int)
res = []
for w_1 in query:
# if the anagram of w_1 was counted before
if w_1 in d:
res.append(d[w_1])
else:
# define num of anagram for the query in dictionary
num_anagram = 0
# loop through all words of the dictionary
for w_2 in dictionary:
if len(w_1) != len(w_2):
continue
# add 1 to num of anagrams in case all letters of word_1 are the same as in word_2.
num_anagram += IsAnagram(w_1) == IsAnagram(w_2)
res.append(num_anagram)
# record the word with its number of anagrams in the dictionary
d[w_1] = num_anagram
return res
The above code has a Time Complexity of O(n*m) where n is the number of words in the query array and m is the number of words in dictionary array. Although it works fine for a small length array, it takes forever to compute the out array for a list of length 5000+. So, how can I improve it or maybe if someone has a different idea?

Answer 1

Here is my solution in Javascript
const dictionary = ['hack', 'rank', 'khac', 'ackh', 'kran', 'rankhacker', 'a', 'ab', 'ba', 'stairs', 'raits']
const query = ['a', 'nark', 'bs', 'hack', 'stairs']
function evaluate (dictionary, query) {
const finalResult = query.reduce((map, obj) => {
map[obj.split('').sort().join('')] = 0
return map
}, {})
dictionary.forEach(item => {
const stringEvaluate = item.split('').sort().join('')
if (finalResult.hasOwnProperty(stringEvaluate)) {
finalResult[stringEvaluate]++
}
})
return Object.values(finalResult)
}
console.log(evaluate(dictionary, query))

Answer 2

Solution in Python returning count of anagram in sequence which was found in dictionary list.
dictionary = ['hack', 'rank', 'khac', 'ackh', 'kran', 'rankhacker', 'a', 'ab', 'ba', 'stairs', 'raits']
query = ['a', 'nark', 'bs', 'hack', 'stairs']
myList=[sorted(element) for element in dictionary]
def anagram_string(query, dictionary):
resList=[]
for element in query:
count = myList.count(sorted(element))
resList.append(count)
return resList
print(anagram_string(query,dictionary))