在Python中从.txt文件向字典添加字谜？

Question

I have a text file which is a long list of words with no spaces eg"neuropsychologicalneuropsychologistneuropsychologyneuropsychopathic" etc I am trying to build a program that will take an inputted word and check the .txt file for any anagrams of the word and then add them to a dictionary.

The code I have come up with so far that prints 1 dictionary with 1 permutation contained in words.txt is;

anagram_dict  = {}

def anagram(word):

    b = open('words.txt', 'r').readlines()
    text = ''.join(line.strip() for line in b)

    #print(len(text))

    for i in range(len(text)-len(word)):
        prop = text[i:i+len(word)]
        if all(char in word for char in prop): #and all(prop.count(char) == prop.count(word) for char in prop):
            anagram_dict[''.join(sorted(word))] = [prop]

    print(anagram_dict)

anagram("demand")

The issue is that the permutation it prints isn't an anagram (I input "demand" expecting "madden"to come out but it prints "amamad" which is in the file but not an anagram) How can I make it so that it only prints the dictionary if the permutation is a true anagram (same letters just rearranged)?

I think the problem is possibly with the "if all(char in word for char in prop): # and all(prop.count(char) == prop.count(word) for char in prop):"

Especially the part that's commented out as when that runs it just prints an empty string.

Apologies for this being so long I just wanted to make sure I explained myself. Thanks for any help.

Answer 1

all(char in word for char in prop) simply checks if all the letters of word are in prop . All the letters of demand are "present" in "madden" , so it would show up in the results.

There is problem with all(prop.count(char) == prop.count(word) for char in prop) (probably a typo).

It should be all(prop.count(char) == word.count(char) for char in prop)

---

There is also a better alternative, check if sorted versions of prop and word are same.

sword = ''.join(sorted(word))
anagram_dict[sword] = []
for i in range(len(text)-len(word)):
    prop = text[i:i+len(word)]
    if sword == ''.join(sorted(prop)):
        anagram_dict[sword].append(prop)