Question

I’m trying to make a program in Python that finds anagrams. Here is my current code:

def anagram(word,checkword):
    for letter in word:  
        if letter in checkword:  
            checkword = checkword.replace(letter, '') 
        else:  
            return False  
    return True  

while True:
    f = open('listofwords.txt', 'r')
    try:
        inputted_word = input('Word? ')
        for word in f:
            word = word.strip()
            if len(word)==len(inputted_word):
                if word == inputted_word:
                    continue
                elif anagram(word, inputted_word):
                    print(word)
                        #try:
                            #if word == 1:
                            #print ('The only anagram for', user_input, 'is', word)
                        #elif word > 1:
                            #print ('The anagrams for', user_input, 'are', word)
                        #except TypeError:
                            #pass
    except:
        break 

I’m having trouble outputting the anagrams. The anagrams should be in one line, and the wording should reflect the amount of anagrams found. Such as…

"There is only one (insert anagram) for (insert word inputted)"

"There are (insert anagrams) for (insert word inputted)"

"There are no anagrams for (insert word inputted)"

"The (insert word inputted) is not in the dictionary")

Answer 1

Here are a few hints:

First, if you have to print the count of anagrams before you print any of them, you need to hold onto a list of them while you’re looping. Something like this:

anagrams = []
for word in f:
    word = word.strip()
    if len(word)==len(inputted_word):
        if word == inputted_word:
            continue
        elif anagram(word, inputted_word):
            anagrams.append(word)

Now you just have to figure out how to print the right text at the end, based on what’s in the anagrams list.

As for what you tried:

#try:
    #if word == 1:
    #print ('The only anagram for', user_input, 'is', word)
#elif word > 1:
    #print ('The anagrams for', user_input, 'are', word)
#except TypeError:
    #pass

This can’t possibly work. First, word is a word, so it can’t possibly be equal to 1, or greater than 1. Also, if you’ve only gone through, say, the first 20 words in the dictionary, and found the first anagram, how could you know that this is the only anagram? There may be 1000 of them in the rest of the dictionary. You can’t decide which sentence to print until you’ve finished the whole dictionary.

Meanwhile, notice that you have different cases for “There is only one” vs. “not in the dictionary”. So, you need some kind of flag for “found inputted_word in the dictionary”, which you set inside that if statement. Or maybe, you could just leave the special case out—for example, at the end, if you have 0 results, you know it was not in the dictionary. It depends on whether you want more logic at the end, or inside the loop.