🔒 Find anagrams

[shell drum]

def find_anagrams(word, candidates):
    exp = []
    for candidate in candidates:
        new = ""

        for letter in candidate:
            if not letter in word:
                break
            
            new += letter
        if new and  new != word:
           if len(new) == len(word):
              exp.append(new)
    return exp


candidates = ["patter"]
print(find_anagrams("tapper", candidates))```

[old yacht]

@shell drum

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[shell drum]

https://exercism.org/tracks/python/exercises/anagram

[lethal ginkgo]

henlo

[shell drum]

hi

[lethal ginkgo]

ok so you take all from candidates that are anagrams of the word?

[shell drum]

yes

[lethal ginkgo]

so the length must be the same, right?

[shell drum]

now the repeating letters of word is causing me problems

[shell drum]

yes

[lethal ginkgo]

!d sorted

[old yacht]

sorted

sorted(iterable, /, *, key=None, reverse=False)```
Return a new sorted list from the items in *iterable*.

Has two optional arguments which must be specified as keyword arguments.

*key* specifies a function of one argument that is used to extract a comparison key from each element in *iterable* (for example, `key=str.lower`). The default value is `None` (compare the elements directly).

*reverse* is a boolean value. If set to `True`, then the list elements are sorted as if each comparison were reversed.

Use [`functools.cmp_to_key()`](https://docs.python.org/3/library/functools.html#functools.cmp_to_key) to convert an old-style *cmp* function to a *key* function.

[lethal ginkgo]

!e

print(sorted("owns"))
print(sorted("snow"))

[old yacht]

@lethal ginkgo :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | ['n', 'o', 's', 'w']
002 | ['n', 'o', 's', 'w']

[shell drum]

am i on right track?

[lethal ginkgo]

see that these would be equal

[shell drum]

oh

so its just one line of code? i dont need looping?

[lethal ginkgo]

well you need to loop through all the candidates

[shell drum]

i mean letters

[lethal ginkgo]

and you should save the sorted word

[shell drum]

i didnt knew sort can solve it

[lethal ginkgo]

do you know list comprehension?

[shell drum]

can i get to the answer following my track am i on right track here?

[shell drum]

yes know them rarely use though

[lethal ginkgo]

I don't know how you would handle duplicate letters that way

because if you have word "aabb", then your code would say that "abbb" would be an anagram, but that's not true

the count of each letter has to be the same too

[shell drum]

ok i will try with sorted thanks

[lethal ginkgo]

👍

[shell drum]

and why?

[lethal ginkgo]

because you can use it very cleanly in this scenario

||

def find_anagrams(word: str, candidates: list[str]) -> list[str]:
    sorted_word = sorted(word)
    return [
        cand
        for cand in candidates
        if sorted(cand) == sorted_word
    ]

||

[shell drum]

yeah ok

thanks againn can i close?

[lethal ginkgo]

I don't have anything else to say, the problem is solved

[shell drum]

!close

[old yacht]

Python help channel closed

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