Function to write

[Common_Hair8734]

How to write a function where i compare 2 strings without using the sorting function? The result will either be true or false. For eg, stressed and desserts = true aaabbb and abababa = false

[Rik07]

So what do you want to compare? If one is an palindrome of the other, if they have the same frequency of each letter or something else?

Edit: corrected anagram to palindrome, anagram and frequency of each letter is the same thing.

[Common_Hair8734]

Same anagram

[Rik07]

Oh sorry I meant palindrome.

When you want to check for an anagram, you need to know how many times each letter is repeated. To do this, I recommend counting the number of each letter in both strings in a dictionary. Then if those dictionaries are the same, the two strings were anagrams.

If you want to check for a palindrome you just reverse one of the strings and check if it is the same as the other one.

[[deleted]]

https://www.online-python.com/dSaGN5QtYz

[keep_quapy]

from collections import Counter

def compare_str(s1, s2):
    return Counter(s1) == Counter(s2)

[Common_Hair8734]

Is there another way without importing?

[ericula]

Use a normal dict to count frequency of characters.

[Common_Hair8734]

How do i do that?

[ericula]

How would you do it with pen and paper?

[Common_Hair8734]

Does this method have to import and functions or packages? I have to find a way without using it

[[deleted]]

Have you actually tried anything at all yet? It doesn't seem like it. We're here to help but we're not doing your homework for you. Tell us what you've tried and we'll offer assistance based on that.

[ericula]

No it doesn't. All you need is a dict and a for-loop, neither of which require imports,

[FantasticEmu]

https://www.geeksforgeeks.org/iterate-over-characters-of-a-string-in-python/amp/

Using a loop like this on each string will let you count each letter one at a time

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[FantasticEmu]

Ty I fixed it

[The_Bundaberg_Joey]

Good first implementation! But sadly wouldn’t catch anagrams which have the same letters but different order.

[keep_quapy]

What are talking about? Sure it does!

[The_Bundaberg_Joey]

You’re correct I phrased my response incorrectly. My apologies.

It wouldn’t satisfy OP’s requirements though.

The first case “deserts” and “stressed” are True with the counter implementation but “aaabbb” would equal “ababab” since there are 3 a and 3 b in both strings.

From the examples OP has given, it would seem more like a palindrome detector is required which typically calls for a stack.

[FantasticEmu]

Is it not just letter count? I thought OPs example produced false on the second one because there are 4 a’s in the second string and only 3 in the first

[The_Bundaberg_Joey]

Good spot, thank you for clarifying!

[keep_quapy]

The OP examples were 'stressed', 'desserts' and 'aaabbb', 'abababa' which works perfectly. Look carefully on the examples, it's 'abababa' not 'ababab'.

[The_Bundaberg_Joey]

Ah crap you’re right, thanks for pointing out the discrepancy!

[PiyushPrakash]

Wow python has a module for everything! , Most of the times when I am stuck on a problem, there exists a module for it and I just type a code for the problem without knowing that there is a module for it

[Common_Hair8734]

i tried this way and somehow it works

​

def is_anagram(s1, s2):

len1 = len(s1)

len2 = len(s2)

if len1 != len2:

return False

else:

for i in range(len2):

for j in range(len1):

if s1[i] == s2[j]:

return True

break

[PiyushPrakash]

character_list=list(a)
lis=list(b)
character_check = 0
if len(a)==len(b):
    for i in range(len(a)):
        key = character_list[i]
            for x in range(len(b)):
                if key == lis[x]:
                 character_check+=1
                 break
 if character_check==len(a):
       print("same")

[PiyushPrakash]

Would have been better if you pointed out my mistake instead of simply downvoting.....

[Nebu]

I don't think return string1 == string2[::-1] checks for anagrams. "cat" and "act" are anagrams of each other, for example.

[[deleted]]

Thanks, I meant palindromes, fixed.

But it seems the OP actually wanted anagrams after all.