python - Checking strings against each other (Anagrams)

Question

The assignment is to write a program that accepts two groups of words from the user and then prints a "True" statement if the two are anagrams (or at least if all the letters of one are present in the other) and a "False" statement if not.

Being very new to programming as a whole, I don't know how to move beyond just indexing a string and comparing all of the pieces of one to another. I stress that I am a beginner; I've read many of the other posts tagged with Python and Anagram, and they are uniformly above my head and reference things I have not been taught. So the simpler the better. Here is my non-working code so far:

s1 = input("Please enter a word:")
s2 = input("Please enter another word:")

for i in range(0, len(s1), 1):
    if i in range (0, len(s2), 1):
        print("The letters in your first word are present in your second word.")

Answer 1

You need to think through your conditional logic a bit more. The loop is on the right track, but if there is a letter in s1 that is NOT in s2, you should break out of this loop and print the "False" statement. Consider using a variable like all_s1_in_s2 = True and then setting that to false if you find a letter that doesn't match.

Some other tips:

• for l in s1 will loop through string s1 giving you access to each letter in sequence as l - you don't need range or len at all

• The if .. in statement can help test whether a letter exists in a string, e.g. if letter in mystring: is a valid statement and this could help you a lot, again not needing range or len

• You should avoid using numbers in variable names where possible - better would be word_one and word_two, as an example

Answer 2

Why not just sort the strings?

>>> sorted('anagram')
['a', 'a', 'a', 'g', 'm', 'n', 'r']
>>> sorted('nagaram')
['a', 'a', 'a', 'g', 'm', 'n', 'r']
>>> sorted('anagram') == sorted('nagaram')
True

Answer 3

You can use the magic Counter from collections library. From documentation:

It is an unordered collection where elements are stored as dictionary keys and their counts are stored as dictionary values

So, you can initialize a Counter object with a string (a iterable) and compare with another Counter from a string

from collections import Counter

def is_anagram(str1, str2):
   return Counter(str1) == Counter(str2)

Answer 4

To check if two strings are anagrams of each other using dictionaries: Note : Even Number, special characters can be used as an input

def anagram(s):
    string_list = []
    for ch in s.lower():
        string_list.append(ch)

    string_dict = {}
    for ch in string_list:
        if ch not in string_dict:
            string_dict[ch] = 1
        else:
            string_dict[ch] = string_dict[ch] + 1

    return string_dict



s1 = "master"
s2 = "stream"

a = anagram(s1)
b = anagram(s2)

if a == b:
    print "Anagram"
else:
    print "Not Anagram"

Answer 5

def is_anagram(w1, w2):
    w1, w2 = list(w1.upper()), list(w2.upper())
    w2.sort()
    w1.sort()
    return w1 == w2

Answer 6

>>> s1 = 'vivid'
>>> s2 = 'dvivi'
>>> s3 = 'vivid'
>>> def is_anagram(s1, s2):
...     if s1.lower() == s2.lower():
...         return False
...     return sorted(s1.lower()) == sorted(s2.lower())
...
>>> is_anagram(s1, s2)
True
>>> is_anagram(s1, s3)
False
>>> s2 = 'dvivii'
>>> is_anagram(s1, s2)
False
>>> s2 = 'evivi'
>>> is_anagram(s1, s2)
False
>>> 

Answer 7

You could use the following code it will not count special characters nor it will count digits and will return "they are anagrams" if the total characters have occurred equally in both strings hence will tell the the strings are anagrams or not .

text = input('Enter a string: ')
text1 = input('Enter a string: ')
text,text1 = text.lower(),text1.lower()
count = 0
count1=0
for i in range(97,123):
    if chr(i) in text and chr(i) in text1:
    count1+=1
        if text.count(chr(i)) == text1.count(chr(i)):
             count +=1
if len(text) >= len(text1):
    num1 = len(text)
else:
    num1 = len(text1)
if count == count1:
    print("they are anagrams")
else :
    print("they are not anagrams")

Answer 8

I think we can get this like this

---

s1 = "listen"
s2 = "silent"
s1=list(s1);s1.sort()
s2 = list(s2);s2.sort()
if s1 == s2:
   print ("Given Strings are Anagram")
else:
   print ("Given String are not anagrams")

---

Answer 9

Not sure if it was proposed up there, but I went with:

def is_anagram(a, b):
    return sorted(a.lower()) == sorted(b.lower())

Answer 10

Just a thought:

def check_(arg):
        mark = hash(str(set(sorted(arg))))
        return mark

def ana(s1, s2):
        if check_(s1) != check_(s2):
                pass
        elif len(s1) != len(s2):
                pass
        else:
             print("{0} could be anagram of  {1}".format(s1, s2))

Answer 11

This worked for me

str1="abcd"
str2="bcad"
word1=[]
word2=[]
for x in range(len(str1)):
    word1.append(str1[x])
for x in range(len(str2)):
    word2.append(str2[x])
if(len(word1)==len(word2)):
    for letter in word1:
        if letter in word2:
            word2.remove(letter)

if len(word2)==0:
    print "anagram"
else:
    print "not anagram"

Answer 12

    #An anagram is the result of rearranging the letters of a word to produce a new word. Anagrams are case insensitive
    #Examples:
    # foefet is an anagram of toffee
    # Buckethead is an anagram of DeathCubeK

    # The shortest my function style *************************************** 
    def is_anagram1(test, original):
        """Сhecks 'test' is anagram of 'original' strings based on:
        1. length of the both string and length of the sets made from the strings is equivalent
        2. then checks equivalents of sorted lists created from test and original strings

        >>> is_anagram1('Same','same')
        False
        >>> is_anagram1('toffee','foeftt')
        False
        >>> is_anagram1('foefet','toffee')
        True
        >>> is_anagram1("Buuckk",'kkkcuB')
        False
        >>> is_anagram1('Buckethead','DeathCubeK')
        True
        >>> is_anagram1('DeathCubeK','Buckethead')
        True
        """
        # check the length of the both string
        if len(test) != len(original):
            return False

        # check is the strings are the same
        t,o = test.lower(), original.lower()
        if t == o:
            return False

        # check the sorted lists
        return sorted(t) == sorted(o)


    # The final my one line code **************************************
    def is_anagram(test, original):
        """Сhecks 'test' is anagram of 'original' in one line of code

        >>> is_anagram('Same','same')
        False
        >>> is_anagram('toffee','foeftt')
        False
        >>> is_anagram('foefet','toffee')
        True
        >>> is_anagram("Buuckk",'kkkcuB')
        False
        >>> is_anagram('Buckethead','DeathCubeK')
        True
        >>> is_anagram('DeathCubeK','Buckethead')
        True
        """
        return False if len(test) != len(original) or test.lower() == original.lower() else sorted(test.lower()) == sorted(original.lower())

    if __name__ == "__main__":
        import doctest
        doctest.testmod(verbose=True)


### 2 items passed all tests:
### 6 tests in __main__.is_anagram
### 6 tests in __main__.is_anagram1
### 12 tests in 3 items.
### 12 passed and 0 failed.
### Test passed

Answer 13

Here's a solution if you are adamant on using Python dictionary and you can't use functional programming:

Create a dictionary using comprehension and compare the dictionaries of the two word with a simple == operator.

def isanagram2(wrd1, wrd2):

    wrd1_dict = {k: 0 for k in wrd1}
    wrd2_dict = {k: 0 for k in wrd2}

    for c1, c2 in zip(wrd1, wrd2):
        wrd1_dict[c1] += 1
        wrd2_dict[c2] += 1

    if wrd1_dict == wrd2_dict:
        return True
    return False

Answer 14

Simplest shortest solution

def anagram(word1, word2):
    return sorted(word1) == sorted(word2)

check

print(anagram("xyz","zyx"))
>>True

print(anagram("xyz","zyy"))
>>False

Answer 15

Here's one typical assignment solution:

def anagram(s1, s2):
""" (str, str) -> bool

Return True if s1 and s2 are anagrams

>>> anagram(s1, s2)
True
"""
    s1 = s1.replace(" ", "")
    s2 = s2.replace(" ", "")

    s1_new = list(s1)
    s2_new = list(s2)

    if len(s1_new) == len(s2_new):
        dupe_found = True
        while dupe_found:
            dupe_found = False
            for i in s1_new:
                for j in s2_new:
                    if i == j:
                        s1_new.remove(i)
                        s2_new.remove(j)
                        dupe_found = True
                        break
                break
    return s1_new == s2_new

Answer 16

def anagram(a,b):
x=[]
y=[]
for i in a:
    if i!=' ': # This will ignore the spaces in the sentence
        x+=[i] # Adds only letters into a list and ignore the spaces
for i in b:
    if i!=' ':
        y+=[i]
if len(x)==len(y): # if length of two lists are not same, They are not anagrams anyway. So it directly returns False.
    for i in range(len(x)):
        for j in range(len(y)):
            if x[i].lower()==y[j].lower(): 
                y.pop(j) # If the letter matched between first and second list, that letter is poped from that list.
                break
    return len(y)==0 # If the given sentences are anagrams, all the letters are poped out from the second list and function returns True(as the lenght of the second list is 0. If not, function will return False.
return False

anagram(a,b)