Making dictionary item frequency

Question

I want to know why it’s not counting the element ?
I made an key then in each iteration it should increase the value.

def isAnagram( s, t):
    if len(s) != len(t):
        return False 
    d_1 = {}
    d_2 = {}
    for i in range(len(s)):
        d_1[s[i]] =+ 1
        d_2[t[i]] =+ 1
    print(d_1)
    print(d_2)
    return True if d_1 == d_2 else False  

s = 'aabb'
t = 'bbaa'
print(isAnagram(s,t))

OUTPUT : 
d_1 = {'a': 1, 'b': 1}
d_2 = {'b': 1, 'a': 1}

Asked By: Mohamed Elhaddad

||

Source

Answer 1

Don’t reinvent the wheel!

from collections import Counter

def isAnagram(s, t):
    return Counter(s) == Counter(t)

s = "aabb"
t = "bbaa"

print(isAnagram(s,t)) # => True

Answered By: Julia

Answer 2

as mentioned in comments. you should use += instead of =+.
+= means adding the value to the previous value stored. while
=+ here means = (assigning the value)

def isAnagram(s, t):
    if len(s) != len(t):
        return False
    d_1 = {}
    d_2 = {}
    for i in range(len(s)):
        d_1[s[i]] += 1
        d_2[t[i]] += 1
    print(d_1)
    print(d_2)
    return True if d_1 == d_2 else False  

s = 'aabb'
t = 'bbaa'
print(isAnagram(s,t))

Answered By: No.BoD

Answer 3

First, you have an error, to add value to the variable you need to do

d_1[s[i]] += 1
d_2[t[i]] += 1

Second, the function fails because you need to define the key and value before modifying the value.
Working function:

def isAnagram( s, t):
    if len(s) != len(t):
        return False 
    d_1 = {}
    d_2 = {}
    for i in range(len(s)):
        try: # checking if the character in the dictionary exists
            d_1[s[i]]
        except KeyError:
            d_1[s[i]] = 0
        try: # checking if the character in the dictionary exists
            d_2[t[i]] 
        except KeyError:
            d_2[t[i]] = 0
        d_1[s[i]] += 1
        d_2[t[i]] += 1
    print(d_1)
    print(d_2)
    return True if d_1 == d_2 else False  

s = 'aabb'
t = 'bbaa'
print(isAnagram(s,t))

Answered By: Domagoj

Answer 4

add a .lower() to Domagoj’s answer to be case-insensitive.
so this will also work:

s = ‘Betrug’.lower()
t = ‘Erbgut’.lower()

{‘b’: 1, ‘e’: 1, ‘t’: 1, ‘r’: 1, ‘u’: 1, ‘g’: 1}
{‘e’: 1, ‘r’: 1, ‘b’: 1, ‘g’: 1, ‘u’: 1, ‘t’: 1}
True

Answered By: aeaeae

Making dictionary item frequency

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