Compare key and values at corresponding index in Python

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There is dictionary in python.key and value spelling is compared.if the mistake is greater than or equal to 2 than print incorrect

input={"their":"thuyr"}

output=incorrect(because t=t,h=h but e!=u,i!=y).

My problem is that i was unable to compare t==t,h==h,e==u,i==y. The below code shows count value 22 but count value must be 2 because only two words mismatches with their

What I have tried:

def find_correct(words_dict):
count=0
for key,value in words_dict.items():
for val in value:
for ky in key:
if(val!=ky):
count+=1
return count

print(find_correct({"their":"thuor"}))

Posted 25-Sep-18 3:50am

Tanya Chaudhary

Updated 25-Sep-18 6:17am

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Richard MacCutchan · Accepted Answer · 2018-09-25T06:17:00

The problem is that you have two loops, so for each character in the value you are comparing against each character in the key. That makes 25 actual tests, and 22 of them will be not equal. Chang to the following:

Python

def find_correct(words_dict):
    count=0
    for key,value in words_dict.items():
        for i in range(len(value)): # this may need adjusting for different length words
            if(value[i]!=key[i]):
                count+=1 
    return count 

print(find_correct({"their":"thuor"}))