CODEWITHSUNDEEP
pythonstring
Christopher Jakob
Christopher Jakob

Reputation: 89

Using str.replace to replace letters in a string at different placeholders of the string

I have a problem that I am working on. The goal of the problem is to take the string placeholder i. If i is an even placeholder, replace the letter at i with the letter at i -1. If the i place holder is odd, then replace the letter i with the letter at i +1.

Here is my code so far:

def easyCrypto (s):
    for i in range (0,len(s)-1):
        if i % 2 == 0:
            str(s).replace(i,((i-1)))
        if i % 2 != 0:
            str(s).replace(i,((i+2)))
    print (s)

My error:

Traceback (most recent call last):
  File "<pyshell#4>", line 1, in <module>
    easyCrypto('abc')
  File "C:/Python/cjakobhomework7.py", line 4, in easyCrypto
    str(s).replace(i,((i-1)))
TypeError: Can't convert 'int' object to str implicitly

update!!

New code based on answers:

def easyCrypto (s):
    for i in range (0,len(s)-1):
        if i % 2 == 0:
            s =  str(s).replace(s(i),(s(i-1)))
        else:
            s = s.replace(s(i), s(i + 1))
    print (s)

However I still have the following errors:

Traceback (most recent call last):
  File "<pyshell#7>", line 1, in <module>
    easyCrypto('abc')
  File "C:/Python/cjakobhomework7.py", line 4, in easyCrypto
    s =  str(s).replace(s(i),(s(i-1)))
TypeError: 'str' object is not callable

Any ideas? thank you

Upvotes: 0

Views: 250

Answers (2)

user2555451
user2555451

Reputation:

There are two things here:

  1. str.replace does not automatically stringify its arguments. You need to manually convert them into strings. Remember that: "Explicit is better than implicit."

  2. str.replace does not work in-place because strings are immutable in Python. You need to reassign s to the new string object returned by str.replace.

Your code should be:

s = s.replace(str(i), str(i-1))

Also, you can replace if i % 2 != 0: with else: since the condition of the second if-statement can only be true if the first is false:

if i % 2 == 0:
   s = s.replace(str(i), str(i-1))
else:
   s = s.replace(str(i), str(i+1))

Regarding your edited question, you are trying to call the string s as a function by placing parenthesis after it. You need to use square brackets to index the string:

>>> 'abcde'[0]
'a'
>>> 'abcde'[3]
'd'
>>>

In your case it would be:

s =  s.replace(s[i], s[i-1])

As a general rule of thumb, parenthesis (...) are for calling functions while square brackets [...] are for indexing sequences/containers.

Upvotes: 0

pts
pts

Reputation: 87271

Use s[i] instead of s(i), and likewise for the other indexes.

Upvotes: 2

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