Coded_correspondence Step 1

Just started with Python 3: String Methods: Coded_Correspondence project. Oh my gosh I so admire those of you who are able to do this project without referring to the solutions. :frowning: I can’t even fully understand the solutions. :frowning: :frowning: Please help me understand my following two questions? Link to the project below …

https://www.codecademy.com/courses/learn-python-3/informationals/python3-coded-communication

In layman’s terms, please explain to me why my code below (circled in blue) outputs a bunch of “j” letters instead of whitespace…

In more layman’s terms, why does my next blue-circled code not work. I thought we were supposed to shift all non-punctuated letters over by 10 letters. So that’s what I coded, which is exactly the same as what was coded correctly in the last line of the “if not” statement. The solution coded simply … message_decoded += letter … but to me, this looks like not moving the letter over by 10 letters which we’re supposed to do!

letter not in punctuation
is not the same as
letter != punctuation

letter not in punctuation is False when letter is contained within punctuation, meaning letter is a substring of punctuation, possible the same as punctuation

letter != punctuation is False when letter and punctuation are exactly the same, meaning letter would have to be ".!? " to make letter != punctuation be False

Notice that
alphabet.find(letter) gives you -1 when letter is not found in alphabet
and when you add 10,
you get 9;
alphabet[9] happens to be "j"

You don’t need to use alphabet at all for a letter/character that is in punctuation, you could just use the original letter/character.
as in

    else:
      message_decoded += letter
1 Like

Thank you Janbazant for your reply.

  1. What do you mean by “letter is a substring of punctuation”?

  2. Ah. Thank you for pointing out to me that letter != punctuation takes in ".!? " as a unit. I see where I went wrong there.

  3. I don’t understand “alphabet.find(letter) gives you -1 when letter is not found in alphabet”. Is that just a concept that I missed in the lesson? And just to clarify, when letter is not found, you mean when the loop lands on whitespace?

  4. Oh I get why the else statement is simply add the letter (which is whitespace) to the message_decoded string. Thank you for explaining that to me.

for the string "abcde",
  "abc" is a substring, and "bc", and "cde" are all substrings of "abcde",
but "ade" is not a substring of "abcde"

str1 = "abcde"
print( str1.find("c") ) # this would be 2, the first index at which "c" is found
print( str1.find("m") )  # this would be -1, since "m" is not a substring

Here’s a link to that in the codecademy docs: python strings .find() method

1 Like

Thank you for your clear explanation and for referring me to the Codecademy docs link on strings.find() method!

Hi again. Moving on to Step 3. I understand everything about the code solution except the placement of the "message_decoded = “” " …

The screenshot above is of the correct placement. But I had originally put it where the blue arrow points to, at the same no-indent level as alphabet and punctuation. I am not fully understanding why my original placement in global is not correct, and why inside the local loop placement is good.

You want to return a string
so you need to declare a variable that’s a string
that you’ll add the letters to
in the function.

1 Like

When we define a function, is it always paired with a return command inside the function?

functions don’t always contain return.
(A function may change something in an object instead of returning something, for example.)

1 Like

Ok I will keep an eye out for this (when function changes something in an object) and remember that functions don’t always need to contain a return. Thank you.

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