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Python

Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points
Posted by Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points

Works one way but not the other

for some reason this code works the way I have it shown, but if I write 

  def __str__(self):
        return "-".join(["dot" if x == "." else "dash" if x == "_" for x in self.pattern])

it won't work... why is that?

morse.py 
class Letter:
    def __init__(self, pattern=None):
        self.pattern = pattern

    def __str__(self):
        return "-".join(["dot" if x == "." else "dash" for x in self.pattern])

class S(Letter):
    def __init__(self):
        pattern = ['.', '.', '.']
        super().__init__(pattern)

3 Answers

Susannah Klanecek
Susannah Klanecek
5,825 Points
Susannah Klanecek
Susannah Klanecek
5,825 Points 
def __str__(self):
    output = []
    for entry in self.pattern:
        if entry == '.':
            output.append('dot')
        else:
            output.append('dash')
    return '-'.join(output)
Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points
Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points

actually, this way works fine

class Letter:
    def __init__(self, pattern=None):
        self.pattern = pattern

    def __str__(self):
        return "-".join(["dot" if x == "." else "dash" for x in self.pattern])

class S(Letter):
    def __init__(self):
        pattern = ['.', '.', '.']
        super().__init__(pattern)

and it's more efficient and 'pythonic' using a list comprehension. in your version you also don't bother testing for " _ " which assumes that the only characters in the list would be either "." or " _ ", I was just curious why it wouldn't work if you tested both. Afterall, what if there's a list with more than just "." or " _ " in it? in both of our solutions anything != "." would evaluate to "dash" even if that element was say... a number or a boolean or any other string character.

Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points
Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points

You make a good point on readability. I think it's readable but maybe most would not see comprehensions as "natural" like me. Second point tho... something is not working on this quiz because if I adjust your solution with an elif then it won't work, so there's something wrong and it's not our answers (I think)

class Letter:
    def __init__(self, pattern=None):
        self.pattern = pattern

    def __str__(self):
    output = []
    for entry in self.pattern:
        if entry == '.':
            output.append('dot')
        elif entry == "_":
            output.append('dash')

    return '-'.join(output)

that won't work for some reason.

Susannah Klanecek
Susannah Klanecek
5,825 Points
Susannah Klanecek
Susannah Klanecek
5,825 Points

Actually I think the list comprehension is ok on second thought, seems pretty readable, and the thing you tried didn't work because of indentation under the def function. Still no idea why your list comprehension didn't work though

David Luo
David Luo
1,215 Points
David Luo
David Luo
1,215 Points

I did the same thing and it didn't work, but created a temporary variable, say morse, and it worked. I.e., keep the list comprehension separate line from the "".join(morse) part. try that...

def __str__(self): 
    morse = ["dot" if x == "." else "dash" for x in self.pattern]
    return ("-".join(morse))
David Luo
David Luo
1,215 Points
David Luo
David Luo
1,215 Points

nevermind, misread your question. I looked up list comprehension and it doesn't support the syntax as you have it on the first section.

Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points
Nicholas Gaerlan
Nicholas Gaerlan
9,501 Points

since researching this, I've come to a guess that comprehensions can't handle the else-if syntax. Everything you pass into a comprehension basically has to resolve in a binary sort of way. In plain language. "this or that" not "this or that or that or that". there's probaby a way around this creating some kind of other variable that abstracts the string in some way before passing it to a list comprehension, but at that point you are creating a solution to fit the tool instead using a tool to find a solution.