Python: Using exception for iteration (beginner)

Question

I just want to know why this doesn't work (I am trying to name the ducklings from a book: Jack, Kack, Lack, Mack, Nack, Ouack, Pack, Quack) Note: Quack and Ouack have a U

prefixes = 'JKLMNOPQ'
suffix = 'ack'

for letter in prefixes:
    if letter != 'O' or 'Q':      #I know this doesn't work, need to know alternative
        print letter + suffix
    else:
        print letter + 'u' + suffix

What does this have to do with exceptions?

Karl Knechtel
– Karl Knechtel

2012-03-25 12:10:57 +00:00
Commented Mar 25, 2012 at 12:10 — Karl Knechtel
– Karl Knechtel, Commented Mar 25, 2012 at 12:10

Rafe Kettler · Accepted Answer · 2012-03-25 06:32:30Z

7

You likely mean this:

if letter != 'O' or letter != 'Q':

The result of your original statement,

if letter != 'O' or 'Q':

compared letter to the result of 'O' or 'Q', which is a boolean (true to be exact) (so you could see why this comparison would always be true as it was).

answered Mar 25, 2012 at 6:32

Rafe Kettler

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3 Comments

Seb Over a year ago

So how can I fix it to add 'u' between prefix and suffix when letter is O or Q?

mechanical_meat Over a year ago

@30trix: Change the or to and. Or else you can use if letter not in 'OQ':; whichever seems clearer to you, I suppose.

mechanical_meat Over a year ago

There are a couple of issues with this answer: 1) as pointed out by @Ondrej Kupka, the inequality comparison is evaluated first, and then the boolean one. or has a lower precedence than !=; 2) letter != 'O' or letter != 'Q' is always True. Operator precedence summary table: docs.python.org/reference/expressions.html#summary

mechanical_meat · Accepted Answer · 2012-03-25 07:19:49Z

5

Please note that

if letter != 'O' or 'Q':

is in fact

if (letter != 'O') or 'Q':

That is probably not what you wanted.

Just a small test on top of it:

>>> True != False or True
True
>>> (True != False) or True
True
>>> True != (False or True)
False

Note: This means that the answer marked on top is not true, letter is not compared to the result of O or Q...

edited Mar 25, 2012 at 7:19

mechanical_meat

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answered Mar 25, 2012 at 6:54

tchap

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Comments

Karl Knechtel · Accepted Answer · 2012-03-25 12:09:56Z

2

Python is not COBOL or some other language which supports this syntax. As a start I would suggest you read Expressions.

Now coming back to your problem, what you expect from the statement

if letter != 'O' or 'Q':

definitely

if letter != 'O' or letter != 'Q':

interestingly Python allows you to think laterally. For example you might also say,

letter not in ['O','Q']

or simply

letter not in 'OQ': #In Python Notation

or can be more expressive like

if all(letter != x for x in 'OQ'):

Just compare the above mentioned syntax and usage with yours

When you wrote

if letter != 'O' or 'Q':

which in Python should be written as

if letter not in 'OQ':

or may even be

if all(letter != x for x in 'OQ'):

edited Mar 25, 2012 at 12:09

Karl Knechtel

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answered Mar 25, 2012 at 6:50

Abhijit

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1 Comment

Karl Knechtel Over a year ago

I slightly adjusted a couple of your examples so that they would actually work.

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