Scott W Dunning wrote:
>
> On Feb 23, 2014, at 2:26 AM, Peter Otten <__pete...@web.de> wrote:
>> If you want to make rows with more or less stars, or stars in other
>> colors you could add parameters:
>>
>> def star_row(numstars, starcolor):
>> for i in range(numstars):
>> fillstar(starcolor)
>> space(25)
>>
>> Your code will then become
>>
>> star_row(6, red)
>> row(25)
>> star_row(5, red)
>> row(25)
>>
> I have a question with the above loop function. Why couldn’t row(25) be
> added into the function so that wouldn’t have to placed in between every
> star_row()?
That's of course possible.
>> which still shows a repetetive pattern and thus you can simplify it with
>> another loop. You should be able to find a way to write that loop with
>> two star_row() calls on a single iteration, but can you do it with a
>> single call too?
> Not sure I understand what you mean in the above paragraph?
What you found out later and put in you next post:
This repetetive pattern
> star_row(5) # oops, typo
> row(25)
> star_row(5)
> row(25)
> star_row(6)
> row(25)
> star_row(5)
> row(25)
> star_row(6)
> row(25)
> star_row(5)
> row(25)
> star_row(6)
> row(25)
> star_row(5)
> row(25)
> star_row(6)
can be turned into this for loop:
> This is what I’m thinking…
>
> for I in range(4)
> star_row(5)
> row(25)
> star_row(6)
> row(25)
>
> Am I at all close?
You hit the jackpot :) Now on to the next challenge:
for row_index in range(9):
row_width = ...
star_row(row_width)
row(25)
Can you replace the ... with a calculation to get row_width=6 for even and
row_width=5 for odd rows?
Hint: look at the % ("modulo") operator.
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