On 04/16/2013 08:37 AM, aaB wrote:
hello,
<snip>
I represent the CA's rule with a list of integers, of value 1 or 0.
Here is the function I use to generate the list:
def get_rule(rulenum):
rule = []
while rulenum > 0:
rule.append(rulenume % 2)
rulenum /= 2
while len(rule) < 8:
rule.append(0)
rule.reverse()
return rule
It appears what you are trying to do is create an eight-element list corresponding to the binary
value of rulenum. Here is a different approach, using some other functions and features of
Python...
First you can use the bin() function to get the binary value of an integer as a string. But
this string has a leading '0b', which can be cut off with slicing. eg. (using your example
value of 8 and bs as the variable name for the binary string)
bin(8) gives the string '0b1000', so slicing off the two leading characters:
bs = bin(8)[2:]
bs becomes the string '1000'
Now, the number of leading zeros needed is (8 - len(bs)). You can create the initial list of
leading zeros with the statement:
rule = [0] * (8 - len(bs)).
This gives rule as the list [0, 0, 0, 0].
Finally use a for loop to append the rest if the binary string as integers
for i in bs:
rule.append(int(i))
So your function becomes (with a few changed variable names):
def get_rule(num):
bs = bin(num)[2:]
rule = [0] * (8 - len(bs))
for i in bs:
rule.append(int(i))
return rule
This can be shortened even more with a list comprehension. You can think of a list
comprehension essentially as a terse variation of a for loop specifically to create a list.
So using a list comprehension you can do it in two lines:
def get_rule(num):
bs = bin(num)[2:]
return [0] * (8 - len(bs)) + [int(i) for i in bs]
Again, I'm assuming you specifically want an 8-element list. But this could easily be
generalized for any length by adding a second function parameter, say size, and change the 8 in
the last statement to size.
<snip>
Any pointers to help me understand this would be appreciated.
I am also running into an other issue with this script, but I'd rather
understand this before asking further questions.
Thanks, and sorry for the rather long post.
Don't worry about the length -- my reply is plenty wordy as well. I'm hoping I went into enough
detail that you can understand it without it being too advanced. Mainly I was just trying to
show you a different approach using other features of Python. There's nothing wrong with your
approach, it is quite straight-forward. In fact I used the same basic approach when I was
writing a C function to commify the display of integers -- that is, to add commas every third
digit.(*) Just about anything in programming can be approached in several different ways. I'm
not saying mine is better, and certainly not the best, just different. I'm not a newbie but FAR
from an expert. Keep learning Python, I think you'll like it.
(*)A special function for this isn't necessary in Python, it's already built in the the
new-style string formatting. Try this statement: '{:,}'.format(1000000)
(That's a colon and comma inside the curly braces.)
-=- Larry -=-
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