Dave Angel davea.name> writes:
>
> > str_num = '1234567890'
> > n = 5
> >
> > strings = [str_num[i:i+5] for i in range(0, len(str_num)) if
> > len(str_num[i:i+5])==5]
>
> If you changed the range() size, you could eliminate the extra if test.
> After all, the only ones that'll be short are t
On 31/05/13 21:49, Nick Shemonsky wrote:
I did stumble upon using reduce ...but I didn't really
understand what it was doing
def product(intlist):
return reduce(operator.mul, intlist)
I explain reduce thusly in my Functional Programming topic:
-
The reduce function
On 05/31/2013 04:49 PM, Nick Shemonsky wrote:
Here's the final code... I kept the if statement that way if I throw
in a random series of numbers that isn't evenly divisible by 5, it'll
always work itself out. And this answered the 1000 digit problem
without issue.
str_num = '1234567890'
n =
Thanks for the responses. I am using python 2.7. I'm not new to
programming but might as well be... I last programmed heavily about a
decade ago in college. I was a MIS major so I did my fair share of
c++, sql, and php work but now I'm a windows sys admin so I haven't
used it much at all in a long
On 31/05/13 19:23, Nick Shemonsky wrote:
or maybe it'd be quicker to compare a to b through each iteration and
just keep the larger product rather than creating a giant list
You are probably right if it is a giant list.
str_num = '1234567890'
n = 5
strings = [str_num[i:i+5] for i in range(0
On 05/31/2013 02:23 PM, Nick Shemonsky wrote:
Hey there everybody. I'm new to python
Welcome. But are you new to programming, or just to Python in
particular? And which Python? I'd guess 2.7
and am attempting to teach
myself to code while brushing up on my math skills via the problems at
Hey there everybody. I'm new to python and am attempting to teach
myself to code while brushing up on my math skills via the problems at
projecteuler.net. My solutions thus far have been mostly brute force
and this is no exception but I'm having an issue with tackling some
iteration in the problem.
Steven D'Aprano wrote:
On Mon, 30 Aug 2010 05:31:00 am bob gailer wrote:
My current reasoning was something of this sort: Find all the
factors of a number, then reduce them to just the prime factors
Very inefficient. IMHO the proper way is to generate a list of all
the prime numbers
"Nick" wrote
I didn't call the functions in the program because I was
calling them myself in the interpreter after running it.
I assume you mean after importing it?
Running a program is generally taken to mean executing the script as
a standalone program. To execute the internal functions
On Mon, 30 Aug 2010 05:31:00 am bob gailer wrote:
> > My current reasoning was something of this sort: Find all the
> > factors of a number, then reduce them to just the prime factors
>
> Very inefficient. IMHO the proper way is to generate a list of all
> the prime numbers up to the square root
"Did you test the program? That is one way to tell whether it works perfectly.
What you showed above will do one visible thing - it will print "Don't forget
to consider primes 2, 3, 5, and 7\n". The rest is a somewhat confusing
collection of function definitions and comments. You never call the
On 8/29/2010 3:08 AM, Nick wrote:
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
#don't forget 2,3,5,7. this function doesn't deliver those as output.
def is_prime(b): #checks a number greater than 7 to see if it is
prime and
"Nick" wrote
What is the largest prime factor of the number 600851475143 ?
For help on the math aspects try Wikipedia.
Look up Prime factors...
Would it be useful for me to buy a book, and if so what are some
easily accessible ones? I feel dive into python is just too
advanced for me.
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
#don't forget 2,3,5,7. this function doesn't deliver those as output.
def is_prime(b): #checks a number greater than 7 to see if it is prime and
returns if is.
if b % 2 != 0 and b
prasad rao schreef:
> hello!
> I got it 266333.
> My code==
>
> t=0
> for x in range(1000):
> if divmod(x,3)[1]==0:t+=x
> if divmod(x,5)[1]==0:t+=x
> t=266333
>
> Am I correct in comprehention of the problem?
Not entirely: you're counting numbers that are multiples of both 3 and 5
double
hello! I got it 266333.
My code==
t=0
for x in range(1000):
if divmod(x,3)[1]==0:t+=x
if divmod(x,5)[1]==0:t+=x
t=266333
Am I correct in comprehention of the problem?
Prasad
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It says: Find the difference between the sum of the squares of the first *
one
hundred* natural numbers and the square of the sum.
You did range(1,111).
On Thu, Apr 24, 2008 at 1:13 PM, kinuthia muchane <[EMAIL PROTECTED]>
wrote:
> Hi,
>
> I am trying to solve problem 6 on the Project Euler, bu
On Thu, Apr 24, 2008 at 11:13:31PM +0300, kinuthia muchane wrote:
> return sum([k*k for k in range(1,111)])# sum of the squares of the
> first one hundred numbers
Wouldn't the first hundred numbers be range(1,101)?
> def aux1():
> inter = sum([k for k in range(1,111))# square of the s
Hi,
I am trying to solve problem 6 on the Project Euler, but when I submit
my answer, I am told that it is wrong. Here is the problem:
The sum of the squares of the first ten natural numbers is,
1² + 2² + ... + 10² = 385
The square of the sum of the first ten natural num
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