Learning Python...

[tiachopvutru]

... that's the name of the book I was learning Python from...

Ah wait, that wasn't my point. :) Anyway, I have a question to ask about what I should do from here. Currently, I'm onto a part (which consists of several chapters) about module and reading a chapter about importing a package of modules. There is still another chapter on module before I get on the two last parts about coding in OOP and Exceptions.

Since the rest of what I still need to learn from the book consist of more than 200 pages, and I really want to code badly, I'm in need of some directions or guidance. I'll continue reading the book as I go, though. Anyway, since the book implies (or rather, what I think it implies) that the knowledge I have so far should be sufficient for some (very) basic coding tasks, I think I am able to at least do something.

I will admit now that I probably didn't go through learning it with as much careful attentions to details as I should have (although I doubt I will be able to remember much anyway without doing something along the way). And back to the original topic, I have tried something like http://www.pythonchallenge.com/ but couldn't even get last Level 1. I'll stop here, though, to prevent any further damage to others' impression on me. :)

It would be sweet if someone says that I can actually code some sorts of desktop applications already... though I doubt it.

[lut4rp]

If you have prior programming experience, read Dive Into Python. Its a much smaller, and faster book, and is freely available as download (in various formats) and to read online also.
ref. http://www.diveintopython.org

[elithrar]

In terms of tackling a project or starting somewhere: what takes your interest? Web development, GUI applications, network programming, CLI apps/scripts?

[tiachopvutru]

If you have prior programming experience, read Dive Into Python. Its a much smaller, and faster book, and is freely available as download (in various formats) and to read online also.
ref. http://www.diveintopython.org

Well, I'm not really looking for a book to learning python right now, as I still have to finish the one that I'm reading...


In terms of tackling a project or starting somewhere: what takes your interest? Web development, GUI applications, network programming, CLI apps/scripts?

If it's only about my interest, then I don't really have a preference yet. Heck, I don't know if I can even create a program right now.

However, out of the ones that you listed, I probably would go for GUI applications (this means making GUI out of already-existed command-line programs, right?). Network programming sounds cool but... complicated. I'm not interested in Web development, and do not know what CLI apps/scripts are.

[WW]

It looks like your book is Learning Python (http://safari.oreilly.com/9780596513986) from O'Reilly. Have you completed the "Brain Builder" exercises for the chapters that you have read so far?

[pmasiar]

If you need bunch of little problems to train on (you did solved all "homework" tasks, right?), try Project Euler. 200 tasks, and you will see how hard it is by how many people succeeded in solving it.

[tiachopvutru]

It looks like your book is Learning Python (http://safari.oreilly.com/9780596513986) from O'Reilly. Have you completed the "Brain Builder" exercises for the chapters that you have read so far?

Yes, somewhat...


If you need bunch of little problems to train on (you did solved all "homework" tasks, right?), try Project Euler. 200 tasks, and you will see how hard it is by how many people succeeded in solving it.

In a way, yes, I do need a bunch of little problems to train on since I'm prone to forget stuffs (mostly methods, since there are so many of them) that I learn.

As for Project Euler, is it designed for beginners? Well, I guess, skill-wise, I fit in the category of "students for whom the basic curriculum is not feeding their hunger to learn (not really)." I just looked at the first question, but it seems like something I will be able to solve (hopefully) rather than something I can't even get past level 1 (like Python Challenge)... Also, do I register there to check my answer?

In addition, may I ask what skills I need to build on before I can actually start coding some sort of general applications?

EDIT: Anyway, I think I have completed Problem #1:

L = [x for x in range(1000) if x % 3 == 0 or x % 5 == 0]

solution = 0
for each in L:
solution += each

print solution

[ghostdog74]

L = [x for x in range(1000) if x % 3 == 0 or x % 5 == 0]

solution = 0
for each in L:
solution += each

print solution




print sum([x for x in range(1000) if x % 3 == 0 or x % 5 == 0])

[tiachopvutru]

print sum([x for x in range(1000) if x % 3 == 0 or x % 5 == 0])


Ah, thank you. I forgot about the sum built-in function.

Anyway, I was able to solve Problem #2 (I think), but it didn't turn out to be as short as I had hoped...

Problem:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

Find the sum of all the even-valued terms in the sequence which do not exceed four million.

My solution:

L = [1, 2]
i = 0

while True:
added = L[i] + L[i+1]
if added <= 4000000:
L.append(added)
i += 1
else:
L = filter((lambda x: x % 2 == 0), L)
print sum(L)
break

[pmasiar]

if x % 3 == 0: # hard way
if not x % 3: # easy way
# read: if no remainder after division
:-)

[-grubby]

Not to highjack this thread, but project Eueler looks interesting (from what I've read from this forum page). Too bad I really suck at math.

[ghostdog74]

if x % 3 == 0: # hard way
if not x % 3: # easy way
# read: if no remainder after division
:-)

Some test runs, taking large numbers. All runs show the hard way runs a bit faster than the easy way. Guess it has to spend time "inverting" the result because of the "not"


# python -c "print sum([x for x in xrange(10000000) if x % 3 == 0 or x % 5 == 0])"
23333331666668
# python -c "print sum([x for x in xrange(10000000) if not x % 3 or not x % 5])"
23333331666668
# time python -c "print sum([x for x in xrange(10000000) if x % 3 == 0 or x % 5 == 0])"
23333331666668

real 0m13.087s
user 0m12.833s
sys 0m0.240s
# time python -c "print sum([x for x in xrange(10000000) if not x % 3 or not x % 5])"
23333331666668

real 0m13.311s
user 0m13.197s
sys 0m0.116s
# time python -c "print sum([x for x in xrange(10000000) if x % 3 == 0 or x % 5 == 0])"
23333331666668

real 0m12.551s
user 0m12.313s
sys 0m0.228s
# time python -c "print sum([x for x in xrange(10000000) if not x % 3 or not x % 5])"
23333331666668

real 0m13.240s
user 0m12.969s
sys 0m0.248s
# time python -c "print sum([x for x in xrange(10000000) if x % 3 == 0 or x % 5 == 0])"
23333331666668

real 0m12.688s
user 0m12.453s
sys 0m0.188s
# time python -c "print sum([x for x in xrange(10000000) if not x % 3 or not x % 5])"
23333331666668

real 0m12.922s
user 0m12.801s
sys 0m0.116s

[tiachopvutru]

Well, I'm getting stuck at Problem #3


The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

The problem is about find prime factors of a number, and here's my code:


num = 600851475143
answer = []

for x in xrange(2, num/2 + 1):
while num % x == 0:
answer.append(x)
num = num/x

print answer


When I run it, I get this error:

OverflowError: long int too large to convert to int


I vaguely remember something about long int being int with an L behind it, but I don't know how to get around this problem...

Also, since the variable num constantly changes, how would it affect the for loop?

@pmasiar: Your example makes sense, but if x%3==0 clicks to me more than if not x%3 :)

[thornmastr]

Not to highjack this thread, but project Eueler looks interesting (from what I've read from this forum page). Too bad I really suck at math.

Instead of Eular, take a look at http://challenge-you.appspot.com/

These are great for people learning Python, or people who know the rudiments of Python and want to build on what they already know.

Since I come from a C++ background and have been doing Python for three weeks, these challenges are ideal and certainly fit my current requirements.

[pmasiar]

I vaguely remember something about long int being int with an L behind it, but I don't know how to get around this problem...

Don't worry about the types, Python will do most typecasting for you. The only need to worry is in integer division: use 5/2.0 instead of 5/2 if you don't want integer division


@pmasiar: Your example makes sense, but if x%3==0 clicks to me more than if not x%3 :)

I know, I mistakenly thought that 'not x%3' is faster. As ghostdog74 found for me, I was wrong.

BTW, I found again that in Python "there is one obvious way to do it right". Thanks, Guido! :-)

[tiachopvutru]

Don't worry about the types, Python will do most typecasting for you. The only need to worry is in integer division: use 5/2.0 instead of 5/2 if you don't want integer division

Actually, my bad. I forgot to post the whole error message.

I got this error:

Traceback (most recent call last):
File "C:\Documents and Settings\Administrator\Desktop\ch06.py", line 4, in <module>
for x in xrange(2, num/2 + 1):
OverflowError: long int too large to convert to int


But eh, I now want to write application even more. =p

[Lster]

print sum([x for x in range(1000) if x % 3 == 0 or x % 5 == 0])

You might be interested to know that this code is faster still:



def add(n, x):
q = n/x
r = n%x

return q*(x+n-r)/2

t = add(10**3-1, 3)
t += add(10**3-1, 5)
t -= add(10**3-1, 15)

print t


Not that it really matters for such small numbers. But for values like 10^12 it does.

[WW]

Actually, my bad. I forgot to post the whole error message.

I got this error:

Traceback (most recent call last):
File "C:\Documents and Settings\Administrator\Desktop\ch06.py", line 4, in <module>
for x in xrange(2, num/2 + 1):
OverflowError: long int too large to convert to int


But eh, I now want to write application even more. =p

From Python Library Reference (http://docs.python.org/lib/built-in-funcs.html):


Note: xrange() is intended to be simple and fast. Implementations may impose restrictions to achieve this. The C implementation of Python restricts all arguments to native C longs ("short" Python integers), and also requires that the number of elements fit in a native C long.

The number 600851475143 is larger than the maximum value that can be represented by a long int.

[Wybiral]

Some test runs, taking large numbers. All runs show the hard way runs a bit faster than the easy way. Guess it has to spend time "inverting" the result because of the "not"

Those results don't look very conclusive to me. Too much fluctuation and not enough tests run (it's also factoring in Python startup/parse time and well, it's parsing two different chunks of code). The difference isn't greater than the overall variance.

[ghostdog74]

Those results don't look very conclusive to me. Too much fluctuation and not enough tests run (it's also factoring in Python startup/parse time and well, it's parsing two different chunks of code). The difference isn't greater than the overall variance.

well, you can try them out on your environment and see if its conclusive. Python startup/parse time is not an issue because the environment are the same for both scenarios.

[tiachopvutru]

You might be interested to know that this code is faster still:



def add(n, x):
q = n/x
r = n%x

return q*(x+n-r)/2

t = add(10**3-1, 3)
t += add(10**3-1, 5)
t -= add(10**3-1, 15)

print t


Not that it really matters for such small numbers. But for values like 10^12 it does.

Ah, thank you. It took me quite a while to figure out the code (although not concrete enough to be able to explain in words), but it's amazing. :)


From Python Library Reference (http://docs.python.org/lib/built-in-funcs.html):

The number 600851475143 is larger than the maximum value that can be represented by a long int.

Well, if I change xrange to range, I receive another error on the resulting list being too long. Does that mean that I ought to use a different and more efficient algorithm in order to solve this problem?

[tiachopvutru]

Well... seem like all I had to do was use a while loop instead of a for loop. I think I got it right...


num = 600851475143
answer = []
i = 2

if num % i == 0:
answer.append(i)
num = num/i

i = 3

while num != 1:
if num % i == 0:
answer.append(i)
num = num/i
else:
i+=2

print answer


I wonder if I would save some performance if I start i from num and slowly decreasing it instead of starting from 2 and slowly increasing it. I would then have to check if the answer is prime, though...

[Lster]

Ah, thank you. It took me quite a while to figure out the code (although not concrete enough to be able to explain in words), but it's amazing.


Let's define add(n, d) to be:

"The sum of all integers, less than or equal to n, that are divisible by d. For example d(10, 3) = 18."

Therefore, add(999, 3) gives the number of numbers less than 1000 (≤ 999) divisible by 3. Similarly, add(999, 5) gives the same but with 5. However, adding them together overestimates the answer for the reason that we have now included multiples of 15 twice (as both 3 and 5 are factors of it). So we have to take off add(999, 15).

The add function is simpler. We calculate the number of divisors (q) and then the midpoint which is the first term (x) plus the last term (n-r) all over 2.

Here is a clearer example:



def add(n, x):
terms = n/x #Number of terms
r = n%x #Remainder from division

first = x #First term
last = n-r #Last term. r is taken to make last divisible by x

#Average = (first+last)/2
return terms*(first+last)/2

t = add(999, 3)
t += add(999, 5)
t -= add(999, 15) #Remove overlap

print t



I wonder if I would save some performance if I start i from num and slowly decreasing it instead of starting from 2 and slowly increasing it. I would then have to check if the answer is prime, though...

You might be interested in SymPy which is a very good symbolic maths library. I think it is an available package in Synaptic so it should be easy to install. Then you can try out the stuff on:

http://docs.sympy.org/

And it also has an isprime() function! :)