Sailor_Spoon posted:

Depends on your user, I guess. I'd much rather get an error then have the function return None, which could mean a lot of things.

Also, does a file that's not seekable still have the seek method? I assumed it does, so an error on seek would presumably not get caught by that except.

tef posted:

tef posted:

code:

with open(...) as fh:
     lines = reversed(fh.readlines())

Rohaq posted:

So, what are my alternatives to using gigantic lists in Pyth 2.7? I'm trying to use matplotlib to produce a pretty graph the results generated by a diff tool I created for large binary files, which produces a text file with the start/end byte of differences, but some as far as I can see, I'd have to create some goddamned massive lists in memory in order for matplotlib to do it's thing. Is there any way to make it use ranges when generating graphs, or am I stuck?

tripwire posted:

Use numpy arrays.

tripwire posted:

Yeah but in python 2.X that has to read the whole file into memory at once, no? I've written a grungy generator before for an environment where I'm forced to use python 2.4; it would spit the file out backwards using seek, while bundling it up into chunks delimited by the line seperator. I'm not really proud of that, and god knows how it would behave on file-like objects which don't seek like a file does.

tripwire posted:

Use numpy arrays.

DirtyDiaperMask posted:

Wouldn't f.readlines() still read the entire file into memory even in Python 3?

Jam2 posted:

In Mathematica, it's simple to test whether an element, n, is an integer. I can type Element[n,Integers] and get a true or false. How do I do a similar test in python?

Jam2 posted:

In Mathematica, it's simple to test whether an element, n, is an integer. I can type Element[n,Integers] and get a true or false. How do I do a similar test in python?

Jam2 posted:

In Mathematica, it's simple to test whether an element, n, is an integer. I can type Element[n,Integers] and get a true or false. How do I do a similar test in python?

MaberMK posted:

code:

>>> x = 0
>>> type(x) == int
True

Jam2 posted:

code:

If type(x) == int == True:
...

This doesn't seem right for Python though.

Lonely Wolf posted:

== : any X any -> bool. You don't need == True. Also If and if are different things as far as Python knows. Are you trying to find out if the type of a variable's value is int or are you trying to determine if a value is integral? Those are two very different things and everyone is answering the former question.

Jam2 posted:

Why is Python turning x = n/15 into a float when n = 15. 1.0 is a goddamn integer Stop making my life difficult.

Thanks for the pitfall prevention.

I need to be able to tell when a number(a) is cleanly divisible by another number(b). Previously, I would test whether a/b == int. Is there a better way to do this in Python?

Janin posted:

Python has three relevant numeric types: 'int', for arbitrary-size integers, 'float', for 64-bit floating-point, and 'decimal.Decimal', for arbitrary-precision rationals.

If you want to test whether a value is of type 'int', use type() or isinstance().

If you want to test whether a value is integral, use the modulus operator %:

code:

def integral(x):
  return x % 1 == 0

print integral(1) # True
print integral(1.0) # True
print integral(1.5) # False

quote:

For (plain or long) integer division, the result is an integer. The result is always rounded towards minus infinity: 1/2 is 0, (-1)/2 is -1, 1/(-2) is -1, and (-1)/(-2) is 0. Note that the result is a long integer if either operand is a long integer, regardless of the numeric value.

Jam2 posted:

Why is Python turning x = n/15 into a float when n = 15. 1.0 is a goddamn integer Stop making my life difficult.

Thanks for the pitfall prevention.

I need to be able to tell when a number(a) is cleanly divisible by another number(b). Previously, I would test whether a/b == int. Is there a better way to do this in Python?

hlfrk414 posted:

Are you using Python 3? Division always returns a float in 3. If you need integer division, you are supposed to use //, which works in both 2 and 3.

Jam2 posted:

Yeah, I'm using 3. If I use //, it turns all results into int's which invalidates my logic test. I want to know when a division operation results in an int naturally. What's a good way to get around this?

I'm solving the Hoppity Hop Facebook puzzle. I already solved it using Mathematica and now I'm trying my hand in Python. I've never used this lang before. It's all really new to me.

Tennis Ball posted:

So I just finished project Euler #3, What is the largest prime factor of the number 600851475143 ?:

It gives the right answer, but is there anything in there that is bad practice? I want to avoid these sorts of things before I form bad habits.

Tennis Ball posted:

So I just finished project Euler #3, What is the largest prime factor of the number 600851475143 ?:

code:

import math
i = 1
k = 1
j = 1
t = 0
list = []
factors = []

while i < math.sqrt(600851475143):
        if 600851475143%i==0:
                list.append(i)
        i = i + 1
while list!=[]:      
        k = list.pop()
        j = 1
        t = 0
        while j<k:
                if k%j==0:
                        t = t + 1
                j = j + 1
        if t > 2:
                print 'skipped'
        if t < 2:
                print k 

print 'finished'

It gives the right answer, but is there anything in there that is bad practice? I want to avoid these sorts of things before I form bad habits.

Stabby McDamage posted:

A few notes:

1. You should define any value you use more than once in a variable, instead of writing the literal number in multiple places. So put "value=600851475143" near the top, and reference that from then on. This makes the code more readable, and lets you adapt the code to take a value as an input later on.

2. Minor issue: putting the math.sqrt call in the while loop means that you're calculating the square root EVERY trip through the loop, even though it doesn't change. (I don't think python is smart enough to realize it's constant.)

3. "while list!=[]" → "while list"

4. Don't use a variable called "list", because list is the name of the type, and you may need to use that name later on.

5. Don't use a while loop when you can use a for. For example, change the first loop to a "for i in xrange(math.sqrt(value))".

5a. Better yet, for simple loops, use a list comprehension instead. For example, you can drop the whole first loop and replace it with "factors = [i for i in xrange(math.sqrt(value)) if value%i==0]"

6. You don't need to pre-define all your variables at the top if you're going to redefine them later. That goes for k, j, and t.

7. The algorithm of the second loop could be improved in a ton of ways. Most simply, the prime checker is very inefficient -- why check for divisibility by 4 when you've already checked 2? Why check 9 when you've checked 3? What's more, you could actually rewrite the whole program to be vastly more efficient by doing prime factorization to begin with on the number...how many small factors could you divide out of that number with very little work?

I'm sure there's a very clever two-line solution to this, but it's 2AM here, so I'm too tired to think about it.

Jam2 posted:

Finally.

code:

def Hoppity(n):
	for i in range(1,n+1):
		x = i % 15
		y = i % 5
		z = i % 3
		if x==0:
			print('Hop')
		elif y==0:
			print('Hophop')
		elif z==0:
			print('Hoppity')

tef posted:

the purpose of that puzzle is to get people comfortable with the submission system, beyond hello world.

tripwire posted:

Jam2, if you are coming from mathematica or matlab, you may want to check out numpy/scipy and matplotlib. They are python libraries for numerical/scientific computation, and graphing/plotting, respectively.