Can you make a more elegant solution? (Python)

[triptoe]

first of all i am not trying to say this solution *IS* elegant.. i am just curious of how i could do it better since I am a begginer. It is problem #11 from the euler site here:

http://projecteuler.net/index.php?section=problems&id=11



# search a grid horizontally, vertically, diag for the largest product of 4 #consecutive values
import re


pat = "[0-9]{2}"
array = []
n = (open("grid.dat").readlines())

def formArray(n):

for line in n:
array.append(re.findall(pat,line))

# convert to int

for i in range(20):
for y in range(20):
array[i][y] = int(array[i][y])





def checkRows(array):
x = 1
for index in range(len(array)):
i = 0
while i < len(array[index]) -3:
y = array[index][i] * array[index][i+1] * array[index][i+2] * array[index][i+3]
if y > x:
x = y
i += 1
print x



def checkCols(array):
x = 1
for index in range(len(array)):
i = 0
while i < len(array) -3:
y = array[i][index] * array[i+1][index] * array[i+2][index] * array[i+3][index]
if y > x:
x = y
i += 1
print x

def checkNeg(array):
newArray = [[]]
p = 16
while p >= 0:
repet = 0
while repet < (len(array)-p):
newArray[0].append(array[p+repet][repet])
repet += 1
p -= 1
p = 0
while p <= 16:
repet = 0
while repet < (len(array)-p):

newArray[0].append(array[repet][repet+p])
repet += 1
p += 1
checkRows(newArray)

def checkPos(array):
newArray = [[]]
p = 19
while p >= 4:
repet = 0
while repet < (p+1):
newArray[0].append(array[p-repet][repet])
repet += 1
print
p -= 1
checkRows(newArray)


formArray(n)
checkRows(array)
checkCols(array)
checkNeg(array)
checkPos(array)

[apetresc]

Yeah, I got the right answer (70600674) in just a few lines:

grid = [[0] * 23 for x in range(23)]
rows = file('data.in','r').read().split("\n")
for i in range(20):
for j in range(20):
grid[i][j] = int(rows[i].split(" ")[j])
currMax = 0
for i in range(20):
for j in range(20):
currMax = max(currMax, grid[i][j] * grid[i+1][j+1] * grid[i+2][j+2] * grid[i+3][j+3],
grid[i][j] * grid[i+1][j] * grid[i+2][j] * grid[i+3][j],
grid[i][j] * grid[i][j+1] * grid[i][j+2] * grid[i][j+3],
grid[i][j] * grid[i-1][j+1] * grid[i-2][j+2] * grid[i-3][j+3])
print currMax
All you need to do is stick the grid data in a file called "data.in" (copy+paste from Euler), and run the script:

adrian@rootbeer:~$ time python problem11.py
70600674

real 0m0.047s
user 0m0.036s
sys 0m0.008s

If you need help understanding this code, let me know, I'd be glad to help you out :)

[triptoe]

Yeah, I got the right answer (70600674) in just a few lines:

grid = [[0] * 23 for x in range(23)]
rows = file('data.in','r').read().split("\n")
for i in range(20):
for j in range(20):
grid[i][j] = int(rows[i].split(" ")[j])
currMax = 0
for i in range(20):
for j in range(20):
currMax = max(currMax, grid[i][j] * grid[i+1][j+1] * grid[i+2][j+2] * grid[i+3][j+3],
grid[i][j] * grid[i+1][j] * grid[i+2][j] * grid[i+3][j],
grid[i][j] * grid[i][j+1] * grid[i][j+2] * grid[i][j+3],
grid[i][j] * grid[i-1][j+1] * grid[i-2][j+2] * grid[i-3][j+3])
print currMax
All you need to do is stick the grid data in a file called "data.in" (copy+paste from Euler), and run the script:

adrian@rootbeer:~$ time python problem11.py
70600674

real 0m0.047s
user 0m0.036s
sys 0m0.008s

If you need help understanding this code, let me know, I'd be glad to help you out :)

I changed your code to print statements to see what it was doing.... very nice!

don't think it can get much better.

[xtacocorex]

AdrianP, the first line in your code just helped me out immensely. I've been trying to take my RK2-Heun (Runge-Kutta 2nd Order Heun Method) FORTRAN 90 code, that I wrote in school to solve differential equations for my Computational Fluid Dynamics course, and put it into Python.

I was having problems because there are no proper arrays like FORTRAN in Python and couldn't figure out how to initialize a nice 2 by X array. Now I can finish it.

[apetresc]

AdrianP, the first line in your code just helped me out immensely. I've been trying to take my RK2-Heun (Runge-Kutta 2nd Order Heun Method) FORTRAN 90 code, that I wrote in school to solve differential equations for my Computational Fluid Dynamics course, and put it into Python.

I was having problems because there are no proper arrays like FORTRAN in Python and couldn't figure out how to initialize a nice 2 by X array. Now I can finish it.

Excellent :D Glad I was helpful. List comprehensions are one of Python's nicest features, and they can do a lot more than what I've used them for here. Check out, for example, this page (http://www.secnetix.de/~olli/Python/list_comprehensions.hawk) for some other things you can do with them :)

[xtacocorex]

Excellent :D Glad I was helpful. List comprehensions are one of Python's nicest features, and they can do a lot more than what I've used them for here. Check out, for example, this page (http://www.secnetix.de/~olli/Python/list_comprehensions.hawk) for some other things you can do with them :)
The problem I have with Python and the code I'm trying to port over is that lists are almost too powerful for what I need; slightly an odd conundrum, but it happens.

Thanks for the link, will definitey look at it for harnessing the power of the list.

[triptoe]

is it just me... or could it have been worded better in list comprehensions built into python?

for instance:


>>> S = [x**2 for x in range(10)]
>>> V = [2**i for i in range(13)]
>>> M = [x for x in S if x % 2 == 0]

Wouldn't it be more readable if it was this:


S=[for x**2 in range(10)]
>>> V = [for 2**i in range(13)]
>>> M = [for x in S if x % 2 = 0]

or am i missing something??

[Klipt]

I was having problems because there are no proper arrays like FORTRAN in Python and couldn't figure out how to initialize a nice 2 by X array.

Not in Python as such, but if you're doing serious numerical workouts you should consider the numpy (numeric python) and scipy (scientific python) packages :-D