Project Euler module

I use Project Euler to teach me programming and not to submit any results. As such I look up the expected return values to double check my solutions.

To organise my files I use the following folder structure:

main.py
\euler # The problem files
     __init__.py # empty
     e001.py
     e002.py
     ...
\input # Additional input files
     8.dat
     11.dat
     ...

My main.py file is the common entry point. It can either run all the solved examples so far or a specific one. This second option is added that I don't need to add an if __name__ == '__main__' guard in every file. The file looks as follows:

TOP_LEVEL = "euler"

def run_module(num):
    """Run specific Problem"""
    mod = importlib.import_module('%s.e%0.3i' % (TOP_LEVEL, num))

    start = time.time()
    ist = mod.run()
    print("  %5i | %6.3f |  %s  | %i" % \
(num, time.time() - start, "ox"[ist == mod.SOLL], ist))

if __name__ == '__main__':
    N_MAX = 67

    # Pre Header
    print('Problem |  Time  | x/o | Solution')
    print("--------+--------+-----+---------")

    global_time = time.time()

    # Run over all problems
    if len(sys.argv) == 2:
        run_module(int(sys.argv[1]))
    else:
        for num in range(1, N_MAX + 1):
            run_module(num)

    # End Header
    print("--------+--------+-----+---------")
    print("Total: %.3f s" % (time.time() - global_time))

I'll show now two example files to show the source files and how old code can be reused. e018.py:

```
"""By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below"""

SOLL = 1074

def run(file = "input/18.dat"):
# Parse File
with open(file) as fid:
tri = [[int(num) for num in line.split(' ')] for line in fid.read().split('\n')]

# From bottom's up find the maximal value
for row

Use .format

Using % is considered old style, so:

print("  %5i | %6.3f |  %s  | %i".format(
    (num, time.time() - start, "ox"[ist == mod.SOLL], ist)))

Be generous with long variables names

For example ist is impossible to understand for me, solution is more natural

Don't abuse Code-Golf techniques

The line:

"ox"[ist == mod.SOLL]

is a well known Code-Golf trick that relies on implicit boolean to integer conversion that is equivalent to:

"x" if ist == mod.SOLL else "o"

please use the latter.

Use argparse

C-Style arguments such as sys.argv[1] should be avoided, I suggest argparse (help to get started here: https://stackoverflow.com/questions/7427101/dead-simple-argparse-example-wanted-1-argument-3-results)

Don't shadow built-ins

def run(file = "input/18.dat"):
    # Parse File
    with open(file) as fid:

file is a built-in, you should use file_ in your code.

Explode long lines

Divide the following lines in two please.

tri = [[int(num) for num in line.split(' ')] for line in fid.read().split('\n')]

StackExchange Code Review Q#78745, answer score: 3