Help building a dictionary of lists

Joshua Landau at
Tue Nov 13 01:04:16 CET 2012

On 12 November 2012 22:26, NJ1706 <nickj1706 at> wrote:

> # List of tests
> TestList = (
>     'Test_1',
>     'Test_2'
> )

Note that TestList is a *tuple*, not a list.

You normally would want to write "test_names" instead of "TestList" for
several reasons:

* Unless it's a class, Python prefers lowercase_names_with_underscores (or
sometimes camelCaseNamesThatLookLikeThis)
* Python tries not to care what type you have - that your names are in a
list is less important than that you can loop over it
* It's definitely not a list. A list would look like this:
TestList = [

Additionally, your comment here is somewhat pointless.  The code says
*exactly* what your comment does, so why write it twice?

# List of tests
Don't need

# Initialise the dictionary of lists
Don't need

# Loop through the list of tests
Don't need

    # Append to the list for each instance
    Not very helpful. If you comment here, you should explain *why* you are
looping, not *that* you are looping. You write "append", but *what* and,
more importantly,*why*?

        # Initialise our string list
        Don't need

        # Build string list
        Again, whilst it is nice to know more, you are not saying *what*
you are building. What is the *purpose* of the list?

        # Convert to string
        Don't need

        # Assign to target list
        Don't need

A good comment might just be one at the start that explains the goal, or
one inside that explains the *point* of a piece of code. Consider these two
commented pieces of code:

# Define a function that takes a, b and c
def f(a, b, c):

    # Set s to ((the square of b) minus four times a times c) all to the
power of 0.5
    s = (B**2 - 4 * a * c) ** 0.5

    # Return the calculated values
    return (- b + s) / (2 * a), (- b - s) / (2 * a)


# Finds x where the equation "ax**2 + bx + c" makes 0
def f(a, b, c):
    # Use the quadratic equation (

    # s is the square root part of the equation (where x**0.5 is the square
root of x)
    s = (B**2 - 4 * a * c) ** 0.5

    # Return two values because there are two possible solutions
    return (- b + s) / (2 * a), (- b - s) / (2 * a)
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