# 3 Common Mistakes that Python Newbies Make

1133 words. Time to Read: About 11 minutes.

Last weekend, I stared mentoring people on exercism.io on the Python track. I wasn’t sure what to expect, but over the last week I have mentored about 50 people, helping them get their solutions from “tests passing” to “tests passing, readable, and Pythonic.” I’m hooked. It’s a total blast. I’m going to write a post specifically on that experience. That’s not this post. This post is to talk about the three most common mistakes I saw over the last week and some possible alternatives that might be better! So let’s start the countdown!

## 1. Deep Nesting of If Statements or Loops

# Calculating whether or not 'year' is a leap year

if year % 4 == 0:
if year % 100 == 0:
if year % 400 == 0:
return True
else:
return False
else:
return True
else:
return False


A lot of times, I’ll pull a line from the Zen of Python to lead off my feedback to a “mentee” (not to be confused with a manitee). When I see this issue, I always lead with

Flat is better than nested.

If you look at your code with your eyes unfocused, looking at the shapes and not reading the words, and you see a bunch of arrows going out and back in again:

\
\
\
\
\
/
/
/
\
\
\
\
/
/
/
/
/
/


It’s not definitely a bad thing, but it is a “code smell,” or a Spidey Sense that something could possibly be refactored.

So, what can you do instead of nest? There are a couple things to try. The first is inverting your logic and using “early returns” to peel off small pieces of the solution space one at a time.

if year % 400 == 0:
return True
if year % 100 == 0:
return False
if year % 4 == 0:
return True
return False


If the number is divisible by 400, then we immediately return true. Otherwise, for the rest of our code, we can know that year is definitely not divisible by 400. So, at that point, any other year that’s divisible by 100 is not a leap year. So we peel off that layer of the onion by returning False.

After that, we can know that year is definitely not a multiple of 400 or 100, and the remainder of the code follows the same pattern.

The other way to avoid nesting is by using “boolean operators:” and, or, and not. We can combine if statements and thus, save ourselves a layer of nesting!

if year % 4 == 0 and (year % 100 != 0 or year % 400 == 0):
return True
else:
return False


Of course, that leads us to our second item…

## 2. Returning Booleans from If Statements

We’ll start with our last example from above:

if year % 4 == 0 and (year % 100 != 0 or year % 400 == 0):
return True
else:
return False


Anytime you find yourself writing:

if something:
return True
else:
return False


You should remember that the clause of an if statement is itself a boolean!

>>> year = 2000
>>> year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
True


So, why not type a little less and return the result of the boolean operation directly?

return (year % 4 == 0 and (year % 100 != 0 or year % 400 == 0))


Granted, at this point, the line may be getting a little long, but the code is a little less redundant now!

## 3. Lists are Like Hammers – Not Everything is a Nail

Here are two possible ways that this could show up:

some_numbers = [1, 2, 5, 7, 8, ...]
other_numbers = [1, 3, 6, 7, 9, ...]
# Let's try to combine these two without duplicates
for number in other_numbers:
if number not in some_numbers:
some_numbers.append(number)


Or:

data = [["apple", 4], ["banana", 2], ["grape", 14]]
# What fruits do we have?
for item in data:
print(item[0])
# => "apple" "banana" "grape"
# How many grapes do we have?
for item in data:
if item[0] == "grape":
print(item[1])
# => 14


In the first case, you’re trying to keep track of some groups of items and you want to combine them without duplicates. This is an ideal candidate for a set. Sets inherently keep track of their items (although not the order, so don’t use a set if the order is important). You can declare them with the built-in set() function or with squiggle braces ({}).

some_numbers = {1, 2, 5, 7, 8}
other_numbers = {1, 3, 6, 7, 9}
# Sets use the 'binary or' operator to do "unions"
# which is where they take all of the unique elements
some_numbers | other_numbers
# => {1, 2, 3, 5, 6, 7, 8, 9}

# You can even add single items in!
some_numbers.add(10)
# => {1, 2, 5, 7, 8, 10}

# But adding a duplicate doesn't change anything
some_numbers.add(1)
# => {1, 2, 5, 7, 8, 10}


In the second case, again, order probably isn’t critical. You want to keep track of some data by a “label” or something, but be able to keep them all together and list them out as necessary. This time, you’re probably looking for a dict. You can create those with either the dict() built-in function or, again, squiggle braces ({}). This time, however, you separate the labels (keys) and the values with a colon.

fruits = {
"apples": 4,
"bananas": 2,
"grapes": 14,
}


You can list out all of the keys (or values!).

list(fruits.keys())
# => ["apples", "bananas", "grapes"]
list(fruits.values())
# => [4, 2, 14]

# Or both!
list(fruits.items())
# => [("apples", 4), ("bananas", 2), ("grapes", 14)]


And you can ask it about (or give it a new value for) specific keys.

# How many grapes are there?
fruits["grapes"]
# => 14

# Not anymore.  I ate some.
fruits["grapes"] = 0

fruits["grapes"]
# => 0


Using a list, the your algorithm loops through every item to find the right one. dict’s are built to have very fast lookups, so, even if your dict is a bazillion fruits long, finding the grapes is still super fast – and easy to type! No loops!

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Author: Ryan Palo | Tags: python beginner |

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