In this article, I will talk about three common errors that Python programmers are just starting to get. Avoiding three common mistakes for this beginner will save you a lot of headaches and debugging time…

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!

```
# 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…

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!

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 `<a href="https://www.geeksforgeeks.org/sets-in-python/" target="_blank">set</a>`

. 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!

#python

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