Replace if-else with dictionary python

Python dictionary comprehension is a method for transforming one dictionary into a new dictionary. Using an if-else in Python dict comprehension is one of the ways to create a new list. It works as a condition to filter out dictionary elements for a new dictionary.

{ (some_key if condition else default_key):(something_if_true if condition
          else something_if_false) for key, value in dict_.items() }

Simple example code adding conditionals to Dictionary comprehension. Replace the value with ‘Even‘ if the value module is zero else replaced it with ‘Odd‘.

dict1 = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6}

# Identify odd and even entries
res = {k: ('Even' if v % 2 == 0 else 'Odd') for (k, v) in dict1.items()}

print(res)

Output:

Do comment if you have any doubts or suggestions on this Python dictionary tutorial.

Note: IDE: PyCharm 2021.3.3 (Community Edition)

Windows 10

Python 3.10.1

All Python Examples are in Python 3, so Maybe its different from python 2 or upgraded versions.

Degree in Computer Science and Engineer: App Developer and has multiple Programming languages experience. Enthusiasm for technology & like learning technical.

If statements are great tools, but as the number of branches grows, they quickly become unwieldy. In this article, we will explore two alternatives to if statements:

  • "match...case", a new Python 3.10 language feature.
  • Dictionaries, particularly useful when dealing with user input.

This article's scenario: a long if statement

Imagine you have an if statement like this one (assume the average function has already been defined):

numbers = [1, 4, 16, 20]
action = input(f"What would you like to do with {numbers}?")  # e.g. add

if action == "add":
    print(sum(numbers))
elif action == "avg":
    print(average(numbers))
elif action == "max":
    print(max(numbers))
else:
    print("Action not recognized")

Using match...case to simplify an if statement chain

In "match...case", we still need to tell Python what the different options are, and what to do in each case:

def average(seq):
	return sum(seq) / len(seq)

numbers = [1, 4, 16, 20]
action = input(f"What would you like to do with {numbers}? ")

match action:
	case "add":
		print(sum(numbers))
	case "avg":
		print(average(numbers))
	case "max":
		print(max(numbers))
	case _:
		print("Operation not recognized.")

While this looks a bit better, there is still much of the same duplication. Each conditional branch has the print() function duplicated, and there are a lot of keywords.

Overall, the length of the code block is the same.

Plus the biggest problem of all remains: that as you add more options, the branching conditional will grow too.

Use a dictionary to simplify a long if statement

Instead of the log if-elif chain or a long match-case chain, you could store the user's options in a dictionary:

options = {
    "add": sum,
    "avg": average,
    "max": max
}

Then you could ask the user for which of the options (from the dictionary) they'd like to use:

options = {
    "add": sum,
    "avg": average,
    "max": max
}
numbers = [1, 4, 16, 20]

action = input(f"What would you like to do with {numbers}?")  # e.g. add

With this, we can retrieve the function from the dictionary directly:

options = {
    "add": sum,
    "avg": average,
    "max": max
}
numbers = [1, 4, 16, 20]

action = input(f"What would you like to do with {numbers}?")  # e.g. add

operation = options.get(action)

Since the dictionary maps strings to functions, the operation variable would now contain the function we want to run.

All that's left is to run operation(numbers) to get our result. If the user entered 'add', then operation will be the sum function.

We should also do some error checking, to make sure we don't try to run a function that doesn't exist if the user entered something that isn't one of the dictionary's keys.

options = {
    "add": sum,
    "avg": average,
    "max": max
}
numbers = [1, 4, 16, 20]

action = input(f"What would you like to do with {numbers}?")  # e.g. add

operation = options.get(action)

if operation:
    operation(numbers)
else:
    print("Action not recognized")

You still need the one if statement just in case the user chooses something that doesn't have a key in the dictionary, but this is a single-branch if statement that won't grow over time.

Another benefit is that you can easily tell the user which options are available to them by using the dictionary keys:

option_texts = '|'.join(options.keys()
action = input(f"What would you like to do with {numbers}? ({option_texts}) ")
# Would show "What would you like to do with [1, 4, 16, 20]? (add|avg|max) "

Conclusion

In this post we've seen two ways you can simplify if statements: by using "match...case" and by using dictionaries.

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