Shorter way to make colliderect in if statement work?


I have two objects:

enemigo=pygame.transform.scale(enemigo, (100, 100))
enemigo_rectangulo=enemigo.get_rect(center=(1000, 100))

enemigo2=pygame.transform.scale(enemigo2, (140, 140))
enemigo_rectangulo2=enemigo2.get_rect(center=(1400, 50))

And I want to make an if statement if they collide with a third object. This code actually works but is too long:

if personaje_rectangulo.colliderect(enemigo_rectangulo) or personaje_rectangulo.colliderect(enemigo_rectangulo2):

And there’s an error when I try to write:

if personaje_rectangulo.colliderect(enemigo_rectangulo, enemigo_rectangulo2):

So, what is wrong? Is there a shorter way to write it?

Asked By: PCandPC



The first thing you can do is use shorter variable names, such as personaje and enemigo.

Additionally, break the statement into smaller pieces by using variables:

hit_enemy1 = personaje_rectangulo.colliderect(enemigo_rectangulo)
hit_enemy2 = personaje_rectangulo.colliderect(enemigo_rectangulo2)

if hit_enemy1 or hit_enemy2:

Alternatively, if you store all of the enemies in a list, you can use any():

if any(personaje_rectangulo.colliderect(enemigo_rectangulo) for enemigo_rectangulo in todos_enemigos):

Now this is also getting long so creating a function to help can shorten the line of code:

def collision(person, enemy):
    return person.colliderect(enemy)

if any(collision(person, enemy) in todos_enemigos):

Another solution is to create a Enemy class which encapsulates the data and behavior of an enemy. This is a bit more advanced, but definitely a good exercise to learn what classes are and how to use them.

Answered By: Code-Apprentice

Use collidelist():

Test whether the rectangle collides with any in a sequence of rectangles. The index of the first collision found is returned. If no collisions are found an index of -1 is returned.

if personaje_rectangulo.collidelist([enemigo_rectangulo, enemigo_rectangulo2]) >= 0:
Answered By: Rabbid76
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