Are all points within a rectangle?

In order to fix the issue, you have to change the if statement to be nested in the for loop you have above. Otherwise, the statement doesn't check every single point of the list but only the last one.

Moreover, if you want your first example of yours to be true you have to check whether the point is less or equal to the secondCorner.

You can also add one more if statement in order to check if the list is empty in order to return false as you would like in the last example of yours.

if not(pointList): return False for i in range(len(pointList)): p_x = pointList[i][0] p_y = pointList[i][1] if not ((p_x >= x1 and p_x <= x2) and (p_y >= y1 and p_y <= y2)): return False return True 

Two things:

• You only return False inside the loop. If the loop never runs (which is the case if the list of points is empty), you will always return true.
• ((p_x >= x1 and p_x < x2) and (p_y >= y1 and p_y < y2)) should be ((p_x >= x1 and p_x <= x2) and (p_y >= y1 and p_y <= y2)) (end coordinates of rectangle should be inclusive, rather than exclusive.)

I would recommend defining one function that determines whether a single point is within a rectangle, then reuse that function to check if all points are within a rectangle:

from collections import namedtuple def point_in_rect(point, rect): return (rect.first.x <= point.x <= rect.second.x) and (rect.first.y <= point.y <= rect.second.y) def points_in_rect(points, rect): return bool(points) and all(point_in_rect(point, rect) for point in points) Point = namedtuple("Point", "x y") Rectangle = namedtuple("Rectangle", "first second") rect = Rectangle(Point(0, 0), Point(5, 5)) print(points_in_rect([Point(1, 1), Point(0, 0), Point(5, 5)], rect)) # True print(points_in_rect([Point(1, 1), Point(0, 0), Point(5, 6)], rect)) # False print(points_in_rect([], rect)) # False 

Just adding another option to the list. You could use numpy here as well, avoiding a for loop.

import numpy as np def allIn(firstCorner=(0,0), secondCorner=(0,0), pointList=[]): # list is empty, return False if not pointList: return False arr = np.array(pointList) return False not in (arr >= firstCorner) & (arr <= secondCorner) print(allIn((0,0), (5,5), [(1,1), (0,0), (5,5)])) # True print(allIn((0,0), (5,5), [(1,1), (0,0), (5,6)])) # False print(allIn((0,0), (5,5), [])) # False 

E.g. for [(1,1), (0,0), (5,6)], the array becomes:

[[1 1] [0 0] [5 6]] 

which is turned into:

[[ True True] [ True True] [ True False]] # i.e. 6 is out of bounds, creating 1 False in the array, returning False for the function.