Faster than if-elif statement

Question

I am receiving Points in large number from a sensor in real-time. However, I just need 4 categories of points, i.e., top_left, top_right, bottom_left, and bottom_right. I have an if-elif statement in Python 2 as follows:

from random import random, randint

# points below are received from sensor. however, 
# here in this post I am creating it randomly.
points = [Point(randint(0, i), random(), random(), random()) for i in range(100)] 

# 4 categories
top_left, top_right, bottom_left, bottom_right = None, None, None, None
for p in points:
    if p.id == 5:
        top_left = p
    elif p.id == 7:
        top_right = p
    elif p.id == 13:
        bottom_left = p
    elif p.id == 15:
        bottom_right = p

print top_left.id, top_left.x, top_left.y, top_left.z # check variable

Each Point has an id and x, y, z parameters. This is an inbuilt class. I am just showing a sample class here.

class Point():
    def __init__(self, id, x, y, z):
        self.id = id
        self.x = x
        self.y = y
        self.z = z

Is there any efficient way considering runtime of achieving the same.

Answer: I am adding the results which I got from the answers. It seems that the answer by Elis Byberi is fastest among all. Below is my test code:

class Point():
    def __init__(self, id, x, y, z):
        self.id = id
        self.x = x
        self.y = y
        self.z = z

from random import random, randint
n = 1000
points = [Point(randint(0, i), random(), random(), random()) for i in range(n)]

def method1():
    top_left, top_right, bottom_left, bottom_right = None, None, None, None
    for p in points:
        if p.id == 5:
            top_left = p
        elif p.id == 7:
            top_right = p
        elif p.id == 13:
            bottom_left = p
        elif p.id == 15:
            bottom_right = p
    #print top_left.id, top_left.x, top_left.y, top_left.z

def method2():
    categories = {
        5: None,  # top_left
        7: None,  # top_right
        13: None,  # bottom_left
        15: None  # bottom_right
    }

    for p in points:
        categories[p.id] = p

    top_left = categories[5]
    #print top_left.id, top_left.x, top_left.y, top_left.z

def method3():
    name_to_id = {'top_left': 5, 'top_right': 7, 'bottom_left': 13, 'bottom_right': 15}
    ids = [value for value in name_to_id.values()]
    bbox = {id: None for id in ids}

    for point in points:
        try:
            bbox[point.id] = Point(point.id, point.x, point.y, point.z)
        except KeyError:  # Not an id of interest.
            pass

    top_left = bbox[name_to_id['top_left']]
    #print top_left.id, top_left.x, top_left.y, top_left.z

from timeit import Timer
print 'method 1:', Timer(lambda: method1()).timeit(number=n)
print 'method 2:', Timer(lambda: method2()).timeit(number=n)
print 'method 3:', Timer(lambda: method3()).timeit(number=n)

See Below the returned output:

ravi@home:~/Desktop$ python test.py 
method 1: 0.174991846085
method 2: 0.0743980407715
method 3: 0.582262039185

Answer 1

You can use a dict to save objects. Dict is very efficient in key lookup.

Using dict is twice as fast as using if else block.

This is the most efficient way in python:

from random import random, randint

class Point():
    def __init__(self, id, x, y, z):
        self.id = id
        self.x = x
        self.y = y
        self.z = z

# points below are received from sensor. however,
# here in this post I am creating it randomly.
points = [Point(randint(0, i), random(), random(), random()) for i in
          range(100)]

# 4 categories
categories = {
    5: None,  # top_left
    7: None,  # top_right
    13: None,  # bottom_left
    15: None  # bottom_right
}

for p in points:
    categories[p.id] = p

>>> print categories[5].id, categories[5].x, categories[5].y, categories[5].z  # check variable
5 0.516239541892 0.935096344266 0.0859987803457

Answer 2

Instead of using list-comprehension:

points = [Point(randint(0, i), random(), random(), random()) for i in range(100)]

use a loop and assign the points during creation:

points = []
for i in range(100):
    p = Point(randint(0, i), random(), random(), random())
    points.append(p)
    if p.id == 5:
        top_left = p
    elif p.id == 7:
       top_right = p
    elif p.id == 13:
        bottom_left = p
    elif p.id == 15:
        bottom_right = p

This way you get all done in one iteration instead of two.

Answer 3

Here's a way that should be faster because it uses a single if to determine whether a Point is one of the ones representing the extremes based on their id attribute, plus the if uses the very fast dictionary in operation for membership testing. Essentially what is done it the bbox dictionary is preloaded with keys that correspond to the four ids sought, which make checking for any of them a single relatively efficient operation.

Note that if there are points with duplicate ids in the Point list, the last one seen will be the one selected. Also note that if no point with a matching id is found, some of the final variables will have a value of None instead of a Point instance.

from random import randint, random
from pprint import pprint
from operator import attrgetter


class Point():
    def __init__(self, id, x, y, z):
        self.id = id
        self.x = x
        self.y = y
        self.z = z


points = [Point(randint(0, 20), random(), random(), random()) for i in range(100)]
name_to_id = {'top_left': 5, 'top_right': 7, 'bottom_left': 13, 'bottom_right': 15}
bbox = {id: None for id in name_to_id.values()}  # Preload with ids of interest.

for point in points:
    if point.id in bbox:  # id of interest?
       bbox[point.id] = point

# Assign bbox's values to variables with meaningful names.
top_left = bbox[name_to_id['top_left']]
top_right = bbox[name_to_id['top_right']]
bottom_left = bbox[name_to_id['bottom_left']]
bottom_right = bbox[name_to_id['bottom_right']]

for point in [top_left, top_right, bottom_left, bottom_right]:
    print('Point({}, {}, {}, {})'.format(point.id, point.x, point.y, point.z))