我正在从传感器实时接收大量积分。但是，我只需要4个类别的点，即top_left，top_right，bottom_left和bottom_right。我在Python 2中有一条if-elif语句，如下所示：
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
每个点都有一个id和x，y，z参数。这是一个内置类。我只是在这里展示示例课程。
class Point():
def __init__(self, id, x, y, z):
self.id = id
self.x = x
self.y = y
self.z = z
是否有任何有效的方法考虑运行时实现相同的目标。
回答：
我正在添加从答案中得到的结果。似乎the answer by Elis Byberi最快。下面是我的测试代码：
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)
请参阅下面返回的输出：
ravi@home:~/Desktop$ python test.py
method 1: 0.174991846085
method 2: 0.0743980407715
method 3: 0.582262039185
最佳答案
您可以使用字典来保存对象。 Dict在键查找中非常有效。
使用dict是使用if else块的两倍。
这是python中最有效的方法：
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
关于python - 比if-elif语句快，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/47002598/