for循环列出python中的理解或映射

问题描述

我试图稍微提高一些python代码的速度，因此尝试将标准for循环移动到列表理解或map调用:

I'm trying to improve the speed of some python code a bit and therefore trying to move a standard for loop to either a list comprehension or map call:

    buf = [0 for i in range(self.numLEDs * 3)]
    temp = [0,0,0]
    for x in range(self.numLEDs):
        r = data[x*3]
        g = data[x*3+1]
        b = data[x*3+2]
        temp[self.c_order[0]] = self.gamma[r]
        temp[self.c_order[1]] = self.gamma[g]
        temp[self.c_order[2]] = self.gamma[b]

        buf[x * 3:x * 3 + 3] = temp

c_order只是另一个列表，在这种情况下为[1,2,0].它控制某些RGB像素的通道顺序. gamma是一个256个元素的列表，其中包含8位通道值中每个值的经过gamma校正的值.

c_order is simply another list, in this case [1,2,0]. It controls the channel order for some RGB pixels. gamma is a list 256 elements long that holds gamma corrected values for each of the 8bit channel values.

我想做的是以某种方式从这段代码中完全删除对循环标准的使用.我已经设法在没有通道交换的情况下做到了这一点，但是通过伽马校正，它的速度快了一倍.像这样:

What'd I'd like to do is somehow completely remove any use of a standard for loop from this bit of code. I've managed to do it without the channel swap, but with the gamma correction and it's twice as fast. Like this:

corrected = [gamma[i] for i in data]
buf[0:len(corrected)] = corrected

我如何在没有for循环的情况下交换列表元素的顺序?

How can I swap the order of list elements as I go without a for loop though?

推荐答案

因此，您需要没有任何扩展库的纯python代码.

So you need pure python code without any extension library.

要加速代码:

1. 在循环中使用局部变量.
2. 更改循环以列出理解.

这是代码:

class Test(object):

    def __init__(self, n):
        self.numLEDs =  n
        self.c_order = [1, 2, 0]
        self.gamma = [i // 2 for i in range(256)]

    def do1(self, data):
        buf = [0 for i in range(self.numLEDs * 3)]
        temp = [0,0,0]
        for x in range(self.numLEDs):
            r = data[x*3]
            g = data[x*3+1]
            b = data[x*3+2]
            temp[self.c_order[0]] = self.gamma[r]
            temp[self.c_order[1]] = self.gamma[g]
            temp[self.c_order[2]] = self.gamma[b]

            buf[x * 3:x * 3 + 3] = temp
        return buf

    def do2(self, data):
        buf = [0] * (self.numLEDs * 3)
        gamma = self.gamma
        for idx, idx2 in enumerate(self.c_order):
            buf[idx2::3] = [gamma[v] for v in data[idx::3]]
        return buf

import random
random.seed(0)
N = 1000
t = Test(N)
data = [random.randint(0, 255) for i in range(3*N)]
r1 = t.do1(data)
r2 = t.do2(data)
print r1 == r2  # check the result

%timeit t.do1(data)
%timeit t.do2(data)

输出，速度提高了6倍:

the output, it's 6x faster:

True
1000 loops, best of 3: 1.1 ms per loop
10000 loops, best of 3: 176 µs per loop

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