使得这款C数组处理code的蟒蛇（甚至numpy的） [英] Making this C array processing code more python (and even numpy)

问题描述

我试图让我的头周围的Python（最终numpy的）的惊人表处理能力。我把一些C $ C $三我写信给蟒蛇。

I'm trying to get my head around the amazing list processing abilities of python (And eventually numpy). I'm converting some C code I wrote to python.

我有一个文本数据文件，其中第一行是一个标题，然后每奇数行是我的输入数据，每个偶数行是我的输出数据。所有数据空间隔开。我很chuffed我设法所有的数据读入使用嵌套列表COM prehensions名单。大放异彩。

I have a text datafile where first row is a header, and then every odd row is my input data and every even row is my output data. All data space separated. I'm quite chuffed that I managed to read all the data into lists using nested list comprehensions. amazing stuff.

with open('data.txt', 'r') as f:
    # get all lines as a list of strings
    lines = list(f)

    # convert header row to list of ints and get info
    header = map(int, lines[0].split(' '))
    num_samples = header[0]
    input_dim = header[1]
    output_dim = header[2]
    del header    

    # bad ass list comprehensions 
    inputs = [[float(x) for x in l.split()] for l in lines[1::2]]
    outputs = [[float(x) for x in l.split()] for l in lines[2::2]]
    del x, l, lines

然后，我想以产生一个新的列表，其中每个元素是一个相应的输入 - 输出对的功能。我无法弄清楚如何与蟒蛇任何特定的优化做到这一点。这是C风格的Python：

Then I want to produce a new list where each element is a function of a corresponding input-output pair. I couldn't figure out how to do this with any python specific optimizations. Here it is in C-style python:

# calculate position
pos_list = [];
pos_y = 0
for i in range(num_samples):
    pantilt = outputs[i];
    target = inputs[i];

    if(pantilt[0] > 90):
        pantilt[0] -=180
        pantilt[1] *= -1
    elif pantilt[0] < -90:
        pantilt[0] += 180
        pantilt[1] *= -1

    tan_pan = math.tan(math.radians(pantilt[0]))
    tan_tilt = math.tan(math.radians(pantilt[1]))

    pos = [0, pos_y, 0]
    pos[2] = tan_tilt * (target[1] - pos[1]) / math.sqrt(tan_pan * tan_pan + 1)
    pos[0] = pos[2] * tan_pan
    pos[0] += target[0]
    pos[2] += target[2]
    pos_list.append(pos)
del pantilt, target, tan_pan, tan_tilt, pos, pos_y

我试着用一个COM prehension做到这一点，或映射但无法弄清楚如何：

I tried to do it with a comprehension, or map but couldn't figure out how to:

• 来自两个不同列表（输入和输出）绘制为pos_list阵列的每个元素


• 把算法的主体在com prehension。它必须是一个单独的功能还是有lambda表达式使用此的一个时髦的方式？


• 将它甚至有可能与没有循环做这所有，只是坚持在numpy的和矢量化整个事情？

推荐答案

一个量化的方法使用的 布尔索引/掩码 -

One vectorized approach using boolean-indexing/mask -

import numpy as np

def mask_vectorized(inputs,outputs,pos_y):
    # Create a copy of outputs array for editing purposes
    pantilt_2d = outputs[:,:2].copy()

    # Get mask correspindig to IF conditional statements in original code
    mask_col0_lt = pantilt_2d[:,0]<-90
    mask_col0_gt = pantilt_2d[:,0]>90

    # Edit the first column as per the statements in original code
    pantilt_2d[:,0][mask_col0_gt] -= 180
    pantilt_2d[:,0][mask_col0_lt] += 180

    # Edit the second column as per the statements in original code
    pantilt_2d[ mask_col0_lt | mask_col0_gt,1] *= -1

    # Get vectorized tan_pan and tan_tilt 
    tan_pan_tilt = np.tan(np.radians(pantilt_2d))

    # Vectorized calculation for: "tan_tilt * (target[1] .." from original code 
    V = (tan_pan_tilt[:,1]*(inputs[:,1] - pos_y))/np.sqrt((tan_pan_tilt[:,0]**2)+1)

    # Setup output numpy array
    pos_array_vectorized = np.empty((num_samples,3))

    # Put in values into columns of output array
    pos_array_vectorized[:,0] = inputs[:,0] + tan_pan_tilt[:,0]*V
    pos_array_vectorized[:,1] = pos_y
    pos_array_vectorized[:,2] = inputs[:,2] + V

    # Convert to list, if so desired for the final output
    # (keeping as numpy array could boost up the performance further)
    return pos_array_vectorized.tolist()

运行测试

In [415]: # Parameters and setup input arrays
     ...: num_samples = 1000
     ...: outputs = np.random.randint(-180,180,(num_samples,5))
     ...: inputs = np.random.rand(num_samples,6)
     ...: pos_y = 3.4
     ...: 

In [416]: %timeit original(inputs,outputs,pos_y)
100 loops, best of 3: 2.44 ms per loop

In [417]: %timeit mask_vectorized(inputs,outputs,pos_y)
10000 loops, best of 3: 181 µs per loop

这篇关于使得这款C数组处理code的蟒蛇（甚至numpy的）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！