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python - Python如何将多个不同长度的数组添加到一个数组中

问题描述

我正在开发一个程序，该程序需要将音频数组与给定的起始索引混合在一起。例如

signal1 = np.array([1,2,3,4])
signal2 = np.array([5,5,5])
signal3 = np.array([7,7,7,7])
sig = np.array([signal1,signal2,signal3])
onset(0, 2, 8)
result = mixing_function(sig,onset)

根据起始点，signal2 将从索引 2 添加到 signal1，signal3 将从索引 8 添加到混合，因此混合部分将被零填充。它应该返回：

[1,2,8,9,5,0,0,0,7,7,7,7]

我不确定为此编写代码的有效方法是什么。现在，我创建了一个最大长度为 maxlen 的零数组。然后我将 sig 中的每个元素添加到结果的相应索引范围：

def mixing_function(sig,onset):
    maxlen = np.max([o + len(s) for o, s in zip(onset, sig)])
    result =  np.zeros(maxlen)
    for i in range(len(onset)):
        result[onset[i]:onset[i] + len(sig[i])] += sig[i] 
    return result

然而，这可能会很慢，尤其是当有许多信号混合在一起时，它们都具有不同的起始点。请告知是否有更有效的方法。

非常感谢

Ĵ

标签： pythonnumpy

解决方案

以下是针对该问题的不同解决方案的一些统计数据。通过对实现进行矢量化以获得 maxlen，我能够获得更多性能，但除此之外，我认为您将不得不尝试 cython 或尝试其他编程语言。

import numpy as np
from numba import jit
from time import time
np.random.seed(42)

def mixing_function(sig, onset):
    maxlen = np.max([o + len(s) for o, s in zip(onset, sig)])
    result =  np.zeros(maxlen)
    for i in range(len(onset)):
        result[onset[i]:onset[i] + len(sig[i])] += sig[i] 
    return result

def mix(sig, onset):
    siglengths = np.vectorize(len)(sig)
    maxlen = max(onset + siglengths)
    result = np.zeros(maxlen)
    for i in range(len(sig)):
        result[onset[i]: onset[i]+siglengths[i]] += sig[i]
    return result

@jit(nopython=True)
def mixnumba(sig, onset):
    # maxlen = np.max([onset[i] + len(sig[i]) for i in range(len(sig))])
    maxlen = -1
    for i in range(len(sig)):
        maxlen = max(maxlen, sig[i].size + onset[i])
    result = np.zeros(maxlen)
    for i in range(len(sig)):
        result[onset[i]: onset[i] + sig[i].size] += sig[i]
    return result

def signal_adder_with_onset(data, onset):
    data = np.array(data)
    # Get lengths of each row of data
    lens = np.array([len(i) for i in data])
    #adjust with offset for max possible lengths
    max_size = lens + onset
    # Mask of valid places in each row
    mask = ((np.arange(max_size.max()) >= onset.reshape(-1, 1)) 
            &  (np.arange(max_size.max()) < (lens + onset).reshape(-1, 1)))

    # Setup output array and put elements from data into masked positions
    out = np.zeros(mask.shape, dtype=data.dtype) #could perhaps change dtype here
    out[mask] = np.concatenate(data)
    return out.sum(axis=0)

sigbig = [np.random.randn(np.random.randint(1000, 10000)) for _ in range(10000)]
onsetbig = np.random.randint(0, 10000, size=10000)
sigrepeat = np.repeat(sig, 500000).tolist()
onsetrepeat = np.repeat(onset, 500000)

assert all(mixing_function(sigbig, onsetbig) == mix(sigbig, onsetbig))
assert all(mixing_function(sigbig, onsetbig) == mixnumba(sigbig, onsetbig))
assert all(mixing_function(sigbig, onsetbig) == signal_adder_with_onset(sigbig, onsetbig))

%timeit result = mixing_function(sigbig, onsetbig)
%timeit result = mix(sigbig, onsetbig)
%timeit result = mixnumba(sigbig, onsetbig)
%timeit result = signal_adder_with_onset(sigbig, onsetbig)
# Output
114 ms ± 1.97 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
108 ms ± 2.53 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
368 ms ± 8.22 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
13.4 s ± 211 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit result = mixing_function(sigrepeat, onsetrepeat)
%timeit result = mix(sigrepeat, onsetrepeat)
%timeit result = mixnumba(sigrepeat, onsetrepeat)
%timeit result = signal_adder_with_onset(sigrepeat.tolist(), onsetrepeat)
# Output
933 ms ± 6.43 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
803 ms ± 21.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
4.07 s ± 85.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
254 ms ± 11.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

TL.DR。np.vectorize通过使用以获得maxlen随机长度的长信号，可以提高边际性能（快 10% 左右） 。请注意，对于许多小信号，@Paritosh Singh 答案的执行速度比其他信号快。

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