Python如何将不同长度的多个数组添加到一个数组中
Python如何将不同长度的多个数组添加到一个数组中
提问于 2019-03-29 03:16:00
我正在工作的程序，需要混合音频阵列与给定的起始索引在一起。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，signal2将从索引2添加到signal1，signal3将从索引8添加到混合，因此混合部分将是零填充。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中：
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，这可能是相当慢的，特别是当有许多信号混合在一起都具有不同的onset时。如果有更有效的方法，请指教。
非常感谢
J
回答 4
Stack Overflow用户
回答已采纳
发布于 2019-03-29 06:27:39
以下是该问题的不同解决方案的一些统计数据。我可以通过对实现进行矢量化来获得maxlen，以获得更高的性能，但除此之外，我认为您必须尝试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。通过使用np.vectorize来获得随机长度的长信号的maxlen，从而实现了性能的边际提升(大约快了10% )。请注意，对于许多小信号，@Paritosh Singh answer比其他答案执行得更快。
Stack Overflow用户
发布于 2019-03-29 04:20:58
这里有一个尝试，应该可以做到这一点。
def signal_adder_with_onset(data, onset):
# 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)
import numpy as np
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 = np.array((0, 2, 8))
result = signal_adder_with_onset(sig, onset)
print(result)
#[1 2 8 9 5 0 0 0 7 7 7 7]编辑:矢量化操作只在数据量较大的情况下生效，而在数据量较小的情况下速度较慢。
添加以供比较
import time
def signal_adder_with_onset(data, onset):
# 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)
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
import numpy as np
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])
sig = np.repeat(sig, 1000000)
onset = np.array((0, 2, 8))
onset = np.repeat(onset, 1000000)
start1 = time.time()
result = signal_adder_with_onset(sig, onset)
end1 = time.time()
start2 = time.time()
result2 = mixing_function(sig,onset)
end2 = time.time()
print(f"Original function: {end2 - start2} \n Vectorized function: {end1 - start1}")
print(result)
#Output:
Original function: 9.28258752822876
Vectorized function: 2.5798118114471436
[1000000 2000000 8000000 9000000 5000000 0 0 0 7000000 7000000 7000000
7000000]Stack Overflow用户
发布于 2019-03-29 04:29:01
如果对信号进行偏移，然后将它们放入数据框中，则会将NaN添加到列中，以使所有行的长度相同。然后你可以做df.sum()。然而，这将返回一个浮点型而不是整型。
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原文链接：
https://stackoverflow.com/questions/55405295
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