Question

我希望我可以使用np.where或一些等效（和高效）的numpy函数构造以下算法：

def generate_signal(r):
    signal = np.zeros(len(r), dtype=int)
    lastSignal = 0
    for i in range(len(r)):
        if r[i] <= 30:
            lastSignal = 1
        elif r[i] >= 60:
            lastSignal = 0
        signal[i] = lastSignal
    return signal

以下是输入/输出的一个示例：

r = np.array([50, 52, 59, 69, 47, 33, 27, 26, 20, 30, 33, 35, 58, 55, 48, 60, 68, 55, 43, 49, 33, 30, 22, 28])
s = generate_signal(r)
print(s) # This is the result: [0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1]
print(list(zip(r, s))) # A zipped result (in case it helps): [(50, 0), (52, 0), (59, 0), (69, 0), (47, 0), (33, 0), (27, 1), (26, 1), (20, 1), (30, 1), (33, 1), (35, 1), (58, 1), (55, 1), (48, 1), (60, 0), (68, 0), (55, 0), (43, 0), (49, 0), (33, 0), (30, 1), (22, 1), (28, 1)]

Answer 1

构建结果信号的正负瞬态：

>>> pos_tran = np.maximum(0, np.diff(np.int8(r <= 30)))
>>> neg_tran = np.maximum(0, np.diff(np.int8(r >= 60)))

在第一次阳性瞬变之前消除负瞬态：

>>> neg_tran[0:np.nonzero(pos_tran)[0][0]] = 0

整合信号（累积和）并重新插入np.diff中丢失的前导0：

>>> np.insert(np.cumsum(pos_tran - neg_tran), 0, 0)
array([0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1])

Answer 2

您可以按如下方式使用np.repeat：

import numpy as np

def f_pp():
    rr = np.concatenate([[70], r, [70]])
    l30, g60 = rr <= 30, rr >= 60
    df, = np.where(l30 | g60)
    reps = np.diff(df)
    reps[0] -= 1
    return l30[df[:-1]].astype(int).repeat(reps)

使用OP的示例和一些较大的（1000元素）随机的计时和验证：

def generate_signal():
    signal = np.zeros(len(r), dtype=int)
    lastSignal = 0
    for i in range(len(r)):
        if r[i] <= 30:
            lastSignal = 1
        elif r[i] >= 60:
            lastSignal = 0
        signal[i] = lastSignal
    return signal

def f_ff():
    pos_tran = np.maximum(0, np.diff(np.int8(r <= 30)))
    neg_tran = np.maximum(0, np.diff(np.int8(r >= 60)))
    neg_tran[0:np.nonzero(pos_tran)[0][0]] = 0
    return np.insert(np.cumsum(pos_tran - neg_tran), 0, 0)

from timeit import timeit

glb = globals()
kwds = dict(globals=glb, number=1000)

r = np.array([50, 52, 59, 69, 47, 33, 27, 26, 20, 30, 33, 35, 58, 55, 48, 60, 68, 55, 43, 49, 33, 30, 22, 28])

print('OP: {:8.4f} ms'.format(timeit(generate_signal, **kwds)))
print('ff: {:8.4f} ms'.format(timeit(f_ff, **kwds)))
print('pp: {:8.4f} ms'.format(timeit(f_pp, **kwds)))

print('ff correct:', np.all(generate_signal() == f_ff()))
print('pp correct:', np.all(generate_signal() == f_pp()))

R = np.random.randint(20, 70, (10, 1000))

print('OP: {:8.4f} ms'.format(sum(timeit(generate_signal, **kwds) for glb['r'] in R) / 10))
print('ff: {:8.4f} ms'.format(sum(timeit(f_ff, **kwds) for glb['r'] in R) / 10))
print('pp: {:8.4f} ms'.format(sum(timeit(f_pp, **kwds) for glb['r'] in R) / 10))

print('ff correct:', all(np.all(generate_signal() == f_ff()) for glb['r'] in R))
print('pp correct:', all(np.all(generate_signal() == f_pp()) for glb['r'] in R))

示例输出：

OP:   0.0138 ms
ff:   0.0405 ms
pp:   0.0121 ms
ff correct: True
pp correct: True
OP:   0.5386 ms
ff:   0.0539 ms
pp:   0.0215 ms
ff correct: False
pp correct: True

使用numpy.where或任何等效函数对迭代算法进行矢量化

2 个答案: