Question

寻找一个快速矢量化函数，它返回连续非零值的滚动数。每当遇到零时，计数应从0开始。结果应该与输入数组具有相同的形状。

给出这样的数组：

x = np.array([2.3, 1.2, 4.1 , 0.0, 0.0, 5.3, 0, 1.2, 3.1])

该函数应该返回：

array([1, 2, 3, 0, 0, 1, 0, 1, 2])

Answer 1

本文列出了一种矢量化方法，基本上包括两个步骤：

初始化与输入向量x相同大小的零向量，并在与x的非零对应的位置设置1。
接下来，在该向量中，我们需要在每个＆＃34;岛＆＃34;的结束/停止位置之后立即减去每个岛的游程长度。目的是稍后再次使用cumsum，这将导致＆＃34;岛屿＃34;和其他地方的零。

以下是实施 -

import numpy as np

#Append zeros at the start and end of input array, x
xa = np.hstack([[0],x,[0]])

# Get an array of ones and zeros, with ones for nonzeros of x and zeros elsewhere
xa1 =(xa!=0)+0

# Find consecutive differences on xa1
xadf = np.diff(xa1)

# Find start and stop+1 indices and thus the lengths of "islands" of non-zeros
starts = np.where(xadf==1)[0]
stops_p1 = np.where(xadf==-1)[0]
lens = stops_p1 - starts

# Mark indices where "minus ones" are to be put for applying cumsum
put_m1 = stops_p1[[stops_p1 < x.size]]

# Setup vector with ones for nonzero x's, "minus lens" at stops +1 & zeros elsewhere
vec = xa1[1:-1] # Note: this will change xa1, but it's okay as not needed anymore
vec[put_m1] = -lens[0:put_m1.size]

# Perform cumsum to get the desired output
out = vec.cumsum()

示例运行

In [116]: x
Out[116]: array([ 0. ,  2.3,  1.2,  4.1,  0. ,  0. ,  5.3,  0. ,  1.2,  3.1,  0. ])

In [117]: out
Out[117]: array([0, 1, 2, 3, 0, 0, 1, 0, 1, 2, 0], dtype=int32)

运行时测试 -

这里有一些运行时测试将建议的方法与其他itertools.groupby based approach进行比较 -

In [21]: N = 1000000
    ...: x = np.random.rand(1,N)
    ...: x[x>0.5] = 0.0
    ...: x = x.ravel()
    ...: 

In [19]: %timeit sumrunlen_vectorized(x)
10 loops, best of 3: 19.9 ms per loop

In [20]: %timeit sumrunlen_loopy(x)
1 loops, best of 3: 2.86 s per loop

Answer 2

您可以使用itertools.groupby和np.hstack：

>>> import numpy as np
>>> x = np.array([2.3, 1.2, 4.1 , 0.0, 0.0, 5.3, 0, 1.2, 3.1])
>>> from itertools import groupby

>>> np.hstack([[i if j!=0 else j for i,j in enumerate(g,1)] for _,g in groupby(x,key=lambda x: x!=0)])
array([ 1.,  2.,  3.,  0.,  0.,  1.,  0.,  1.,  2.])

我们可以基于非零元素对数组进行分组，然后使用列表推导并枚举用这些索引替换非零子数组，然后用np.hstack展平列表。

Answer 3

这个子问题出现在 Kick Start 2021 年的 A 轮中。我的解决方案：

let user = document.getElementsByClassName("username")[0].value;
let pass = document.getElementsByClassName("password")[0].value;

事实上，这个问题还需要一个你向下连续计数的版本。因此，这里是另一个带有可选关键字参数的版本，它对 def current_run_len(a): a_ = np.hstack([0, a != 0, 0]) # first in starts and last in stops defined d = np.diff(a_) starts = np.where(d == 1)[0] stops = np.where(d == -1)[0] a_[stops + 1] = -(stops - starts) # +1 for behind-last return a_[1:-1].cumsum() 执行相同的操作：

rev=False

结果：

def current_run_len(a, rev=False):
    a_ = np.hstack([0, a != 0, 0])  # first in starts and last in stops defined
    d = np.diff(a_)
    starts = np.where(d == 1)[0]
    stops = np.where(d == -1)[0]
    if rev:
        a_[starts] = -(stops - starts)
        cs = -a_.cumsum()[:-2]
    else:
        a_[stops + 1] = -(stops - starts)  # +1 for behind-last
        cs = a_.cumsum()[1:-1]
    return cs

a = np.array([1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1])
print('a                             = ', a)
print('current_run_len(a)            = ', current_run_len(a))
print('current_run_len(a, rev=True)  = ', current_run_len(a, rev=True))

对于仅由 0 和 1 组成的数组，您可以将 a = [1 1 1 1 0 0 0 1 1 0 1 0 0 0 1] current_run_len(a) = [1 2 3 4 0 0 0 1 2 0 1 0 0 0 1] current_run_len(a, rev=True) = [4 3 2 1 0 0 0 2 1 0 1 0 0 0 1] 简化为 [0, a != 0, 0]。但发布的版本也适用于任意非零数字。

Numpy sum运行长度为非零值

3 个答案: