用于多个开始和停止值的矢量化 NumPy linspace [英] Vectorized NumPy linspace for multiple start and stop values

问题描述

我需要创建一个二维数组，其中每一行都可以以不同的数字开始和结束.假设给定了每行的第一个和最后一个元素，并且所有其他元素只是根据行的长度进行插值在一个简单的情况下，假设我想创建一个 3X3 数组，其起点相同，但结尾不同，由 W 下面给出:

I need to create a 2D array where each row may start and end with a different number. Assume that first and last element of each row is given and all other elements are just interpolated according to length of the rows In a simple case let's say I want to create a 3X3 array with same start at 0 but different end given by W below:

array([[ 0.,  1.,  2.],
       [ 0.,  2.,  4.],
       [ 0.,  3.,  6.]])

有没有比以下更好的方法来做到这一点:

Is there a better way to do this than the following:

D=np.ones((3,3))*np.arange(0,3)
D=D/D[:,-1] 
W=np.array([2,4,6]) # last element of each row assumed given
Res= (D.T*W).T  

推荐答案

这是使用 广播 -

Here's an approach using broadcasting -

def create_ranges(start, stop, N, endpoint=True):
    if endpoint==1:
        divisor = N-1
    else:
        divisor = N
    steps = (1.0/divisor) * (stop - start)
    return steps[:,None]*np.arange(N) + start[:,None]

样品运行 -

In [22]: # Setup start, stop for each row and no. of elems in each row
    ...: start = np.array([1,4,2])
    ...: stop  = np.array([6,7,6])
    ...: N = 5
    ...: 

In [23]: create_ranges(start, stop, 5)
Out[23]: 
array([[ 1.  ,  2.25,  3.5 ,  4.75,  6.  ],
       [ 4.  ,  4.75,  5.5 ,  6.25,  7.  ],
       [ 2.  ,  3.  ,  4.  ,  5.  ,  6.  ]])

In [24]: create_ranges(start, stop, 5, endpoint=False)
Out[24]: 
array([[ 1. ,  2. ,  3. ,  4. ,  5. ],
       [ 4. ,  4.6,  5.2,  5.8,  6.4],
       [ 2. ,  2.8,  3.6,  4.4,  5.2]])

让我们利用多核！

我们可以利用 多核与numexpr模块用于大数据并获得内存效率和性能 -

Let's leverage multi-core!

We can leverage multi-core with numexpr module for large data and to gain memory efficiency and hence performance -

import numexpr as ne

def create_ranges_numexpr(start, stop, N, endpoint=True):
    if endpoint==1:
        divisor = N-1
    else:
        divisor = N
    s0 = start[:,None]
    s1 = stop[:,None]
    r = np.arange(N)
    return ne.evaluate('((1.0/divisor) * (s1 - s0))*r + s0')

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