用于多个起始值和终止值的矢量化NumPy Linspace [英] Vectorized NumPy linspace for multiple start and stop values

问题描述

我需要创建一个2D数组，其中每一行都可以以不同的数字开头和结尾.假设给出了每行的第一个和最后一个元素，并且仅根据行的长度对所有其他元素进行插值.在一个简单的情况下，假设我要创建一个3X3数组，该数组的起点从0开始，但结尾由W指定，如下所示:

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.]])

是否有比以下方法更好的方法:

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

以下是使用 broadcasting -

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 multi-core使用numexpr模块处理大数据并获得内存效率，从而提高性能-

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')

这篇关于用于多个起始值和终止值的矢量化NumPy Linspace的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！