numpy插值以增加向量大小

问题描述

我必须扩大向量内部的点数，才能将向量扩大到固定大小.例如:

Hi I have to enlarge the number of points inside of vector to enlarge the vector to fixed size. for example:

为此简单向量

>>> a = np.array([0, 1, 2, 3, 4, 5])
>>> len(a)
# 6

现在，我想获得一个大小为11的向量，以a向量为基础，结果将是

now, I want to get a vector with size of 11 taken the a vector as base the results will be

# array([ 0. ,  0.5,  1. ,  1.5,  2. ,  2.5,  3. ,  3.5,  4. ,  4.5,  5. ])

编辑1

我需要的是一个函数，该函数将输入基本向量和必须为结果向量的值的数量，然后返回一个新的向量，其大小等于该参数.像

what I need is a function that will enter the base vector and the number of values that must be the resultant vector, and I return a new vector with size equal to the parameter. something like

def enlargeVector(vector, size):
    .....
    return newVector

使用方式如下:

>>> a = np.array([0, 1, 2, 3, 4, 5])
>>> b = enlargeVector(a, 200):
>>> len(b)
# 200

和b包含线性，三次或任何插值方法的数据结果

and b contains data results of linear, cubic, or whatever interpolation methods

推荐答案

在scipy.interpolate中有很多方法可以做到这一点.我最喜欢的是UnivariateSpline，它产生一个有序的k样条，保证可微分的k次.

There are many methods to do this within scipy.interpolate. My favourite is UnivariateSpline, which produces an order k spline guaranteed to be differentiable k times.

要使用它:

from scipy.interpolate import UnivariateSpline
old_indices = np.arange(0,len(a))
new_length = 11
new_indices = np.linspace(0,len(a)-1,new_length)
spl = UnivariateSpline(old_indices,a,k=3,s=0)
new_array = spl(new_indices)

在这种情况下，s是一个平滑因子，应将其设置为0(因为数据准确).

The s is a smoothing factor that you should set to 0 in this case (since the data are exact).

请注意，对于您指定的问题(由于a只是单调增加1)，这是过大的，因为第二个np.linspace已经提供了所需的输出.

Note that for the problem you have specified (since a just increases monotonically by 1), this is overkill, since the second np.linspace gives already the desired output.

阐明了长度是任意的

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