插入一个numpy数组以适合另一个数组 [英] Interpolating a numpy array to fit another array

问题描述

说我的形状为(1, n)的some_data.我有形状为(1, n±x)的新incoming_data，其中x是一些远小于n的正整数.我想挤压或拉伸incoming_data，使其长度与n相同.使用SciPy堆栈该怎么做?

Say I have some_data of shape (1, n). I have new incoming_data of shape (1, n±x), where x is some positive integer much smaller than n. I would like to squeeze or stretch incoming_data such that it is of the same length as n. How might this be done, using the SciPy stack?

这是我要完成的事情的一个例子.

Here's an example of what I'm trying to accomplish.

# Stretch arr2 to arr1's shape while "filling in" interpolated value
arr1 = np.array([1, 5, 2, 3, 7, 2, 1])
arr2 = np.array([1, 5, 2, 3, 7, 1])
result
> np.array([1, 5, 2, 3, 6.x, 2.x 1])  # of shape (arr1.shape)

另一个例子:

# Squeeze arr2 to arr1's shape while placing interpolated value.
arr1 = np.array([1, 5, 2, 3, 7, 2, 1])
arr2 = np.array([1, 5, 2, 3, 4, 7, 2, 1])
result
> np.array([1, 5, 2, 3.x, 7.x, 2.x, 1])  # of shape (arr1.shape)

推荐答案

您可以使用

You can implement this simple compression or stretching of your data using scipy.interpolate.interp1d. I'm not saying it necessarily makes sense (it makes a huge difference what kind of interpolation you're using, and you'll generally only get a reasonable result if you can correctly guess the behaviour of the underlying function), but you can do it.

这个想法是将原始数组的索引作为x值插值，然后使用稀疏的x网格执行插值，同时保持其端点不变.因此，从本质上讲，您必须对离散数据进行连续近似，并在必要的点对其进行重新采样:

The idea is to interpolate your original array over its indices as x values, then perform interpolation with a sparser x mesh, while keeping its end points the same. So essentially you have to do a continuum approximation to your discrete data, and resample that at the necessary points:

import numpy as np
import scipy.interpolate as interp
import matplotlib.pyplot as plt

arr_ref = np.array([1, 5, 2, 3, 7, 1])  # shape (6,), reference
arr1 = np.array([1, 5, 2, 3, 7, 2, 1])  # shape (7,), to "compress"
arr2 = np.array([1, 5, 2, 7, 1])        # shape (5,), to "stretch"
arr1_interp = interp.interp1d(np.arange(arr1.size),arr1)
arr1_compress = arr1_interp(np.linspace(0,arr1.size-1,arr_ref.size))
arr2_interp = interp.interp1d(np.arange(arr2.size),arr2)
arr2_stretch = arr2_interp(np.linspace(0,arr2.size-1,arr_ref.size))

# plot the examples, assuming same x_min, x_max for all data
xmin,xmax = 0,1
fig,(ax1,ax2) = plt.subplots(ncols=2)
ax1.plot(np.linspace(xmin,xmax,arr1.size),arr1,'bo-',
         np.linspace(xmin,xmax,arr1_compress.size),arr1_compress,'rs')
ax2.plot(np.linspace(xmin,xmax,arr2.size),arr2,'bo-',
         np.linspace(xmin,xmax,arr2_stretch.size),arr2_stretch,'rs') 
ax1.set_title('"compress"')
ax2.set_title('"stretch"')

结果图:

在图中，蓝色圆圈是原始数据点，红色正方形是内插的数据点(这些边界重叠).如您所见，我所谓的压缩和拉伸实际上是对基础(默认为线性)函数的上采样和下采样.这就是为什么我说您必须非常小心地进行插值:如果期望值与数据不匹配，您可能会得到非常错误的结果.

In the plots, blue circles are the original data points, and red squares are the interpolated ones (these overlap at the boundaries). As you can see, what I called compressing and stretching is actually upsampling and downsampling of an underlying (linear, by default) function. This is why I said you must be very careful with interpolation: you can get very wrong results if your expectations don't match your data.

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