使用numpy.interp进行线性插值

3 回答

• 13

  你可以使用scipy.interpolate.interp1d：

  >>> from scipy.interpolate import interp1d
  >>> import numpy as np
  >>> x = np.array([10., 20., np.nan, 40., 50., np.nan, 30.])
  >>> not_nan = np.logical_not(np.isnan(x))
  >>> indices = np.arange(len(x))
  >>> interp = interp1d(indices[not_nan], x[not_nan])
  >>> interp(indices)
  array([ 10.,  20.,  30.,  40.,  50.,  40.,  30.])
  

  EDIT ：我花了一段时间才弄明白 np.interp 的工作原理，但这也可以完成这项工作：

  >>> np.interp(indices, indices[not_nan], x[not_nan])
  array([ 10.,  20.,  30.,  40.,  50.,  40.,  30.])
  

  回复于 2024-04-27T16:20:31+08:00
• 7

  我会选择 pandas . 使用oneliner的简约方法：

  from pandas import *
  a=np.array([10.,20.,nan,40.,50.,nan,30.])
  Series(a).interpolate()   
  
  Out[219]:
  0    10
  1    20
  2    30
  3    40
  4    50
  5    40
  6    30
  

  或者，如果您想将其保留为数组：

  Series(a).interpolate().values
  
  Out[221]:
  array([ 10.,  20.,  30.,  40.,  50.,  40.,  30.])
  

  回复于 2024-04-27T16:20:31+08:00
• 0

  要在每次要插入数据时不在系列中创建新的Series对象或新项，请使用RedBlackPy . 请参阅以下代码示例：

  import redblackpy as rb
  
  # we do not include missing data
  index = [0,1,3,4,6]
  data = [10,20,40,50,30]
  # create Series object
  series = rb.Series(index=index, values=data, dtype='float32',
                     interpolate='linear')
  
  # Now you have access at any key using linear interpolation
  # Interpolation does not creates new items in Series
  print(series[2]) # prints 30
  print(series[5]) # prints 40
  # print Series and see that keys 2 and 5 do not exist in series
  print(series)
  

  最后一个输出如下：

  Series object Untitled
  0: 10.0
  1: 20.0
  3: 40.0
  4: 50.0
  6: 30.0
  

  回复于 2024-04-27T16:20:31+08:00