在Python中，筛选小于给定公差的列表元素。

huangapple
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英文:

List elements less than tolerance in Python

问题

  1. if (Test[i][j]<tol):
英文:

I have a list Test containing numpy arrays. I want to see if each array element is less than the tol value. If it is less, it should return empty list. But I am getting an error. I present the expected output.

  1. import numpy as np
  2. tol=1e-12
  3. Test=[[np.array([9.75016872e-15])], [np.array([9.75016872e-15]), np.array([0.00019793]), np.array([0.0001007])]]
  5. for i in range(0,len(Test)):
  6.     for j in range(0,len(Test[i])): 
  7.         if (Test[j][i]&lt;tol): 
  8.             Test[j][i]=[] 
  9.         else: 
  10.             Test=Test[j][i]
  11. print(Test)

The error is

  1. in &lt;module&gt;
  2.     if (CB[j][i]&lt;tol):
  4. IndexError: list index out of range

The expected output is

  1. [[[]], [[], array([0.00019793]), array([0.0001007])]]

答案1

得分: 2

  1. # 列表推导式怎么样？
  2. out = [[[] if (a < tol).all() else a for a in l] for l in Test]
  4. # 修复你的代码：
  5. for i in range(len(Test)):
  6.     for j in range(len(Test[i])): 
  7.         if Test[i][j] < tol:
  8.             Test[i][j] = []
英文:

What about a list comprehension?

  1. out = [[[] if (a&lt;tol).all() else a for a in l] for l in Test]
  3. # [[[]], [[], array([0.00019793]), array([0.0001007])]]

Fix of your code:

  1. for i in range(len(Test)):
  2.     for j in range(len(Test[i])): 
  3.         if Test[i][j] &lt; tol:
  4.             Test[i][j] = []

huangapple
  • 本文由 发表于 2023年2月14日 00:42:25
  • 转载请务必保留本文链接：https://go.coder-hub.com/75438803.html
  • list
  • numpy
  • python
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