使用python列表理解根据条件查找元素的索引 [英] Finding the index of elements based on a condition using python list comprehension

问题描述

当来自 Matlab 背景时，以下 Python 代码似乎很冗长

<预><代码>>>>a = [1, 2, 3, 1, 2, 3]>>>[索引的索引，枚举中的值(a)如果值>2][2, 5]

在 Matlab 中我可以写:

<代码>>>a = [1, 2, 3, 1, 2, 3];>>找到(a>2)答案 =3 6

是否有用 Python 编写此代码的速记方法，还是我只使用长版本?

<小时>

感谢您对 Python 语法原理的所有建议和解释.

在numpy网站上找到以下内容后，我想我找到了一个我喜欢的解决方案:

http://docs.scipy.org/doc/numpy/user/basics.indexing.html#boolean-or-mask-index-arrays

将来自该网站的信息应用于我上面的问题，将给出以下内容:

<预><代码>>>>从 numpy 导入数组>>>a = 数组([1, 2, 3, 1, 2, 3])>>>b＝a＞2数组([假，假，真，假，假，真]，dtype=bool)>>>r = 数组(范围(len(b)))>>>r(b)[2, 5]

接下来应该可以工作(但我手头没有 Python 解释器来测试它):

class my_array(numpy.array):def find(self, b):r = 数组(范围(len(b)))返回 r(b)>>>a = my_array([1, 2, 3, 1, 2, 3])>>>a.find(a>2)[2, 5]

解决方案

• 在 Python 中，您根本不会为此使用索引，而只是处理值——[value for value in a if value >2].通常处理索引意味着你没有以最好的方式做某事.

• 如果您确实需要一个类似于 Matlab 的 API，您可以使用 numpy，一个Python 中的多维数组和数值数学包，深受 Matlab 的启发.您将使用 numpy 数组而不是列表.

  <预><代码>>>>导入 numpy>>>a = numpy.array([1, 2, 3, 1, 2, 3])>>>一种数组([1, 2, 3, 1, 2, 3])>>>numpy.where(a > 2)(数组([2, 5]),)>>>＞2数组([假，假，真，假，假，真]，dtype=bool)>>>[numpy.where(a > 2)]数组([3, 3])>>>[a＞2]数组([3, 3])

The following Python code appears to be very long winded when coming from a Matlab background

>>> a = [1, 2, 3, 1, 2, 3]
>>> [index for index,value in enumerate(a) if value > 2]
[2, 5]

When in Matlab I can write:

>> a = [1, 2, 3, 1, 2, 3];
>> find(a>2)
ans =
     3     6

Is there a short hand method of writing this in Python, or do I just stick with the long version?

---

Thank you for all the suggestions and explanation of the rationale for Python's syntax.

After finding the following on the numpy website, I think I have found a solution I like:

http://docs.scipy.org/doc/numpy/user/basics.indexing.html#boolean-or-mask-index-arrays

Applying the information from that website to my problem above, would give the following:

>>> from numpy import array
>>> a = array([1, 2, 3, 1, 2, 3])
>>> b = a>2 
array([False, False, True, False, False, True], dtype=bool)
>>> r = array(range(len(b)))
>>> r(b)
[2, 5]

The following should then work (but I haven't got a Python interpreter on hand to test it):

class my_array(numpy.array):
    def find(self, b):
        r = array(range(len(b)))
        return r(b)


>>> a = my_array([1, 2, 3, 1, 2, 3])
>>> a.find(a>2)
[2, 5]

解决方案

• In Python, you wouldn't use indexes for this at all, but just deal with the values—[value for value in a if value > 2]. Usually dealing with indexes means you're not doing something the best way.

• If you do need an API similar to Matlab's, you would use numpy, a package for multidimensional arrays and numerical math in Python which is heavily inspired by Matlab. You would be using a numpy array instead of a list.

  >>> import numpy
  >>> a = numpy.array([1, 2, 3, 1, 2, 3])
  >>> a
  array([1, 2, 3, 1, 2, 3])
  >>> numpy.where(a > 2)
  (array([2, 5]),)
  >>> a > 2
  array([False, False,  True, False, False,  True], dtype=bool)
  >>> a[numpy.where(a > 2)]
  array([3, 3])
  >>> a[a > 2]
  array([3, 3])
  

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