Python中的MATLAB风格的find（）函数 [英] MATLAB-style find() function in Python

问题描述

在MATLAB中，很容易找到符合特定条件的值的索引：

$ $ $ $ $ $ $ $ $ $ $& a = [1,2,3,1,2,3,1,2,3];
>> find（a> 2）％找到这个条件为真的indecies
[3,6,9]％（MATLAB使用基于1的索引）
>> a（find（a> 2））％得到这些位置的值
[3,3,3]

在Python中这样做的最好方法是什么？

到目前为止，我已经提出了以下内容。要获取值：

 >>> a = [1,2,3,1,2,3,1,2,3] 
>>> val中val的值，如果val> 2] 
 [3，3，3] 
  

但是如果我想要每个这些值有点复杂：

 >>> a = [1,2,3,1,2,3,1,2,3] 
>>> inds = [i for（i，val）in enumerate（a）if val> 2] 
>>> inds 
 [2,5,8] 
>>> （a）如果我在工作中] 
 [3，3，3] 
  

有没有更好的方法来做到这一点在Python中，尤其是对于任意条件（而不是'val> 2'）？

我发现函数相当于NumPy中的MATLAB'find'，但我目前无法访问那些库。 解决方案

您可以创建一个函数它需要一个可调用的参数，这个参数将在列表理解的条件部分使用。然后，您可以使用 lambda 或其他函数对象来传递任意条件：
$ b $ pre $ code $ def $ index $ $ b $ return ）if func（val）]

a = [1,2,3,1,2,3,1,2,3]

inds = indices（a，lambda x：x> 2）

>>> inds
[2，5，8]

接近你的Matlab例子，而不必加载所有的numpy。

In MATLAB it is easy to find the indices of values that meet a particular condition:

>> a = [1,2,3,1,2,3,1,2,3];
>> find(a > 2)     % find the indecies where this condition is true
[3, 6, 9]          % (MATLAB uses 1-based indexing)
>> a(find(a > 2))  % get the values at those locations
[3, 3, 3]

What would be the best way to do this in Python?

So far, I have come up with the following. To just get the values:

>>> a = [1,2,3,1,2,3,1,2,3]
>>> [val for val in a if val > 2]
[3, 3, 3]

But if I want the index of each of those values it's a bit more complicated:

>>> a = [1,2,3,1,2,3,1,2,3]
>>> inds = [i for (i, val) in enumerate(a) if val > 2]
>>> inds
[2, 5, 8]
>>> [val for (i, val) in enumerate(a) if i in inds]
[3, 3, 3]

Is there a better way to do this in Python, especially for arbitrary conditions (not just 'val > 2')?

I found functions equivalent to MATLAB 'find' in NumPy but I currently do not have access to those libraries.

解决方案

You can make a function that takes a callable parameter which will be used in the condition part of your list comprehension. Then you can use a lambda or other function object to pass your arbitrary condition:

def indices(a, func):
    return [i for (i, val) in enumerate(a) if func(val)]

a = [1, 2, 3, 1, 2, 3, 1, 2, 3]

inds = indices(a, lambda x: x > 2)

>>> inds
[2, 5, 8]

It's a little closer to your Matlab example, without having to load up all of numpy.

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