user344226 - 7 months ago 92
Python Question

# MATLAB-style find() function in Python

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.

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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