Finding the index of elements based on a condition using python list comprehension

Question

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]

Answer 1

Another way:

>>> [i for i in range(len(a)) if a[i] > 2]
[2, 5]

In general, remember that while find is a ready-cooked function, list comprehensions are a general, and thus very powerful solution. Nothing prevents you from writing a find function in Python and use it later as you wish. I.e.:

>>> def find_indices(lst, condition):
...   return [i for i, elem in enumerate(lst) if condition(elem)]
... 
>>> find_indices(a, lambda e: e > 2)
[2, 5]

Note that I'm using lists here to mimic Matlab. It would be more Pythonic to use generators and iterators.