Apply numpy.where on condition with either list or numpy.array

Question

I discovered that numpy.where behaves differently when applied on a condition such as foo==2 when foo is a list or foo is a numpy.array

foo = ["a","b","c"]
bar = numpy.array(["a","b","c"])
numpy.where(foo == "a") # Returns array([])
numpy.where(bar == "a") # Returns array([0])

I want the same command to make this applicable to either list or numpy.array, and I am concerned about how to perform this efficiently. Is the following ok ?

numpy.where(numpy.array(foo, copy=False) == "a") # Returns array([0])
numpy.where(numpy.array(bar, copy=False) == "a") # Returns array([0])

Result is as expected, but is this the best way to answer my need ? Using each time numpy.array constructor is the best way to ensure object type ?

Thanks !

Answer 1

To me your solution is already the best:

numpy.where(numpy.array(foo, copy=False) == "a")

It is concise, very clear and totaly efficient thanks to copy=False .

Answer 2

If you are really looking for the most numpy -esque solution, use np.asarray :

numpy.where(numpy.asarray(foo) == "a")

And if you also want your code to work with subclasses of numpy.ndarray , without "downconverting" them to their base class of ndarray , then use np.asanyarray :

numpy.where(numpy.asanyarray(foo) == "a")

This works for np.matrix for instance, without converting it to an array. I suppose that this would also ensure that the np.matrix instance doesn't get copied or reconstructed into an array prior to checking.

Note: I think that copies are made by np.array for lists, because it needs to construct the array object. This can be seen in the documentation for np.array :

copy : bool, optional
    If true (default), then the object is copied.
    Otherwise, a copy will only be made if __array__ returns a copy, if
    obj is a nested sequence, or if a copy is needed to satisfy any of the
    other requirements (dtype, order, etc.).

np.asarray would also make a copy in this case.