using numpy_where on a 1D array

Question

I am trying to use numpy_where to find the index of a particular value. Though I have searched quite a bit on the web including stackoverflow I did not find a simple 1D example.

ar=[3,1,4,8,2,1,0]
>>> np.where(ar==8)
(array([], dtype=int64),)

I expected np.where(ar==8) to return me the index/location of 8 in the the array. What am I doing wrong? Is it something in my array? Thanks

Answer 1

This is a really good example of how the range of variable types in Python and numpy can be confusing for a beginner. What's happening is [3,1,4,8,2,1,0] returns a list, not an ndarray. So, the expression ar == 8 returns a scalar False, because all comparisons between list and scalar types return False. Thus, np.where(False) returns an empty array. The way to fix this is:

arr = np.array([3,1,4,8,2,1,0])
np.where(arr == 8)

This returns (array([3]),). There's opportunity for further confusion, because where returns a tuple. If you write a script that intends to access the index position (3, in this case), you need np.where(arr == 8)[0] to pull the first (and only) result out of the tuple. To actually get the value 3, you need np.where(arr == 8)[0][0] (although this will raise an IndexError if there are no 8's in the array).

This is an example where numeric-specialized languages like Matlab or Octave are simpler to use for newbies, because the language is less general and so has fewer return types to understand.