Reputation: 167
I am using the where function in numpy to look for a one letter string in an array of strings.
For example:
I will look for 'U'
in ['B' 'U' 'A' 'M' 'R' 'O']
and get the index of 'U'
.
letter = 'U'
row = ['B', 'U', 'A', 'M', 'R', 'O']
letter_found = np.where(row == letter)
However when I am looking for a letter that isn't present in the array of strings I get an empty tuple that looks like this:
(array([], dtype=int64),)
I need to be able to detect when it does not find the letter I am looking for in the array.
I've tried the following:
if not letter_found:
print 'not found'
But this does not work. How can I detect that the tuple
returned from the where function in numpy
is empty? Is it because one of my variables is possibly the wrong type? I am new at python
and programming in general.
Upvotes: 9
Views: 13980
Reputation: 35980
The nomeclature:
if some_iterable:
#only if non-empty
only works when something is empty. In your case, the tuple isn't actually empty. The thing the tuple contains is empty. So you might want to do the following:
if any(map(len, my_tuple)):
#passes if any of the contained items are not empty
as len
on an empty iterable will yield 0
and thus will be converted to False
.
Upvotes: 16
Reputation: 7592
Your test is failing because letter_found
is actually a tuple containing one element, so it's not empty. numpy.where
returns a tuple of index values, one for each dimension in the array that you're testing. Typically when using this for searching in one-dimensional arrays, I use Python's tuple unpacking to avoid just this sort of situation:
letter = 'U'
row = ['B', 'U', 'A', 'M', 'R', 'O']
letter_found, = np.where(row == letter)
Note the comma after letter_found
. This will unpack the result from numpy.where
and assign letter_found
to be the first element of that tuple.
Note also that letter_found
will now refer to a numpy array, which cannot be used in a boolean context. You'll have to do something like:
if len(letter_found) == 0:
print('not found!')
Upvotes: 4