Find the minimum and maximum indices of a list given a condition

Question

I have a list, let's say:

list_A = [0,0,0,1.0,2.0,3.0,2.0,1.0,0,0,0]

I would like to find the minimum and maximum indices of this list where list_A > 0, i.e. in the above example, it would be 3 and 7.

For other lists, which increase monotonically, I have been using np.searchsorted, like np.searchsorted(list,[0.5,1.0]) to find the indices wherein the list is between 0.5 and 1.0 respectively.

But this case is quite different and the np.searchsorted doesn't work here, or maybe it does in a way which I don't know !

Answer 1

Filter the zipped list with its indixes and take the min and the max:

>>> list_A = [0,0,0,1.0,2.0,3.0,2.0,1.0,0,0,0]
>>> filtered_lst = [(x,y) for x,y in enumerate(list_A) if y > 0]
>>> max(filtered_lst)
(7, 1.0)
>>> min(filtered_lst)
(3, 1.0)

If you just need the index, unpack the returned value:

>>> maX,_ =  max(filtered_lst)
>>> maX
7

Answer 2

An alternative would be to use next():

list_A = [0,0,0,1.0,2.0,3.0,2.0,1.0,0,0,0]

print(next(idx for idx, item in enumerate(list_A) if item>0))
print(next(len(list_A)-1-idx for idx, item in enumerate(list_A[::-1]) if item>0))

Output

3
7

Using next() to find the first item in the list > 0 is an elegant solution.

To find the last item in the list > 0 is trickier with this method. I use next() to iterate over and find the first item > 0 in the reversed list using list_A[::-1]. I then convert the index generated to the correct index by subtracting it from len(list)-1, using len(list)-1-idx .

Answer 3

You can use the np.where function to return indices of all elements > 0

In [116]: list_A = [0,0,0,1.0,2.0,3.0,2.0,1.0,0,0,0]

In [117]: arr = np.array(list_A)

In [118]: indx = np.where(arr > 0)[0]

In [119]: mini = indx[0]

In [120]: mini
Out[120]: 3

In [121]: maxi = indx[-1]

In [122]: maxi
Out[122]: 7