Good numpy implementation of the algorithm. If, for whatever reason, numpy isn't available, you can also pull it with a good comprehension:

    def count_doubles2(val):
        return sum(1 for c1, c2 in zip(val, val[1:]) if c1 == c2)

Which will also allow you to avoid a function call entirely, if it was useful in some way:

    In [56]: %timeit count_doubles(val)
    198 ms ± 13.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

    In [57]: %timeit count_doubles2(val)
    189 ms ± 21.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

    In [58]: %timeit sum(1 for c1, c2 in zip(val, val[1:]) if c1 == c2)
    135 ms ± 3.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

    In [59]: %timeit count_double_chars_np(val)
    6.95 ms ± 782 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

(Numpy still beats it, for long strings).