Fastest way to create strictly increasing lists in Python

Running a version of @juanpa.arrivillaga's function with numba

import numba

def psi(A):
    a_cummax = np.maximum.accumulate(A)
    a_new, idx = np.unique(a_cummax, return_index=True)
    return idx

def foo(arr):
    aux=np.maximum.accumulate(arr)
    flag = np.concatenate(([True], aux[1:] != aux[:-1]))
    return np.nonzero(flag)[0]

@numba.jit
def f(A):
    m = A[0]
    a_new, idx = [m], [0]
    for i, a in enumerate(A[1:], 1):
        if a > m:
            m = a
            a_new.append(a)
            idx.append(i)
    return idx

---

timing

%timeit f(a)
The slowest run took 5.37 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.83 µs per loop

%timeit foo(a)
The slowest run took 9.41 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 6.35 µs per loop

%timeit psi(a)
The slowest run took 9.66 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 9.95 µs per loop

You can calculate the cumulative max of a and then drop duplicates with np.unique with which you can also record the unique index so as to subset b correspondingly:

a = np.array([2,1,2,3,4,5,4,6,5,7,8,9,8,10,11])
b = np.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15])

a_cummax = np.maximum.accumulate(a)    
a_new, idx = np.unique(a_cummax, return_index=True)

a_new
# array([ 2,  3,  4,  5,  6,  7,  8,  9, 10, 11])

b[idx]
# array([ 1,  4,  5,  6,  8, 10, 11, 12, 14, 15])