Fastest way to create strictly increasing lists in Python

Question

I would like to find out what is the most efficient way to achieve the following in Python:

Suppose we have two lists a and b which are of equa

Answer 1

Here is a vanilla Python solution that does one pass:

>>> a = [2,1,2,3,4,5,4,6,5,7,8,9,8,10,11]
>>> b = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
>>> a_new, b_new = [], []
>>> last = float('-inf')
>>> for x, y in zip(a, b):
...     if x > last:
...         last = x
...         a_new.append(x)
...         b_new.append(y)
...
>>> a_new
[2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
>>> b_new
[1, 4, 5, 6, 8, 10, 11, 12, 14, 15]

I'm curious to see how it compares to the numpy solution, which will have similar time complexity but does a couple of passes on the data.

Here are some timings. First, setup:

>>> small = ([2,1,2,3,4,5,4,6,5,7,8,9,8,10,11], [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15])
>>> medium = (np.random.randint(1, 10000, (10000,)), np.random.randint(1, 10000, (10000,)))
>>> large = (np.random.randint(1, 10000000, (10000000,)), np.random.randint(1, 10000000, (10000000,)))

And now the two approaches:

>>> def monotonic(a, b):
...     a_new, b_new = [], []
...     last = float('-inf')
...     for x,y in zip(a,b):
...         if x > last:
...             last = x
...             a_new.append(x)
...             b_new.append(y)
...     return a_new, b_new
...
>>> def np_monotonic(a, b):
...     a_new, idx = np.unique(np.maximum.accumulate(a), return_index=True)
...     return a_new, b[idx]
...

Note, the approaches are not strictly equivalent, one stays in vanilla Python land, the other stays in numpy array land. We will compare performance assuming you are starting with the corresponding data structure (either numpy.array or list):

So first, a small list, the same from the OP's example, we see that numpy is not actually faster, which isn't surprising for small data structures:

>>> timeit.timeit("monotonic(a,b)", "from __main__ import monotonic, small; a, b = small", number=10000)
0.039130652003223076
>>> timeit.timeit("np_monotonic(a,b)", "from __main__ import np_monotonic, small, np; a, b = np.array(small[0]), np.array(small[1])", number=10000)
0.10779813499539159

Now a "medium" list/array of 10,000 elements, we start to see numpy advantages:

>>> timeit.timeit("monotonic(a,b)", "from __main__ import monotonic, medium; a, b = medium[0].tolist(), medium[1].tolist()", number=10000)
4.642718859016895
>>> timeit.timeit("np_monotonic(a,b)", "from __main__ import np_monotonic, medium; a, b = medium", number=10000)
1.3776302759943064

Now, interestingly, the advantage seems to narrow with "large" arrays, on the order of 1e7 elements:

>>> timeit.timeit("monotonic(a,b)", "from __main__ import monotonic, large; a, b = large[0].tolist(), large[1].tolist()", number=10)
4.400254560023313
>>> timeit.timeit("np_monotonic(a,b)", "from __main__ import np_monotonic, large; a, b = large", number=10)
3.593393853981979

Note, in the last pair of timings, I only did them 10 times each, but if someone has a better machine or more patience, please feel free to increase number

Answer 2

unique with return_index uses argsort. With maximum.accumulate that isn't needed. So we can cannibalize unique and do:

In [313]: a = [2,1,2,3,4,5,4,6,5,7,8,9,8,10,11]
In [314]: arr = np.array(a)
In [315]: aux = np.maximum.accumulate(arr)
In [316]: flag = np.concatenate(([True], aux[1:] != aux[:-1])) # key unique step
In [317]: idx = np.nonzero(flag)
In [318]: idx
Out[318]: (array([ 0,  3,  4,  5,  7,  9, 10, 11, 13, 14], dtype=int32),)
In [319]: arr[idx]
Out[319]: array([ 2,  3,  4,  5,  6,  7,  8,  9, 10, 11])
In [320]: np.array(b)[idx]
Out[320]: array([ 1,  4,  5,  6,  8, 10, 11, 12, 14, 15])

In [323]: np.unique(aux, return_index=True)[1]
Out[323]: array([ 0,  3,  4,  5,  7,  9, 10, 11, 13, 14], dtype=int32)

---

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

In [330]: timeit foo(arr)
....
100000 loops, best of 3: 12.5 µs per loop
In [331]: timeit np.unique(np.maximum.accumulate(arr), return_index=True)[1]
....
10000 loops, best of 3: 21.5 µs per loop

With (10000,) shape medium this sort-less unique has a substantial speed advantage:

In [334]: timeit np.unique(np.maximum.accumulate(medium[0]), return_index=True)[1]
1000 loops, best of 3: 351 µs per loop
In [335]: timeit foo(medium[0])
The slowest run took 4.14 times longer ....
10000 loops, best of 3: 48.9 µs per loop

[1]: Use np.source(np.unique) to see code, or ?? in IPython

Answer 3

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])

Answer 4

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

Answer 5

a = [2, 1, 2, 3, 4, 5, 4, 6, 5, 7, 8, 9, 8,10,11]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14,15]
print(sorted(set(a)))
print(sorted(set(b)))
#[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
#[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]