Develop Reference

Numba is not enhancing the performance

I am testing numba performance on some function that takes a numpy array, and compare:
import numpy as np
from numba import jit, vectorize, float64
import time
from numba.core.errors import NumbaWarning
import warnings
warnings.simplefilter('ignore', category=NumbaWarning)
#jit(nopython=True, boundscheck=False) # Set "nopython" mode for best performance, equivalent to #njit
def go_fast(a): # Function is compiled to machine code when called the first time
trace = 0.0
for i in range(a.shape[0]): # Numba likes loops
trace += np.tanh(a[i, i]) # Numba likes NumPy functions
return a + trace # Numba likes NumPy broadcasting
class Main(object):
def __init__(self) -> None:
super().__init__()
self.mat = np.arange(100000000, dtype=np.float64).reshape(10000, 10000)
def my_run(self):
st = time.time()
trace = 0.0
for i in range(self.mat.shape[0]):
trace += np.tanh(self.mat[i, i])
res = self.mat + trace
print('Python Diration: ', time.time() - st)
return res
def jit_run(self):
st = time.time()
res = go_fast(self.mat)
print('Jit Diration: ', time.time() - st)
return res
obj = Main()
x1 = obj.my_run()
x2 = obj.jit_run()
The output is:
Python Diration: 0.2164750099182129
Jit Diration: 0.5367801189422607
How can I obtain an enhance version of this example ?

The slower execution time of the Numba implementation is due to the compilation time since Numba compile the function at the time it is used (only the first time unless the type of the argument change). It does that because it cannot know the type of the arguments before the function is called. Hopefully, you can specify the argument type to Numba so it can compile the function directly (when the decorator function is executed). Here is the resulting code:
#njit('float64[:,:](float64[:,:])')
def go_fast(a):
trace = 0.0
for i in range(a.shape[0]):
trace += np.tanh(a[i, i])
return a + trace
Note that njit is a shortcut for jit+nopython=True and that boundscheck is already set to False by default (see the doc).
On my machine this result in the same execution time for both Numpy and Numba. Indeed, the execution time is not bounded by the computation of the tanh function. It is bounded by the expression a + trace (for both Numba and Numpy). The same execution time is expected since both implement this the same way: they create a temporary new array to perform the addition. Creating a new temporary array is expensive because of page faults and the use of the RAM (a is fully read from the RAM and the temporary array is fully stored in RAM). If you want a faster computation, then you need to perform the operation in-place (this prevent page faults and expensive cache-line write allocations on x86 platforms).

Compile Scipy function with Cython

I am running a simulation in Python 3.4 - that involves a lot of dot products between a sparse array (in csr format) and a dense vector. I am using Scipy for the sparse matrix, numpy for everything else.
Using Cython gave me a massive boost (~x6 speed increase), after making sure that I cdef everything properly and after minimizing Python interaction (bt going through the html file that Cython gives me and modifying my code).
Now, I profile the code and 50% of the simulation time is spent on the line with the dot product. I am wondering if it is possible to somehow accelerate this line, say by complining this one dot function in Cython?
I know I could write my own implementation for (csr sprase 2d matrix) dot (dense vector), but I am trying to avoid that.
Edit: I have included a minimal example of the code. I am sorry, I can't see how to make it smaller. It is a textbook exercise in statistical mechanics. Place marbles in pots until one of the pots exceeds capacity. Then, start a cascade which propagates according to a (here sparse) matrix. I am using batch sampling.
Please focus on the line towards the end.
from __future__ import division
import numpy as np
import cython
cimport numpy as np
cimport cpython.array
import scipy.sparse as sps
#cython.cdivision(True)
#cython.nonecheck(False)
#cython.boundscheck(False)
#cython.wraparound(False)
def simulate(long[:] capacity_vec,
int random_array_size,
long n,
int seed,
int[:] A_col,
int[:] A_row,
long[:] A_data):
#### Initialise ####################################################################################################
# Initialise states
cdef int[:] damage = np.random.randint(0, int(np.min(capacity_vec)/2), n).astype(np.int32)
cdef int[:] dr_list = np.random.choice(n, 1000).astype(np.int32)
cdef int[:] states = np.zeros(n).astype(np.int32)
cdef int[:] states_ = np.zeros(n).astype(np.int32)
cdef int[:] change = np.zeros(n).astype(np.int32)
# Initialise counters
cdef int k, violations, violations_, counter= 0, dr_id=0, increment_index = 0
# Build Sparse Adjecency Matrix
cA_sps = sps.csr_matrix( (A_data, (A_row, A_col) ), shape=(n,n) ).astype(np.int32)
while counter < 1000:
#### Place damage until a cascade starts #######################################################################
while damage[increment_index] <= capacity_vec[increment_index]:# Check for violations
increment_index = dr_list[dr_id] # Where do we place the marble?
damage[increment_index] = damage[increment_index] + 1 # place the marble
dr_id = dr_id + 1 # another random number used
if dr_id == random_array_size - 1: # Check if we run out of random numbers
dr_list = np.random.choice(n, random_array_size).astype(np.int32) # if so, pick new increment_index
dr_id = 0 # and reset the counter
#### Initialise cascade ########################################################################################
violations, violations_ = 1, 0
states[increment_index] = 1
#### Propagate cascade #########################################################################################
while violations > violations_: # check for fixed point, propagate cascade
for k in range(n): change[k] = states[k] - states_[k]
### THIS LINE IS THE PROBLEM. It takes up half of all simulation time.
damage = damage + cA_sps.dot(change).astype(np.int32) # spread violations
states_ = states.copy() # store previous states
# Determine previous and current violations
violations, violations_ = 0 , violations
for k in range(n):
states_[k] = 0
if damage[k] > capacity_vec[k]:
violations = violations + 1
states[k] = 1 # deactivate any node that has a violation
for k in range(n): damage[k] = 0
counter = counter + 1 # progress cascade id after storing

I'd discourage from writing your own matrix multiplication. SciPy is done by smart people who know what they are doing, and unless your confident in numerical computing, just don't. Most of SciPy code is already compiled.
However, what you might look at is code for sparse.csr_matrix.dot. Getting into definition directly here and then here, you'll see that there are few checks done in Scipy. If you know what exact form you want, you could write your own method (modify your SciPy copy) and compute your product directly. Not sure how much that would help, though.
If you want to build Scipy yourself it is as easy as checking out whole project from GitHug and then running
python setup.py build
python setup.py install
For more direct instructions check build documentation.

Reading hdf5 file quickly with cython and h5py

I'm trying to speed up a python3 function that takes some data, which is an array of indexes and saves them if they meet a certain criterion. I have tried to speed it up by using "cython -a script.py", but the bottle neck seems to be the h5py I/O slicing datasets.
I'm relatively new to cython, so I was wondering whether there is anyway to speed this up or am I just limited by the h5py I/O here?
Here is the function I'm trying to improve:
import numpy as np
import h5py
cimport numpy as np
cimport cython
from libc.math cimport sqrt
DTYPE64 = np.int64
ctypedef np.int64_t DTYPE64_t
DTYPE32 = np.int32
ctypedef np.int32_t DTYPE32_t
#cython.boundscheck(False)
#cython.wraparound(False)
def tag_subhalo_branch(np.ndarray[DTYPE64_t] halos_z0_treeindxs,
np.ndarray[DTYPE64_t] tree_pindx,
np.ndarray[DTYPE32_t] tree_psnapnum,
np.ndarray[DTYPE64_t] tree_psnapid,
np.ndarray[DTYPE64_t] tree_hsnapid, hf,
int size):
cdef int i
cdef double radial, progen_x, progen_y, progen_z
cdef double host_x, host_y, host_z, host_rvir
cdef DTYPE64_t progen_indx, progen_haloid, host_id
cdef DTYPE32_t progen_snap
cdef int j = 0
cdef int size_array = size
cdef np.ndarray[DTYPE64_t] backsplash_ids = np.zeros(size_array,
dtype=DTYPE64)
for i in range(0, size_array):
progen_indx = tree_pindx[halos_z0_treeindxs[i]]
if progen_indx != -1:
progen_snap = tree_psnapnum[progen_indx]
progen_haloid = tree_psnapid[progen_indx]
while progen_indx != -1 and progen_snap != -1:
# ** This is slow **
grp = hf['Snapshots/snap_' + str('%03d' % progen_snap) + '/']
host_id = grp['HaloCatalog'][(progen_haloid - 1), 2]
# **
if host_id != -1:
# ** This is slow **
progen_x = grp['HaloCatalog'][(progen_haloid - 1), 6]
host_x = grp['HaloCatalog'][(host_id - 1), 6]
progen_y = grp['HaloCatalog'][(progen_haloid - 1), 7]
host_y = grp['HaloCatalog'][(host_id - 1), 7]
progen_z = grp['HaloCatalog'][(progen_haloid - 1), 8]
host_z = grp['HaloCatalog'][(host_id - 1), 8]
# **
radial = 0
radial += (progen_x - host_x)**2
radial += (progen_y - host_y)**2
radial += (progen_z - host_z)**2
radial = sqrt(radial)
host_rvir = grp['HaloCatalog'][(host_id - 1), 24]
if radial <= host_rvir:
backsplash_ids[j] = tree_hsnapid[
halos_z0_treeindxs[i]]
j += 1
break
# Find next progenitor information
progen_indx = tree_pindx[progen_indx]
progen_snap = tree_psnapnum[progen_indx]
progen_haloid = tree_psnapid[progen_indx]
return backsplash_ids

As described here: http://api.h5py.org/, h5py uses cython code to interface with the HDF5 c code. So your own cython code might be able to access that directly. But I suspect that will require a lot more study.
Your code is using the Python interface to h5py, and cythonizing isn't going to touch that.
cython code is best used for low level actions, especially iterative things that can't be expressed as array operations. Study and experiment with the numpy examples first. You are diving into cython at the deep end of the pool.
Have you tried to improve that code just with Python and numpy? Just at glance I'm seeing a lot of redundant h5py calls.
====================
Your radial calculation accesses the h5py indexing 6 times when it could get by with 2. Maybe you wrote it that way in hopes that cython would preform the following calculation faster than numpy?
data = grp['HaloCatalog']
progen = data[progen_haloid-1, 6:9]
host = data[host_id-1, 6:9]
radial = np.sqrt((progren-host)**2).sum(axis=1))
Why not load all data[progen_haloid-1,:] and data[host_id-1,:]? Even all of data? I'd have to review when h5py switches from working directly with the arrays on the file and when they become numpy arrays. In any case, math on arrays in memory will be a lot faster than file reads.

Fast Numpy Loops

How do you optimize this code (without vectorizing, as this leads up to using the semantics of the calculation, which is quite often far from being non-trivial):
slow_lib.py:
import numpy as np
def foo():
size = 200
np.random.seed(1000031212)
bar = np.random.rand(size, size)
moo = np.zeros((size,size), dtype = np.float)
for i in range(0,size):
for j in range(0,size):
val = bar[j]
moo += np.outer(val, val)
The point is that such kind loops correspond quite often to operations where you have double sums over some vector operation.
This is quite slow:
>>t = timeit.timeit('foo()', 'from slow_lib import foo', number = 10)
>>print ("took: "+str(t))
took: 41.165681839
Ok, so then let's cynothize it and add type annotations likes there is no tomorrow:
c_slow_lib.pyx:
import numpy as np
cimport numpy as np
import cython
#cython.boundscheck(False)
#cython.wraparound(False)
def foo():
cdef int size = 200
cdef int i,j
np.random.seed(1000031212)
cdef np.ndarray[np.double_t, ndim=2] bar = np.random.rand(size, size)
cdef np.ndarray[np.double_t, ndim=2] moo = np.zeros((size,size), dtype = np.float)
cdef np.ndarray[np.double_t, ndim=1] val
for i in xrange(0,size):
for j in xrange(0,size):
val = bar[j]
moo += np.outer(val, val)
>>t = timeit.timeit('foo()', 'from c_slow_lib import foo', number = 10)
>>print ("took: "+str(t))
took: 42.3104710579
... ehr... what? Numba to the rescue!
numba_slow_lib.py:
import numpy as np
from numba import jit
size = 200
np.random.seed(1000031212)
bar = np.random.rand(size, size)
#jit
def foo():
bar = np.random.rand(size, size)
moo = np.zeros((size,size), dtype = np.float)
for i in range(0,size):
for j in range(0,size):
val = bar[j]
moo += np.outer(val, val)
>>t = timeit.timeit('foo()', 'from numba_slow_lib import foo', number = 10)
>>print("took: "+str(t))
took: 40.7327859402
So is there really no way to speed this up? The point is:
if I convert the inner loop into a vectorized version (building a larger matrix representing the inner loop and then calling np.outer on the larger matrix) I get much faster code.
if I implement something similar in Matlab (R2016a) this performs quite well due to JIT.

Here's the code for outer:
def outer(a, b, out=None):
a = asarray(a)
b = asarray(b)
return multiply(a.ravel()[:, newaxis], b.ravel()[newaxis,:], out)
So each call to outer involves a number of python calls. Those eventually call compiled code to perform the multiplication. But each incurs an overhead that has nothing to do with the size of your arrays.
So 200 (200**2?) calls to outer will have all that overhead, whereas one call to outer with all 200 rows has one overhead set, followed by one fast compiled operation.
cython and numba don't compile or otherwise bypass the Python code in outer. All they can do is streamline the iteration code that you wrote - and that isn't consuming much time.
Without getting into details, the MATLAB jit must be able to replace the 'outer' with faster code - it rewrites the iteration. But my experience with MATLAB dates from a time before its jit.
For real speed improvements with cython and numba you need to use primitive numpy/python code all the way down. Or better yet focus your effort on slow inner pieces.
Replacing your outer with a streamlined version cuts run time about in half:
def foo1(N):
size = N
np.random.seed(1000031212)
bar = np.random.rand(size, size)
moo = np.zeros((size,size), dtype = np.float)
for i in range(0,size):
for j in range(0,size):
val = bar[j]
moo += val[:,None]*val
return moo
With the full N=200 your function took 17s per loop. If I replace the inner two lines with pass (no calculation), time drops to 3ms per loop. In other words, the outer loop mechanism is not a big time consumer, at least not compared to many calls to outer().

Memory permitting, you can use np.einsum to perform those heavy calculations in a vectorized manner, like so -
moo = size*np.einsum('ij,ik->jk',bar,bar)
One can also use np.tensordot -
moo = size*np.tensordot(bar,bar,axes=(0,0))
Or simply np.dot -
moo = size*bar.T.dot(bar)

Many tutorials and demonstrations of Cython, Numba, etc. make it seem as if these tools can speed up your code automagically, but in practice, this is often not the case: You'll need to modify your code a little to extract the best performance. If you had already implemented some degree of vectorization, it usually means writing out ALL the loops. Reasons Numpy array operations are non-optimal include:
Lots of temporary arrays are created and looped over;
Significant per-call overhead if the arrays are small;
Short-circuiting logic can't be implemented, because arrays are processed as a whole;
Sometimes the optimal algorithm can't be expressed using array expressions and you settle for an algorithm with a worse time complexity.
Using Numba or Cython wont optimize these problems away! Instead, these tools allow you to write loopy code that is much faster than plain Python.
Also, for Numba specifically, you should be aware of the difference between "object mode" and "nopython mode". The tight loops from your example have to run in nopython mode to provide any significant speedup. However, numpy.outer is not yet supported by Numba, resulting in the function to be compiled in object mode. Decorate with jit(nopython=True) to let such cases throw an exception.
Example to demonstrate a speedup is indeed possible:
import numpy as np
from numba import jit
#jit
def foo_nb(bar):
size = bar.shape[0]
moo = np.zeros((size, size))
for i in range(0,size):
for j in range(0,size):
val = bar[j]
moo += np.outer(val, val)
return moo
#jit
def foo_nb2(bar):
size = bar.shape[0]
moo = np.zeros((size, size))
for i in range(size):
for j in range(size):
for k in range(0,size):
for l in range(0,size):
moo[k,l] += bar[j,k] * bar[j,l]
return moo
size = 100
bar = np.random.rand(size, size)
np.allclose(foo_nb(bar), foo_nb2(bar))
# True
%timeit foo_nb(bar)
# 1 loop, best of 3: 816 ms per loop
%timeit foo_nb2(bar)
# 10 loops, best of 3: 176 ms per loop

The example you show us is kind of inefficient algorithm, since you calculate the same outer product multiple times. The resulting time complexity is O(n^4). It can be reduced to n^3.
for i in range(0,size):
val = bar[i]
moo += size * np.outer(val, val)

Vectorize a Newton method in Python/Numpy

I am trying to figure out if Python/Numpy is a viable alternative to develop my numerical software which is already available in C++. In order to get performance in Python/Numpy, one need to "vectorize" the code. But it turns out that as soon as I move away from very simple examples, I struggle to vectorize the code (I am not talking about SIMD instructions but "efficient Numpy code" without loops). Here is an algorithm that I want to get efficiently in Python/Numpy.
Create an numpy array containing: 1.0, 1.0 + 1/n, 1.0 + 2/n, ..., 2.0
For every u in the array, compute the root of x^2 - u, using a Newton method, stopping when |dx| <= 1.0e-7. Store the result in an array result.
Sum all the elements of the result array
Here is the algorithm in Python I want to speed up
import numpy as np
n = 1000000
data = np.arange(1.0, 2.0, 1.0 / n)
def newton(u):
x = 2.0
while True:
f = x**2 - u
df_dx = 2 * x
dx = f / df_dx
if (abs(dx) <= 1.0e-7):
break
x -= dx
return x
result = map(newton, data)
print result[n - 1]
Here is a version of the algorithm in C++11
#include <iostream>
#include <vector>
#include <cmath>
int main (int argc, char const *argv[]) {
auto n = std::size_t{100000000};
auto v = std::vector<double>(n + 1);
for(size_t k = 0; k < v.size(); ++k) {
v[k] = 1.0 + static_cast<double>(k) / n;
}
auto result = std::vector<double>(n + 1);
for(size_t k = 0; k < v.size(); ++k) {
auto x = double{2.0};
while(true) {
auto f = double{x * x - v[k]};
auto df_dx = double{2 * x};
auto dx = double{f / df_dx};
if (std::abs(dx) <= 1.0e-7) {
break;
}
x -= dx;
}
result[k] = x;
}
auto somme = double{0.0};
for(size_t k = 0; k < result.size(); ++k) {
somme += result[k];
}
std::cout << somme << std::endl;
return 0;
}
It takes 2.9 seconds to run on my machine. Is there a way to make a fast Python/Numpy algorithm that does the same thing (I am willing to get something that is less than 5 times slower).
Thanks.

You can do step 1. with numpy efficiently:
1.0 + np.arange(n + 1) / n
however I think you would need the np.vectorize() method to feed back x into your calculated values and it's not an efficient function (basically a wrapper for a python loop). If you can use scipy then there are built in methods that might do what you want http://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.optimize.newton.html
EDIT: Having thought a bit more about this I followed up on #ev-br's point and tried some alternatives. The masking uses too much processing but the abs().max() is pretty fast so a compromise might be to "divide the problem into blocks" both in the 1st dimension of the array and in iteration direction. The following doesn't do too badly (< 20s) on my pretty low power laptop - certainly much faster than np.vectorize() or any of the scipy solving systems I could find. (If I set m too big it runs out of something (memory?) and grinds to a complete halt!)
n = 100000000
m = 5000000
block = 3
u = 1.0 + np.arange(n + 1) / n
x = np.full(u.shape, 2.0)
dx = np.ones(u.shape)
for i in range(0, n, m):
while np.abs(dx[i:i+m]).max() > 1.0e-7:
for j in range(block):
dx[i:i+m] = (x[i:i+m] ** 2 - u[i:i+m]) / (2 * x[i:i+m])
x[i:i+m] -= dx[i:i+m]

Here's a toy example. Notice that often vectorization means writing your code as if you're manipulating numbers, and letting numpy do its magic:
>>> import numpy as np
>>> a = np.array([1., 2., 3.])
>>> def f(x):
... return x**2 - a, 2.*x # function and derivative
>>>
>>> def newt(f, x0):
... x = np.asarray(x0)
... for _ in range(5): # hardcode the number of iterations (I know)
... v, dv = f(x)
... x -= v / dv
... return x
>>>
>>> newt(f, [1., 1., 1.])
array([ 1. , 1.41421356, 1.73205081])
If this is a performance bottleneck, this is unlikely to be competetive with hand-written C++ code: First of all, you're manipulating python objects with all the overhead; then numpy is likely doing a bunch of array allocations under the hood.
An often viable strategy is to start by writing things in python/numpy, and then move bottlenecks into a compiled code --- eg Cython or C++ wrapped by Cython. In this particular case since you already have the C++ code, just wrapping it with Cython is likely easiest but YMMV.

I'm not looking to wave small snippets of code as a solution, but here's something to get you started. I have a strong suspicion that you're having troubles just declaring such an array in python without spending too much time on it, so I'll mostly help you out there.
As far as the square roots come in, please add your example python code and I'll see what I can help optimize from that point on. In my example roots and sums are found with the default numpy functions/methods.
def summing():
n = 1000000
ar = np.arange(0, n)
ar = ar/float(n)
ar = ar + np.ones(n)
sqrt = np.sqrt(ar)
return np.sum(ar)
In short, to get the starting array it's best to use a "workaround".
initialize an array ar with values `[1,2,3,....n]
divide ar with n. This gets us the 1/n, 2/n ... members
add to that an array of same dimensions that contain just the number 1.0
This gets us the full array [ 1., 1.000001, 1.000002, ..., 1.999998, 1.999999]) we're after. If I understood you right.
find square roots, sum it
Average of 10 sequential execution times is 0.018786 seconds.

Obviously I'm 6 years late to this party, but this question is a common stumbling block for people in effectively using numpy for real scientific work. The basic idea is covered in #ev-br's answer. The OP points out that the solution offered there (even modified to stop iterating when a convergence criterion is met rather than after a fixed number of iterations) takes the same number of passes for each element of u. I want to show how you can avoid that objection using pure numpy code, making explicit the mask suggestion in #ev-br's comment.
However, I also want to point out that in many real world situations, the number of passes for Newton-like iteration to converge varies so little that this general technique I illustrate here will actually slow numpy code down significantly. If the average number of iterations will be within a factor of two or three of the maximum number of iterations, you should stick with something closer to #ev-br's answer (including his first comment).
The numpy performance numbers you need to understand are these: Loops over array indices will run 200 to 500 times slower in pure numpy code than in compiled code. On the other hand, if you manage to use numpy's array syntax to avoid all index loops, you can get within about a factor of 5 of compiled speed. (The factor of 5 is partly because of memory management as #ev-br mentions, but also because optimized compiled code overlaps many different arithmetical operations inside each index loop, while numpy just performs a single arithmetic operation, storing everything back to memory after each operation.) The point is that factor of 100 difference means that it often pays to do substantial amounts of "extra" work in numpy code: Even if you do 10 times the number of floating point operations in vectorized numpy code, it will still run 10 times faster than the index-loop code that avoids the "extra" work. (Incidentally, the python map function is implemented as an interpreted index loop - it has nothing to do with numpy array operations.)
from numpy import asfarray, broadcast_arrays, arange
# Begin by defining the function to be inverted by Newton's method.
def f_dfdx(x):
x = asfarray(x) # always avoid repeated type conversions
return x**2, 2.*x
# First, the simplest algorithm to find x such that f(x)=y.
# We must supply a starting guess x0 for x.
def f_inverse0(f_dfdx, y, x0, tol=1.e-7):
y, x = broadcast_arrays(asfarray(y), asfarray(x0))
x = x.copy() # without this may clobber input x0
for npass in range(20):
f, dfdx = f_dfdx(x)
dx = (f - y) / dfdx
if (abs(dx) <= tol).all():
break # iterate all x until all have converged
x -= dx
else:
raise RuntimeError("failed to converge")
return x
# A frequently slower algorithm that avoids extra iterations.
def f_inverse1(f_dfdx, y, x0, tol=1.e-7):
y, x = broadcast_arrays(asfarray(y), asfarray(x0))
shape = x.shape
y, x = y.ravel(), x.flatten() # avoid clobbering x0
unconverged = arange(y.size)
for npass in range(20):
f, dfdx = f_dfdx(x[unconverged])
dx = (f - y[unconverged]) / dfdx
unc = abs(dx) > tol
unconverged = unconverged[unc]
if not unconverged.size:
break # iterate all x until all have converged
x[unconverged] -= dx[unc]
else:
raise RuntimeError("failed to converge")
return x.reshape(shape)
On my machine, the OP's C++ program runs in 2.03 s (1.64+0.38 user+sys). For n=100 million as for the C++ program, f_inverse0 runs in 20.4 s (4.7+15.6 user+sys). As expected, f_inverse1 is slower, 51.3 s (11.5+39.8 user+sys). Again, don't automatically try to minimize total operation count when you are writing numpy code. The high system overhead is probably due to heavy memory management - every vector temporary is 0.8 GB and the memory manager is struggling.
Cutting the array size to n = 1 million elements (8 MB), then multiplying the runtime by 100 brings the system time down by a large factor, f_inverse0 now takes 16.1 s (12.5+3.6), while f_inverse1 takes 22.3 s (16.2+5.1). This factor of 8 to 10 slower than compiled code is not unreasonable to expect for numpy performance.