# Numba njit compiler causes computes different numbers compared to plain Python code?

Hello everyone,

I faced an issue with Numba’s njit tool in Python. I noticed that the function gives different results when run with @numba.njit and when run as plain Python code. Particularly, after debugging, I noticed that the discrepancy in calculation occurs when performing matrix inversion using numpy. Please see my test code below. The value for matrix A and vector b are in the following csv files that can be accessed via the following links: A.csv and b.csv [links removed by moderator. please use gist.github.com so users can more easily inspect the content before downloading]

The results from the plain Python function is the correct one. Please help me solve this issue! Do I need use a Numba wrapper function around the numpy matrix inversion function to resolve what seems to be a numerical issue?

Kind regards, and I look forward to hearing from you guys soon

``````@numba.njit
def cal_Test_jit(A,b):
c = np.linalg.inv(A)@b
return c, np.linalg.inv(A)

def cal_Test(A,b):
c = np.linalg.inv(A)@b
return c, np.linalg.inv(A)

c_jit, Ai_jit = cal_Test_jit(A,b)
c, Ai = cal_Test(A,b)
err_c = abs(c-c_jit)
err_A = abs(Ai_jit-Ai)

# ploting the error in the parameters
plt.figure()
plt.plot(err_c)

# only ploting the error in first three columns of
fig, ax = plt.subplots(1,3)
ax[0].plot(err_A[:,0])
ax[1].plot(err_A[:,1])
ax[2].plot(err_A[:,2])
``````

Hi aabuaish,

I didn’t download/open you csv files (security risk), but I ran the following matrix inversion test and the jitted version returns the same values (within 16 decimal places) as the non-jitted version.

``````>>> import numpy as np
>>> from numba import njit
>>> a = np.random.rand(5, 5)
>>> a
array([[0.20280058, 0.50347831, 0.45402395, 0.37740858, 0.7230904 ],
[0.16618721, 0.71158419, 0.10857761, 0.46165089, 0.25024504],
[0.2720345 , 0.0191336 , 0.96012383, 0.23156634, 0.37785698],
[0.86418734, 0.70872044, 0.3755512 , 0.95231224, 0.58039226],
[0.40051125, 0.66755511, 0.82855679, 0.99395518, 0.67244711]])
>>> @njit
... def jit_inv(x):
...     return np.linalg.inv(x)
...
>>> jit_inv
CPUDispatcher(<function jit_inv at 0xffff91cae310>)
>>> b = jit_inv(a)
>>> c = np.linalg.inv(a)
>>> b
array([[-0.23710366,  0.21047698,  1.0576892 ,  1.72162595, -1.90363977],
[-0.09010521,  2.90454158,  0.94189871, -0.22096339, -1.32255824],
[-0.65481468,  1.00827047,  1.53481102, -0.46018126, -0.13633473],
[-0.92644321, -1.60702481, -1.61053186, -0.20810057,  2.67884687],
[ 2.40689122, -1.87574404, -1.07557502,  0.06856096,  0.14219256]])
>>> c
array([[-0.23710366,  0.21047698,  1.0576892 ,  1.72162595, -1.90363977],
[-0.09010521,  2.90454158,  0.94189871, -0.22096339, -1.32255824],
[-0.65481468,  1.00827047,  1.53481102, -0.46018126, -0.13633473],
[-0.92644321, -1.60702481, -1.61053186, -0.20810057,  2.67884687],
[ 2.40689122, -1.87574404, -1.07557502,  0.06856096,  0.14219256]])

>>> np.isclose(b, c, atol=1e-16)
array([[ True,  True,  True,  True,  True],
[ True,  True,  True,  True,  True],
[ True,  True,  True,  True,  True],
[ True,  True,  True,  True,  True],
[ True,  True,  True,  True,  True]])

``````

Running numba 0.53.0.rc1 on Python 3.8.3 on ARM.

BR,
Ryan

It’s not uncommon for us see differences way out in the digits

Many thanks for the feedback, @ryanchien and @nelson2005.

I took @nelson2005 comment and I started testing. I rounded up the numbers in matrix A to four digits after the decimal, and the error decreased significantly. If I round up to a lower number of digits, the matrix A becomes singular. I am just curious, why does Numba have this numerical issue? This issue will make me think twice when developing and testing code and make things harder to spot errors. Also, do you guys know of any other nuances that you have faced when using Numba that users should be cautious of?

Can you please provide a reproducible example that does not require me to download csv files? I’d like to try it out. BR.

I suspect this all comes down to NumPy and Numba do slightly different things for `np.linalg.inv`.

1. Numba uses LU factorization to compute the inverse, whereas NumPy uses the Singular Value Decomposition (SVD) via solving `Ax=B`, where `A` is the system to invert and `B` is the identity matrix. If the system to invert is well conditioned then there’s probably little to no difference in the result, but LU factorization will run quicker than SVD. If on the other hand the system is poorly conditioned but not quite singular to machine precision, then I’d expect there to be a difference between the Numba and the NumPy result. To compute the condition number of your input system you can use `np.linalg.cond`. Not that it matters in this case, but SciPy’s `scipy.linalg.inv` also uses a LU factorization (this can be seen in the results of the script below). I suppose technically Numba should use the SVD, but there’s also the argument that in the most common case there’s perhaps less value in inverting a near singular system and performance is more desirable.
2. NumPy uses uses double precision routines even in the case of single precision input:
numpy/linalg.py at maintenance/1.21.x · numpy/numpy · GitHub
3. Numba uses the LAPACK routines (the library that does the actual linear algebra work) from SciPy. Differences in the way NumPy and SciPy are linked to their required numerical libraries could make a difference (e.g. potentially different LAPACKs!).

Demo of 1.:

``````from numba import njit
import numpy as np
from scipy.linalg import inv as scipy_inv

np.random.seed(0) # ensure reproducible

@njit
def inverse(A):
return np.linalg.inv(A)

@njit
def inverse_via_svd(A):
b = np.diag(np.ones(len(A)))
return np.linalg.solve(A, b)

# Case 1: Poorly conditioned system, 64bit precision
print("Case 1:")
n = 4
# A = U * S * V**T
U = np.random.random((n, n))
S = np.diag([1,] * (n - 2) + [0,] * 2)
A = (U @ S @ U.T)
print("Matrix condition:", np.linalg.cond(A))
print(A)

# Numba uses the LAPACK bindings from SciPy
numba_inv = inverse(A)
numpy_inv = inverse.py_func(A)
sp_inv = scipy_inv(A)
numba_solve = inverse_via_svd(A)
print("abs diff against NumPy:\n", np.abs(numba_inv - numpy_inv))
print("abs diff against SciPy:\n", np.abs(numba_inv - sp_inv))
print("abs diff solve against NumPy:\n", np.abs(numba_solve - numpy_inv))
``````

gives me:

``````Case 1:
Matrix condition: 3.5983733242678336e+16
[[0.81269209 0.69444408 0.80310443 0.9737274 ]
[0.69444408 0.59666259 0.65592297 0.83849222]
[0.80310443 0.65592297 1.07567331 0.90231557]
[0.9737274  0.83849222 0.90231557 1.17940376]]
abs diff against NumPy:
[[12.        0.25      1.5       6.      ]
[ 4.        0.        1.        4.      ]
[ 0.5       0.078125  0.125     0.5     ]
[ 0.        0.5       0.        4.      ]]
abs diff against SciPy:
[[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]]
abs diff solve against NumPy:
[[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]]
``````

Thanks for the follow up. Below is the 76 by 76 matrix. Inverting it will reproduce the issue. I posted it earlier but it was removed by the auto-system.

``````A=np.array([[1.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,1.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,2.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,6.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[1.0,0.100000,0.010000,0.001000,0.000100,0.000010,0.000001,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.0,0.100000,0.010000,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,1.0,0.200000,0.030000,0.004000,0.000500,0.000060,0.000007,0.0,-1.0,-0.200000,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,2.0,0.600000,0.120000,0.020000,0.003000,0.000420,0.0,0.0,-2.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,6.0,2.400000,0.600000,0.120000,0.021000,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,0.0,24.0,12.0,3.600000,0.840000,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.0,0.900000,0.810000,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.0,0.900000,0.810000,0.729000,0.656100,0.590490,0.531441,0.478297,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,1.0,1.800000,0.0,-1.0,-1.800000,-2.430000,-2.916000,-3.280500,-3.542940,-3.720087,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,2.0,0.0,0.0,-2.0,-5.400000,-9.720000,-14.580000,-19.683000,-24.800580,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,-6.0,-21.600000,-48.600000,-87.480000,-137.781000,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0],
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``````

Thanks @stuartarchibald for the follow up. I will take a closer look at your post to understand the mathematical techniques used by each library. I also got an answer on StackOverflow that recommended that I would use np.linalg.solve(A,b) if my goal is to solve a system of linear equations, which was my case.

Again thank you so much for your detailed feedback. I will keep your comments in mind if my code involves inverting matrices.