I've a very basic issue.
import numpy as np
def u(x):
return 1 - x x**2 - x**3 x**4 - x**5 x**6 - x**7 x**8 - x**9 x**10
X = np.array([8, 9])
Y = u(X)
print("u(8) = ", u(X)[0], "or", u(8))
print("u(9) = ", u(X)[1], "or", u(9))
I create an array containing 8 and 9, and then appy the function "u" to this array. But for some reason u(X)[1] != u(9) (even though X[1] == 9):
u(8) = 954437177 or 954437177
u(9) = -1156861335 or 3138105961
Weirdly enough, I don't have this problem for n < 9. What is wrong here? (and with me...)
CodePudding user response:
Numpy does not use Python arbitrary-precision number types, internally, but (usually) machine numbers. In this case, the array X
you created automatically got 32 bit integers as their type. For those, your function overflows. I couldn't easily demonstrate this in Python, but here's the same effect in Julia:
julia> u(Int32(9))
-1156861335
julia> u(Int32(8))
954437177
julia> u(Int64(9))
3138105961
julia> u(Int64(8))
954437177
julia> typemax(Int32)
2147483647
The solution is to tell np.array
what types you want to use explicitely:
In [2]: X = np.array([8, 9], dtype=float)
In [4]: u(X)
Out[4]: array([9.54437177e 08, 3.13810596e 09])
In [5]: X = np.array([8, 9], dtype=np.int64)
In [6]: u(X)
Out[6]: array([ 954437177, 3138105961])
# you can use Python integers, but it's going to lose efficiency!
In [8]: X = np.array([8, 9], dtype=object)
In [9]: u(X)
Out[9]: array([954437177, 3138105961], dtype=object)
# for comparison
In [22]: X = np.array([8, 9], dtype=np.int32)
In [25]: u(X)
Out[25]: array([ 954437177, -1156861335], dtype=int32)