Two differents values when evaluating a numpy array

Question

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...)

phipsgabler · Accepted Answer · 2021-11-20 13:00:25Z

2

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)

edited Nov 20, 2021 at 13:00

answered Nov 20, 2021 at 12:52

phipsgabler

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3 Comments

President James K. Polk Over a year ago

You can get OPs result with dtype=np.int32.

phipsgabler Over a year ago

@PresidentJamesK.Polk Yeah, but that's really what they already have. What I wanted to show was u(np.int32(9)) -- which to my surprise doesn't overflow, though.

phipsgabler Over a year ago

type(np.int32(2147483647) + 1) is np.int64... that's good to know.

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