计算的 Numpy 数组问题，int 类型，int32 与 int64

Question

I don't know why there is a discrepancy between the two lines of code below. The Numpy array (ie, "load_values") for some reason is slightly off in the calculation? When I replace the Numpy array index with the actual value that is stored (in this case, 1800), I get the correct answer. Also, I replaced the Numpy array with a regular List and it also got the correct answer. It's only when I use the Numpy array that I am getting the calculation slightly off. Any reason why this would be the case? Is it obvious and I'm just not seeing it?

In [27]: alpha[3] + alpha[2] * 1800 + alpha[1] * (1800 ** 2) + alpha[0] * (1800 ** 3)

Out[27]: 1.2057057142857146

In [28]: alpha[3] + alpha[2] * load_values[2] + alpha[1] * (load_values[2] ** 2) + alpha[0] * (load_values[2] ** 3)

Out [28]: 1.2048772097918872

Edit: Here are the alpha and load_values:

In[54]: alpha

Out[54]: array([ 4.24382716e-13, -1.18055556e-09, -6.69194444e-04,  1.64000000e-03])

In[55]: load_values

Out[55]: array([ 600, 1200, 1800, 2400, 3000])

Answer 1

As mentioned in the comments, the root of the issue is an overflow when doing the operation: load_values[2] ** 3 . The default int-type for numpy appears to be int32 while the standard python int appears to be at least the numpy.int64 equivalent. You need to use int64 to properly compute your equation. This can be seen here:

# load_values as int32 array (numpy default):
print(type(load_values[2]))  # <class 'numpy.int32'>
print(type(1800))            # <class 'int'>

answer_1 = 1800 ** 3
answer_2 = load_values[2] ** 3

print(answer_1)              # 5832000000
print(answer_2)              # 1537032704


# load_values as int64 array:
load_values = np.array(load_values, dtype=np.int64)

print(type(load_values[2]))  # <class 'numpy.int64'>
print(type(1800))            # <class 'int'>

answer_1 = 1800 ** 3
answer_2 = load_values[2] ** 3

print(answer_1)              # 5832000000
print(answer_2)              # 5832000000

I will leave the bit-by-bit analysis up to you.