精度差异：NumPy对象数组与浮点数组

Question

I understand that the precision of a NumPy float array's element is limited by the machine epsilon.

However, I'm struggling to understand why specifying the array's datatype as a Python object, instead of as the default float, results in the array storing the precise value I feed it. Can someone please explain this behaviour?

The code below illustrates the rounding error associated with the float datatype, and the change in precision when using the object datatype.

import numpy as np

np.set_printoptions(precision=64)

MyArray = np.empty(2)
MyArray.fill(0.442)
print(MyArray)

# [ 0.442000000000000003996802888650563545525074005126953125
#   0.442000000000000003996802888650563545525074005126953125]

MyArray_precise = np.empty(2, dtype = object)
MyArray_precise.fill(0.442)
print(MyArray_precise)

# [0.442 0.442]

I'm running a 32-bit Python 2.7.12 installation on 64-bit Windows.

Answer 1

That's just a matter of display formatting you're seeing. You're not actually getting a more precise number either way; it's just that the precision=64 display setting you set doesn't apply to object arrays. It only applies to arrays of floating-point dtype.

If you print more digits of the contents of MyArray_precise :

print(format(MyArray_precise[0], '.64'))
# 0.442000000000000003996802888650563545525074005126953125

You'll see that it's not actually any better than the other array.

Answer 2

I agree that the issue you with floats is a matter of display, not precision.

But there is a different issue with long integers. Python has a long integer type that does not have a numpy dtype.

In [87]: x=12312312312311231231241241242342
In [88]: x
Out[88]: 12312312312311231231241241242342

This is Py3. Py2 shows it as 12312312312311231231241241242342L

In [90]: np.array([x])
Out[90]: array([12312312312311231231241241242342], dtype=object)
In [91]: np.array([x],int)
....
OverflowError: Python int too large to convert to C long