python numpy machine epsilon

Question

I am trying to understand what is machine epsilon. According to the Wikipedia, it can be calculated as follows:

def machineEpsilon(func=float):
    machine_epsilon = func(1)
    while func(1)+func(machine_epsilon) != func(1):
        machine_epsilon_last = machine_epsilon
        machine_epsilon = func(machine_epsilon) / func(2)
    return machine_epsilon_last

However, it is suitable only for double precision numbers. I am interested in modifying it to support also single precision numbers. I read that numpy can be used, particularly numpy.float32 class. Can anybody help with modifying the function?

That function is general enough to work with all precisions. Just pass a `numpy.float32` as an argument to the function! — David Zwicker, Oct 02 '13 at 16:08

ali_m · Accepted Answer · 2015-01-07T15:59:42.827

242

An easier way to get the machine epsilon for a given float type is to use np.finfo():

print(np.finfo(float).eps)
# 2.22044604925e-16

print(np.finfo(np.float32).eps)
# 1.19209e-07

edited Jan 07 '15 at 15:59

answered Oct 02 '13 at 16:16

ali_m

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just to be 100% confident, the first one provides python "standard" precision of innate floats while the second one the precision of numpy's floats? – Charlie Parker Nov 01 '17 at 16:47
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note that numpy's standard accuracy is 64 (in a 64 bit computer): `>>> print(np.finfo(np.float).eps) = 2.22044604925e-16` and `>>> print(np.finfo(np.float64).eps) = 2.22044604925e-16` – Charlie Parker Nov 01 '17 at 17:24
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@CharlieParker I could have used `np.float` instead, since it's just an alias of Python's builtin `float`. Python floats are 64-bit (C `double`) on almost all platforms. `float` and `np.float64` therefore usually have equivalent precision, and for most purposes you can use them interchangeably. However they aren't identical - `np.float64` is a numpy-specific type, and an `np.float64` scalar has different methods to a native `float` scalar. As you'd expect, `np.float32` is a 32-bit float. – ali_m Nov 01 '17 at 18:46

score 109 · Answer 2 · edited Jan 06 '15 at 17:08

109

Another easy way to get epsilon is:

In [1]: 7./3 - 4./3 -1
Out[1]: 2.220446049250313e-16

edited Jan 06 '15 at 17:08

ali_m

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answered Aug 06 '14 at 08:24

Ullen

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Yeah, and why does `8./3 - 5./3 - 1` yield `-eps`, and `4./3 - 1./3 - 1` yields zero, and `10./3 - 7./3 - 1` yields zero? – Steve Tjoa Jul 08 '15 at 18:20
22

Ah, the answer is here, Problem 3: http://rstudio-pubs-static.s3.amazonaws.com/13303_daf1916bee714161ac78d3318de808a9.html Basically, if you subtract the binary representation of 4/3 from 7/3, you get the definition of machine epsilon. So I suppose this should hold for any platform. – Steve Tjoa Jul 08 '15 at 18:24
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This is too esoteric of an answer which requires too much knowledge of Python and `numpy` internals when there's an existing `numpy` function to find the epsilon. – Olga Botvinnik Oct 26 '15 at 20:14
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This answer does not require any knowledge of Python or numpy internals. – GuillaumeDufay Oct 12 '17 at 22:16
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That would need to be `abs(7./3 - 4./3 - 1)`, because otherwise one gets `-1.192093e-07` on an [ESP32 microcontroller](https://en.wikipedia.org/wiki/ESP32). – Serge Stroobandt Dec 25 '17 at 13:25
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Indeed, it asserts that the reader is aware about Python running on computers that aren't using underlying base-3 computation. – kokociel Jul 24 '18 at 08:02
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The only reason this is upvoted is because it's cool. But it is a bad answer. I hope no code base has to deal with something looking like this. – Gulzar May 13 '21 at 12:15

score 16 · Answer 3 · answered Oct 02 '13 at 16:12

16

It will already work, as David pointed out!

>>> def machineEpsilon(func=float):
...     machine_epsilon = func(1)
...     while func(1)+func(machine_epsilon) != func(1):
...         machine_epsilon_last = machine_epsilon
...         machine_epsilon = func(machine_epsilon) / func(2)
...     return machine_epsilon_last
... 
>>> machineEpsilon(float)
2.220446049250313e-16
>>> import numpy
>>> machineEpsilon(numpy.float64)
2.2204460492503131e-16
>>> machineEpsilon(numpy.float32)
1.1920929e-07

answered Oct 02 '13 at 16:12

Claudiu

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btw your function will raise `NameError` if condition in `while` will be satisfied on the first check, so it probably makes sense to do `machine_epsilon = machine_epsilon_last = func(1)` in the first statement – Azat Ibrakov Sep 22 '19 at 10:13

python numpy machine epsilon

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