Python floating-point precision format specifier

Question

Let's say I have some 32-bit numbers and some 64-bit numbers:

>>> import numpy as np
>>> w = np.float32(2.4)
>>> x = np.float32(4.555555555555555)
>>> y = np.float64(2.4)
>>> z = np.float64(4.555555555555555)

I can print them out with %f but it has extra, unneeded decimals:

>>> '%f %f %f %f' % (w, x, y, z)
'2.400000 4.555555 2.400000 4.555556'

I can use %g but it seems to have a small default precision:

>>> '%g %g %g %g' % (w, x, y, z)
'2.4 4.55556 2.4 4.55556'

I was thinking I should use something like .7 for 32-bit values and .15 for 64-bit values:

>>> '%.7g %.7g %.15g %.15g' % (w, x, y, z)
'2.4 4.555555 2.4 4.55555555555556'

This seems to work reasonably well, but the precision number is also used up for numbers in front of the decimal place too, e.g. 34567.375768.

In summary, what is the correct way to serialize floating-point values to text such that it preserves appropriate precision for 32-bit and 64-bit values but doesn't use any unnecessary space?

Update:

Examples of what I think the output should be:

number                float32     float64
5                     5           5
0.1                   0.1         0.1
2.4                   2.4         2.4
4.555555555555555     4.5555553   4.5555555555555554
12345678.92345678635  12345679.0  12345678.923456786

What I get with .7/.16. This actually looks okay:

>>> v32 = np.array([5, 0.1, 2.4, 4.555555555555555, 12345678.92345678635], dtype=np.float32)
>>> v64 = np.array([5, 0.1, 2.4, 4.555555555555555, 12345678.92345678635], dtype=np.float64)
>>> ('%.7g ' * len(v32)) % tuple(v32)
'5 0.1 2.4 4.555555 1.234568e+07 '
>>> ('%.16g ' * len(v64)) % tuple(v64)
'5 0.1 2.4 4.555555555555555 12345678.92345679 '

Answer 1

You could try to use the np.finfo function to get the precision corresponding to your float

finfo32 = np.finfo(np.float32)
finfo64 = np.finfo(np.float64)

finfo32.resolution = 1e-6
finfo64.resolution = 1e-15

Now that you know how many decimals you want, say, 6, just use a rstrip("0") to get rid of the unnecessary 0s:

print ("%.6f" % your_float).strip("0")

If you're leaning towards %g, perhaps you may want to use a dynamic format such as:

>>> strf = lambda v: ("%%.%ig" % max(np.ceil(np.log10(v)), 7)) % v
>>> strf(123.456789)
'123.45679'
>>> strf(123456789.12345)
'123456789'

Answer 2

I am not exactly sure what you are trying to accomplish. However, this might help. You wrote

This seems to work reasonably well, but the precision number is also used up for numbers in front of the decimal place too, e.g. 34567.375768.

You can use %f instead of g. From the Python docs:

The precision is a decimal number indicating how many digits should be displayed after the decimal point for a floating point value formatted with 'f' and 'F', or before and after the decimal point for a floating point value formatted with 'g' or 'G'.

>>> z = np.float64(4.555555555555555)
>>> "%.3f" % z
'4.556'
>>> q = np.float64(2131234.555555555555555)
>>> "%.3f" % q
'2131234.556'
>>>