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(4.555         


        

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'