How can I impose the total number of digits in pyt

问题:

I would like to keep the total number of digits (before and after the decimal point) of a float constant in python.

For example, if I want to impose a fixed width of 7: 1234.567890123 would become 1234.567 but 12345.678901234 would become 12345.67

Fixing the number of decimals does not work in this case since it depends on how many digits I have before the decimal point. I also tried the [width] option but it impose a minimum width and I need a maximum.

Thanks for your input!

回答1:

Just by using your example,

a = 1234.567890123 
b = 12345.678901234
str(a)[:8] # gives '1234.567'
str(b)[:8] # gives '12345.67'

回答2:

The simplest solution is probably to use the exponential format with one less than the number of digits.

"{0:.6e}".format(1234.457890123) = '1.234568e+03'

I ended up writing this solution that can print floats as well as exponentials but it's probably unnecessarily long for most needs.

 import numpy as np

 def sigprint(number,nsig):
     """
     Returns a string with the given number of significant digits.
     For numbers >= 1e5, and less than 0.001, it does exponential notation
     This is almost what ":.3g".format(x) does, but in the case
     of '{:.3g}'.format(2189), we want 2190 not 2.19e3. Also in the case of
     '{:.3g}'.format(1), we want 1.00, not 1
     """

     if ((abs(number) >= 1e-3) and (abs(number) < 1e5)) or number ==0:
         place = decplace(number) - nsig + 1
         decval = 10**place
         outnum = np.round(np.float(number) / decval) * decval
         ## Need to get the place again in case say 0.97 was rounded up to 1.0
         finalplace = decplace(outnum) - nsig + 1 
         if finalplace >= 0: finalplace=0
         fmt='.'+str(int(abs(finalplace)))+'f'
     else:
         stringnsig = str(int(nsig-1))
         fmt = '.'+stringnsig+'e'
         outnum=number
     wholefmt = "{0:"+fmt+"}"

     return wholefmt.format(outnum)

 def decplace(number):
     """
     Finds the decimal place of the leading digit of a number. For 0, it assumes
     a value of 0 (the one's digit)
     """
     if number == 0:
         place = 0
     else:
         place = np.floor(np.log10(np.abs(number)))
     return place

回答3:

You can set the precision when using decimal

It sounds like you also want to round down, but could choose other rounding options if you like. You create a context that includes the precision, rounding logic, and a few other options. You can apply the context to all future operations with setcontext, a single number using normalize, or with a context manager using localcontext.

import decimal
ctx = decimal.Context(prec=7, rounding=decimal.ROUND_DOWN)
print(decimal.Decimal.from_float(1234.567890123).normalize(ctx))
print(decimal.Decimal.from_float(12345.678901234).normalize(ctx))