在 python 中四舍五入到有效数字

Question

I want to round like this 42949672 -> 42949700, 2147483647 -> 2147480000, 4252017622 -> 4252020000 etc.

I tried to use following, but only works for the first one. How can I make a more general one? thanks

round(42949672, -2)

Answer 1

I want to round like this 42949672 -> 42949700, 2147483647 -> 2147480000, 4252017622 -> 4252020000 etc.

I tried to use following , but only works for the first one. How can I make a more general one? thanks

round(42949672, -2)

Answer 2

I want to round like this 42949672 -> 42949700, 2147483647 -> 2147480000, 4252017622 -> 4252020000 etc.

I tried to use following , but only works for the first one. How can I make a more general one? thanks

round(42949672, -2)

Answer 3

I want to round like this 42949672 -> 42949700, 2147483647 -> 2147480000, 4252017622 -> 4252020000 etc.

I tried to use following , but only works for the first one. How can I make a more general one? thanks

round(42949672, -2)

Answer 4

I want to round like this 42949672 -> 42949700, 2147483647 -> 2147480000, 4252017622 -> 4252020000 etc.

I tried to use following , but only works for the first one. How can I make a more general one? thanks

round(42949672, -2)

Answer 5

Expanded on @david-joy's function. Indeed, it did not work properly in the else clause, and actually not all cases were covered in the if clause. So here is my function.

def round_to_nsf(number, nsf=2):
    integer_part = math.floor(number)
    if integer_part > 0:
        integer_part_len = len(str(integer_part))
        return round(number, nsf-integer_part_len)
    else:
        st = {'1', '2', '3', '4', '5', '6', '7', '8', '9'}
        index = next((i for i, ch in enumerate(str(number)) if ch in st), None)
        return round(number, index)

Answer 6

These work for me, for ints and floats, both positive and negative. If desired they may be modified to preserve the original int or float type by applying isinstance() to x and then return int() or float().

Note that format beyond around 14 digits can give unexpected results.

# Option 1: use math
import math
def rnd1(x, sigdig=14):
    """Round x to sigdig significant digits"""
    # n = digits to the left of decimal, can be negative
    n = math.floor(math.log10(abs(x))) + 1
    return round(x, sigdig - n)

# Option 2: use format which also rounds
def rnd2(x, sigdig=14):
    """Round x to sigdig significant digits"""
    f = f'{{:.{sigdig - 1}E}}'
    s = f.format(x)
    return float(s)

# Bonus: use format to truncate
def sig(x, sigdig=14):
    """Truncate x to sigdig significant digits"""
    f = f'{{:.{sigdig}E}}'
    s = f.format(x)
    decimal= 1
    minus = 1 if x < 0 else 0
    s = ''.join( (s[:sigdig + minus + decimal], s[s.index('E'):]) )
    return float(s)

# Unexpected format results with large number of significant digits
>>> '{:.20E}'.format(10**-1)
'1.00000000000000005551E-01'

>>> '{:.20E}'.format(10**-10)
'1.00000000000000003643E-10'

>>> '{:.20E}'.format(10**-20)
'9.99999999999999945153E-21'

>>> n = .123456789012345678901234567890
>>> '{:.20E}'.format(n)
'1.23456789012345677370E-01'