如何在Python中加一个浮点数？

Question

Is there a simple and direct way to add 'one' at float number in Python?

I mean this:

if a == 0.0143:
    a = plus(a)
    assert a == 0.0144

def plus(a):
    sa = str(a)
    index = sa.find('.')
    if index<0:
        return a+1
    else:
        sb = '0'*len(sa)
        sb[index] = '.'
        sb[-1] = 1
        return a+float(sb)

This is not what I want, because it gives me 0.0144000000001.

Answer 1

As you've noticed, not all decimal numbers can be represented exactly as floats:

>>> Decimal(0.1)
Decimal('0.1000000000000000055511151231257827021181583404541015625')
>>> Decimal(0.2)
Decimal('0.200000000000000011102230246251565404236316680908203125')
>>> Decimal(0.3)
Decimal('0.299999999999999988897769753748434595763683319091796875')
>>> Decimal(0.4)
Decimal('0.40000000000000002220446049250313080847263336181640625')
>>> Decimal(0.5)
Decimal('0.5')

Since you're working with the properties of decimal numbers, use the decimal module, which implements them exactly:

from decimal import Decimal

def plus(n):
    return n + Decimal('10') ** n.as_tuple().exponent

And a demo:

>>> n = Decimal('0.1239')
>>> plus(n)
Decimal('0.1240')

You have to represent the number as a string, as representing it as a float will lose precision.

The downside is that using Decimal will make your plus function about 20-30 times slower than if you used floating point operations, but that's the cost of precision.

Answer 2

Blender's answer is definitely a good answer, but if you insist to use floats I believe the simple way to do this is:

1. Find out x for 10 ** x which can multiply your float into an integer.

2. Add one to the enlarged number.

3. Divide your previous multiplier.

So it looks like:

n = 0.125
e = len(str(n)) - 2
temp_n = n * 10 ** e
temp_n += 1
n = temp_n / 10 ** e
print n

EDIT:

In the previous script, things went wrong when the number was very long. Results are truncated by str() and print , so I changed the script a little:

n = 0.1259287345982795
e = len(repr(n)) - 2
temp_n = n * 10 ** e
temp_n += 1
n = temp_n / 10 ** e
print repr(n)

Answer 3

n = n + 1/10**(len(repr(n)) - 2)