如何四舍五入浮点数到一定的小数位？

Use the decimal module: http://docs.python.org/library/decimal.html

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This is normal (and has nothing to do with Python) because 8.83 cannot be represented exactly as a binary float, just as 1/3 cannot be represented exactly in decimal (0.333333... ad infinitum).

If you want to ensure absolute precision, you need the decimal module:

>>> import decimal
>>> a = decimal.Decimal("8.833333333339")
>>> print(round(a,2))
8.83

If you round 8.8333333333339 to 2 decimals, the correct answer is 8.83, not 8.84. The reason you got 8.83000000001 is because 8.83 is a number that cannot be correctly reprecented in binary, and it gives you the closest one. If you want to print it without all the zeros, do as VGE says:

print "%.2f" % 8.833333333339   #(Replace number with the variable?)

If you want to round, 8.84 is the incorrect answer. 8.833333333333 rounded is 8.83 not 8.84. If you want to always round up, then you can use math.ceil. Do both in a combination with string formatting, because rounding a float number itself doesn't make sense.

"%.2f" % (math.ceil(x * 100) / 100)

8.833333333339 (or 8.833333333333334, the result of 106.00/12) properly rounded to two decimal places is 8.83. Mathematically it sounds like what you want is a ceiling function. The one in Python's math module is named ceil:

import math


v = 8.8333333333333339
print(math.ceil(v*100)/100)  # -> 8.84

Respectively, the floor and ceiling functions generally map a real number to the largest previous or smallest following integer which has zero decimal places — so to use them for 2 decimal places the number is first multiplied by 10² (or 100) to shift the decimal point and is then divided by it afterwards to compensate.

If you don't want to use the math module for some reason, you can use this (minimally tested) implementation I just wrote:

def ceiling(x):
n = int(x)
return n if n-1 < x <= n else n+1

How all this relates to the linked Loan and payment calculator problem:

From the sample output it appears that they rounded up the monthly payment, which is what many call the effect of the ceiling function. This means that each month a little more than ¹⁄₁₂ of the total amount is being paid. That made the final payment a little smaller than usual — leaving a remaining unpaid balance of only 8.76.

It would have been equally valid to use normal rounding producing a monthly payment of 8.83 and a slightly higher final payment of 8.87. However, in the real world people generally don't like to have their payments go up, so rounding up each payment is the common practice — it also returns the money to the lender more quickly.

You want to use the decimal module but you also need to specify the rounding mode. Here's an example:

>>> import decimal
>>> decimal.Decimal('8.333333').quantize(decimal.Decimal('.01'), rounding=decimal.ROUND_UP)
Decimal('8.34')
>>> decimal.Decimal('8.333333').quantize(decimal.Decimal('.01'), rounding=decimal.ROUND_DOWN)
Decimal('8.33')
>>>

Just for the record. You could do it this way:

def roundno(no):
return int(no//1 + ((no%1)/0.5)//1)

There, no need for includes/imports

A much simpler way is to simply use the round() function. Here is an example.

total_price = float()
price_1 = 2.99
price_2 = 0.99
total_price = price_1 + price_2

If you were to print out total_price right now you would get

3.9800000000000004

But if you enclose it in a round() function like so

print(round(total_price,2))

The output equals

3.98

The round() function works by accepting two parameters. The first is the number you want to round. The second is the number of decimal places to round to.

Here is my solution for the round up/down problem

< .5  round down


> = .5  round up

import math


def _should_round_down(val: float):
if val < 0:
return ((val * -1) % 1) < 0.5
return (val % 1) < 0.5


def _round(val: float, ndigits=0):
if ndigits > 0:
val *= 10 ** (ndigits - 1)
is_positive = val > 0
tmp_val = val
if not is_positive:
tmp_val *= -1
rounded_value = math.floor(tmp_val) if _should_round_down(val) else math.ceil(tmp_val)
if not is_positive:
rounded_value *= -1
if ndigits > 0:
rounded_value /= 10 ** (ndigits - 1)
return rounded_value


# test
# nr = 12.2548
# for digit in range(0, 4):
#     print("{} decimals : {} -> {}".format(digit, nr, _round(nr, digit)))


# output
# 0 decimals : 12.2548 -> 12
# 1 decimals : 12.2548 -> 12.0
# 2 decimals : 12.2548 -> 12.3
# 3 decimals : 12.2548 -> 12.25

The easiest way to do this is by using the below function, which is built in:

format()

For example:

format(1.242563,".2f")

The output would be:

1.24

Similarly:

format(9.165654,".1f")

would give:

9.2

Here is a simple function to do this for you:

def precision(num,x):
return "{0:.xf}".format(round(num))

Here, num is the decimal number. x is the decimal up to where you want to round a floating number.

The advantage over other implementation is that it can fill zeros at the right end of the decimal to make a deciaml number up to x decimal places.

Example 1:

precision(10.2, 9)

will return

10.200000000 (up to 9 decimal points)

Example 2:

precision(10.2231, 2)

will return

10.22 (up to two decimal points)

I have this code:

tax = (tax / 100) * price

and then this code:

tax = round((tax / 100) * price, 2)

round worked for me