Question

我有一个浮点数，小数点后有300个数字。我该如何输入并打印该数字直到小数位？

示例：我有 pi_300 = 3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273

我接受输入，例如-15 我需要3.141592653589793 我接受输入，例如-20 我现在需要3.14159265358979323846 或

another_num = 3345.1234567890123456789012345678901234567890

我接受输入，例如-15 我需要3345.123456789012345 我接受输入，例如-18 我现在需要3345.123456789012345678

我尝试了format方法，但是没有用

Answer 1

浮点数的精度有限。当您执行操作时，会立即丢失大多数小数。看到这里：

>>> pi_300 = 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594
08128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273
>>> print(pi_300)
3.141592653589793

无法从该数字中获取更多数字，因为它们已经丢失。

现在，如果要提高精度，可以使用Decimal：

>>> from decimal import Decimal
>>> pi_300 = Decimal('3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822
3172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914
1273')
>>> print(pi_300)
3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745
0284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273

您现在可以将其舍入为15位小数：

>>> print(round(pi_300, 15))
3.141592653589793

或更多：

>>> print(round(pi_300, 25))
3.1415926535897932384626434

请注意，如果它超出了decimal模块所定义的精度，则位数可能会失败：

>>> print(round(pi_300, 45))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
decimal.InvalidOperation: [<class 'decimal.InvalidOperation'>]

但这也可以配置：

>>> import decimal
>>> decimal.setcontext(decimal.Context(prec=100))
>>> print(round(pi_300, 45))
3.141592653589793238462643383279502884197169399

Answer 2

两次使用格式迷你语言-一次创建格式字符串，一次进行打印：

from decimal import Decimal

pi_300 = Decimal("3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273")

def printWith(num, wholeNum, decimNum):
    """Creates a format string and prints it"""
    # create format string
    f = "{{:{}.{}f}}".format(wholeNum,decimNum+1) # 1 decimal more then needed
    t = f.format(num)[:-1] # remove the last (maybe rounded) digit
    # print outout (or better return it)
    print ( t )


printWith(pi_300,2,15)

输出：

3.141592653589793

Answer 3

格式似乎对我有用。这是我尝试过的。

>>>'{0:.6f}'.format(pi_300)
>>>'3.141593'
>>>'{0:.2f}'.format(pi_300)
>>>'3.14'

根据小括号中的点后的数字更改小数。它的格式语法比%稍新。请注意，格式化它会自动舍入数字。

>>> round(pi_300, 3)
3.142
>>> '{0:.3f}'.format(pi_300)
'3.142'

所以，这里：

>>> def print_rounded(num, to):
...     return '{}'.format(round(num, to))
... 
>>> print_rounded(pi_300, 3)
'3.142'

如何使用数字作为输入并在Python中打印浮点数直到小数点

3 个答案: