Question

def isinteger(x):
    while x > 0.0:
        x = x - 1.0
    if x < 0.0:
        return 0.0
    elif x == 0:
        return 1.0

d = input(  )
print isinteger(d)

守则有点不言自明。 I m 用它作一片断层口译。我的问题是: 我投入了一部分,如22/7,我获得一份产出。它是否与伊斯兰组织做些什么?

Answer 1

22.0/7<>/code>或22/7.0。这两条均导致<代码>3.146......。

Answer 2

If you re using python 2, inputting "22/7" directly leads to integer division, i.e. it rounds your input down to 3, thus the algorithm returns 1. Try inputting 22.0/7 instead.

而且,如果你重新使用这一算法,你可能希望考虑更快捷的替代办法。可能性:

def isinteger(x):
    return int(x) == x

def isinteger(x):
    return isinstance(x, int)

Answer 3

除此以外,还有另一个问题:由于

In [37]: isinteger(190.000000000000000001)
Out[37]: 1.0

使用海标<代码>int(x)=x建议:

In [39]: x = 190.000000000000000001
In [40]: int(x) == x
Out[40]: True

I realize it doesn t directly address the question, but I hope it will prevent a future one :)

Answer 4

如果你想要检查ger,例如发现大精炼的广场,那么你必须考虑双亲代表不准确(不要求正面数字,而是需要消极数字):

x = -190.00000000001
print int(x) == x

epsilon = 1E-10

def isinteger(n):
    " near enoungh to pass as integer considering round of errors "
    return abs(n-int(n)) < epsilon

print isinteger(x)

据认为,捆绑的Eval,2 过于强大,无法安全。如果你想要输入数字,而不是让用户给出任何公式(而且每例都需要增加......用户投入的处理除外):

number = raw_input( Give number:  )
number = int(number) if all(c.isdigit() for c in number) else float(number)
print number

Answer 5

你们的投入都是分类的,因此可以提供3项投入,从而产生1项产出。

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