問題描述

我試圖找到最大的立方根，它是一個(gè)整數(shù)，小于 12,000.

I am trying to find the largest cube root that is a whole number, that is less than 12,000.

processing = True
n = 12000
while processing:
    n -= 1
    if n ** (1/3) == #checks to see if this has decimals or not

我不知道如何檢查它是否是一個(gè)整數(shù)！我可以將其轉(zhuǎn)換為字符串，然后使用索引來檢查最終值并查看它們是否為零，但這似乎相當(dāng)麻煩.有沒有更簡(jiǎn)單的方法?

I am not sure how to check if it is a whole number or not though! I could convert it to a string then use indexing to check the end values and see whether they are zero or not, that seems rather cumbersome though. Is there a simpler way?

推薦答案

要檢查浮點(diǎn)值是否為整數(shù)，請(qǐng)使用 float.is_integer() 方法:

To check if a float value is a whole number, use the float.is_integer() method:

>>> (1.0).is_integer()
True
>>> (1.555).is_integer()
False

該方法在 Python 2.6 中被添加到 float 類型中.

The method was added to the float type in Python 2.6.

考慮到在 Python 2 中，1/3 是 0(整數(shù)操作數(shù)的地板除法！)，并且浮點(diǎn)運(yùn)算可能不精確(a float 是使用二進(jìn)制分?jǐn)?shù)的近似值，不是精確的實(shí)數(shù)).但是稍微調(diào)整一下你的循環(huán)會(huì)給出:

Take into account that in Python 2, 1/3 is 0 (floor division for integer operands!), and that floating point arithmetic can be imprecise (a float is an approximation using binary fractions, not a precise real number). But adjusting your loop a little this gives:

>>> for n in range(12000, -1, -1):
...     if (n ** (1.0/3)).is_integer():
...         print n
... 
27
8
1
0

這意味著任何超過 3 的立方(包括 10648)由于上述不精確性而被遺漏:

which means that anything over 3 cubed, (including 10648) was missed out due to the aforementioned imprecision:

>>> (4**3) ** (1.0/3)
3.9999999999999996
>>> 10648 ** (1.0/3)
21.999999999999996

您必須檢查數(shù)字 close 來代替整數(shù)，或者不使用 float() 來查找您的數(shù)字.就像向下取整 12000 的立方根:

You'd have to check for numbers close to the whole number instead, or not use float() to find your number. Like rounding down the cube root of 12000:

>>> int(12000 ** (1.0/3))
22
>>> 22 ** 3
10648

如果您使用的是 Python 3.5 或更新版本，則可以使用 math.isclose() 函數(shù) 查看浮點(diǎn)值是否在可配置的邊距內(nèi):

If you are using Python 3.5 or newer, you can use the math.isclose() function to see if a floating point value is within a configurable margin:

>>> from math import isclose
>>> isclose((4**3) ** (1.0/3), 4)
True
>>> isclose(10648 ** (1.0/3), 22)
True

對(duì)于舊版本，該函數(shù)的簡(jiǎn)單實(shí)現(xiàn)(跳過錯(cuò)誤檢查并忽略無窮大和 NaN)為 :

For older versions, the naive implementation of that function (skipping error checking and ignoring infinity and NaN) as mentioned in PEP485:

def isclose(a, b, rel_tol=1e-9, abs_tol=0.0):
    return abs(a - b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)

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