python 函数在公式中的奇怪行为

Question

我注意到我的函数在放入公式时出现了一些奇怪的行为。这是其中没有函数的原始方程式：

>>> (.39 * 4.0) + (11.8 * 1.5) - 15.59
3.6700000000000017    # the answer im looking for

我用 return 相同数字的函数替换了其中一些数字。

def average_words(astring):
    word_count = len(re.findall(r'\w+', astring))
    period_count = len(re.findall(r'\.+', astring))
    period_count = period_count or 1
    return word_count/period_count

>> average_words(astring)
4.0


def average_syllables(astring):
    words = re.findall(r'\w+', astring)
    vowels_count = []
    for word in words:

        vowels = len(re.findall(r'[aeiou]+', word))
        vowels_count.append(vowels)
    return  sum(vowels_count)/len(vowels_count)

>>>average_syllables(astring)
1.5

但是现在当我用函数替换数字时，它在 return

中给了我一些奇数

>>> (.39 * average_words(astring)) + (11.8 * average_words(astring)) - 15.59
33.17

为什么会这样？

编辑：

完整代码：

import re

def average_words(astring):
    word_count = len(re.findall(r'\w+', astring))
    period_count = len(re.findall(r'\.+', astring))
    period_count = period_count or 1
    return word_count/period_count

def average_syllables(astring):
    words = re.findall(r'\w+', astring)
    vowels_count = []
    for word in words:

        vowels = len(re.findall(r'[aeiou]+', word))
        vowels_count.append(vowels)
    return  sum(vowels_count)/len(vowels_count)

def flesch_kincaid(s):
    """ Takes a string and returns grade level rounded off to two decimal points."""
    return (.39 *  average_words(s)) + (11.8 *  average_syllables(s)) - 15.59


print flesch_kincaid("The turtle is leaving.")

Answer 1

在 REPL 会话中，您正在做 (11.8 * average_words(astring))，而不是 (11.8 * average_syllables(astring))。但它在完整代码中是正确的..

python 函数在公式中的奇怪行为

python functions in a formula odd behavior

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