如何检查列表中的位置是否为ascii

Question

我一直在寻找一种方法来做到这一点，我只找到了用字符串做的方法我想知道它是否可以用列表来完成。

#----------------------------
#flesch reading score program
#----------------------------

#----getting file to read----
myfile = open('project2test.txt')
txt = myfile.read()
words = txt.split()
txt = list(txt)
txt.append(' ')

#----global variables----
numWords, numberOfWords, numSentences, numSyllables = 0,0,0,0

punctuation = '.?;:!'
extendedPunctuation = '.?;:!,/<>]}\|)(*&^%$#@'
abc = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
syllables = 'aeiouAEIOU'
numbers = '1234567890'

#----counting functions----
def sentenceEnd(i,numSentences):
    if txt[i] in punctuation:
        numSentences += 1
    return numSentences

def wordEnd(i,numWords):
    if txt[i] in abc:
        if txt[i+1] == ' ' or txt[i+1] == '\n':
            numWords += 1
        if txt[i+1] in extendedPunctuation:
            numWords += 1
        if txt[i+1] in numbers:
            numWords += 1
    return numWords

def syllableEnd(i,numSyllables):
    if txt[i] in syllables:
        if txt[i+1] not in syllables:
            numSyllables += 1
    return numSyllables

#----main loop----
for i in range(len(txt)):
    numWords = wordEnd(i, numWords)
    numSentences = sentenceEnd(i,numSentences)
    numSyllables = syllableEnd(i,numSyllables)

for word in words:
    numberOfWords += 1


flesh = 206.835 - 1.015 * (numberOfWords/numSentences) - 84.6 * ((numSyllables)/numberOfWords)
print("The flesch redability of your document is: ", flesh) 
print(numSyllables, ' ', numSentences, ' ' , numWords, ' ', numberOfWords)

现在我必须定义一切。 我低声想知道如果不是坐着puntuation =，abs = ... list [i] .ascii == true

Answer 1

punctuations = [' ',',']
original_string = "Hello,World! How Clean is Python!"

words = []
words.append(original_string)

for x in punctuations:
    temp = []
    for word in words:
        temp += word.split(x)
    words = temp

这应该有用。