我如何计算一个被随机字母包围的特定字符

Question

我正在尝试使用字典来计算：postrophe('(和连字符( - (的标点。我想看看是否可以使用list/dictionary/for loops and boolean表达式来实现此功能。这些标点只有在被任何其他字母包围时才能计算出来！例如。千斤顶箱(即3个连字符(，不应该(1个撇号(。这些字母可以是从A到Z的任何东西。另外，由于这是分配的一部分，因此不能使用模块/库。我没有想法，不知道该怎么办。任何帮助将不胜感激。

这是我尝试的：但是我得到了一个KeyError：0

def countpunc2():
filename = input("Name of file? ")
text = open(filename, "r").read()
text = text.lower()  #make all the words lowercase (for our convenience)
for ch in '!"#$%&()*+./:<=>?@[\]^_`{|}~':
    text = text.replace(ch, ' ')
for ch in '--':
    text = text.replace(ch, ' ')
words = text.split('n')       #splitting the text for words
wordlist = str(words)
count = {}                 #create dictionary; the keys/values are added on
punctuations = ",;'-"
letters = "abcdefghijklmnopqrstuvwxyz"
for i, char in enumerate(wordlist):
    if i < 1:
        continue
    if i > len(wordlist) - 2:
        continue
    if char in punctuations:
        if char not in count:
            count[char] = 0
        if count[i-1] in letters and count[i+1] in letters:
            count[char] += 1
print(count)

更新：我将代码更改为：

def countpunc2():
filename = input("Name of file? ")
text = open(filename, "r").read()
text = text.lower()  #make all the words lowercase (for our convenience)
for ch in '!"#$%&()*+./:<=>?@[\]^_`{|}~':
    text = text.replace(ch, ' ')
for ch in '--':
    text = text.replace(ch, ' ')
words = text.split('n')       #splitting the text for words
wordlist = str(words)
count = {}                 #create dictionary; the keys/values are added on
punctuations = ",;'-"
letters = "abcdefghijklmnopqrstuvwxyz"
for i, char in enumerate(wordlist):
    if i < 1:
        continue
    if i > len(wordlist) - 2:
        continue
    if char in punctuations:
        if char not in count:
            count[char] = 0
        if wordlist[i-1] in letters and wordlist[i+1] in letters:
            count[char] += 1
print(count)

当它给我输出时，这是不正确的。示例文件：https：//www.dropbox.com/s/kqwvudflxnmldqr/sample1.txt?dl = 0预期的结果必须是：{'，'：27，' - '：10，';'：5，"'"：1}

Answer 1

我可能会比这更简单。

#!/usr/bin/env python3
sample = "I'd rather take a day off, it's hard work sitting down and writing a code. It's amazin' how some people find this so easy. Bunch of know-it-alls."
punc = "!"#$%&'()*+,-./:;<=>?@[\]^_`{|}~" 
letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
d = {}
for i, char in enumerate(sample):
    if i < 1:
        continue
    if i > len(sample) - 2:
        continue
    if char in punc:
        if char not in d:
            d[char] = 0
        if sample[i - 1] in letters and sample[i + 1] in letters:
            d[char] += 1
print(d)

输出：

{"'": 3, ',': 0, '.': 0, '-': 2}

dunno在哪里得到"；从。另外，您的逗号旁边有一个空间..所以它不算在这里..如果确实计算为字母变量增加一个空间。

解释发生的事情：

我们使用 enumerate启动dict并在示例文本中读取示例文本，并按字符迭代字符，并使用 CC_2播放索引。如果它离末端太近或开始有资格，我们会跳过。

我在我们使用枚举的i变量之前和之后检查字符。并添加计算是否有资格。

注意：尽管有Shebang，但此代码在Python2

Answer 2

您可以将输入字符串的字符映射到3个类别：字母(a(，标点符号(P(和空格(S(。然后将它们分成三倍(3个字符的序列(。从中隔离A-P-A三元组并计算不同标点字符的数量。

例如：

string="""jack-in-a-box (that is 3 hyphens) and shouldn't (1 apostrophe)."""
categ   = [ "pa"[c.isalpha()] if c != " " else "s" for c in string ]
triples = [ triple for triple   in zip(categ,categ[1:],categ[2:]) ]
pChars  = [ p      for p,triple in zip(s[1:],triples) if triple==("a","p","a") ]
result  = { p:pChars.count(p) for p in set(pChars) }
print(result) # {"'": 1, '-': 3}

如果您不允许使用isAlpha()或zip()，则可以使用in操作员和for循环进行编码。

Answer 3

这是一个以非常拼写的方式进行的示例：

end_cap_characters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
special_characters = [";", ":", "'", "-", ","]
def count_special_characters(in_string):
    result = {}
    for i in range(1, len(in_string) - 1):
        if in_string[i - 1] in end_cap_characters:
            if in_string[i + 1] in end_cap_characters:
                if in_string[i] in special_characters:
                    if in_string[i] not in result:
                        result[in_string[i]] = 1
                    else:
                        result[in_string[i]] +=1
    return result
print(count_special_characters("jack-in-the-box"))
print(count_special_characters("shouldn't"))
print(count_special_characters("jack-in-the-box, shouldn't and a comma that works,is that one"))

输出：

{'-': 3}
{"'": 1}
{'-': 3, "'": 1, ',': 1}

显然可以凝结，但我会把它作为您的练习；(。

update

基于您编辑的问题和已发布的代码，您需要更新以下行：

if count[i-1] in letters and count[i+1] in letters:

to：

if wordlist[i-1] in letters and wordlist[i+1] in letters: