counting punctuation in a given text

问题

I have a text file with a huge text written in forms of paragraphs.
I need to count certain pieces of punctuation: , and ; from that text without using any module, not even regex.
In addition, my program also needs to count ' and -, but only under certain circumstances.
Specifically, it should count ' marks, but only when they appear as apostrophes surrounded by letters, i.e. indicating a contraction such as "shouldn't" or "won't". (Apostrophe is being included as an indication of more informal writing, perhaps direct speech.) and also, it should count - signs, but only when they are surrounded by letters, indicating a compound-word, such as "self-esteem".

Any other punctuation or letters, e.g. digits, should be regarded as white space, so serve to end words.
Note: Some of the texts we will use include double hyphen, i.e. --. This is to be regarded as a space character.

I first created a string and stored some punctuations inside it for example punctuation_string = ";./'-" but it is giving me the total; what I need is count for individual punctuation.
Because of that I have to change certain_cha variable number of times.

with open("/Users/abhishekabhishek/downloads/l.txt") as f:
    text_lis = f.read().split()
punctuation_count = {}
certain_cha = "/"
freq_coun = 0
for word in text_lis:
    for char in word:
       if char in certain_char:
        freq_coun += 1
 punctuation_count[certain_char] = freq_count

I need values to be displayed like this

etc. but what I get is total (71).

回答1:

You will need to create a dictionary where each entry stores the count of each of those punctuation characters.
For commas and semicolons, we can simply do a string search to count the number of occurences in a word. But we'll need to handle ' and - slightly differently.

This should take care of all the cases:

with open("/Users/abhishekabhishek/downloads/l.txt") as f:
    text_words = f.read().split()
punctuation_count = {}
punctuation_count[','] = 0
punctuation_count[';'] = 0
punctuation_count["'"] = 0
punctuation_count['-'] = 0


def search_for_single_quotes(word):
    single_quote = "'"
    search_char_index = word.find(single_quote)
    search_char_count = word.count(single_quote)
    if search_char_index == -1 and search_char_count != 1:
        return
    index_before = search_char_index - 1
    index_after = search_char_index + 1
    # Check if the characters before and after the quote are alphabets,
    # and the alphabet after the quote is the last character of the word.
    # Will detect `won't`, `shouldn't`, but not `ab'cd`, `y'ess`
    if index_before >= 0 and word[index_before].isalpha() and \
            index_after == len(word) - 1 and word[index_after].isalpha():
        punctuation_count[single_quote] += 1


def search_for_hyphens(word):
    hyphen = "-"
    search_char_index = word.find(hyphen)
    if search_char_index == -1:
        return
    index_before = search_char_index - 1
    index_after = search_char_index + 1
    # Check if the character before and after hyphen is an alphabet.
    # You can also change it check for characters as well as numbers
    # depending on your use case.
    if index_before >= 0 and word[index_before].isalpha() and \
            index_after < len(word) and word[index_after].isalpha():
        punctuation_count[hyphen] += 1


for word in text_words:
    for search_char in [',', ';']:
        search_char_count = word.count(search_char)
        punctuation_count[search_char] += search_char_count
    search_for_single_quotes(word)
    search_for_hyphens(word)


print(punctuation_count)

回答2:

following should work:

text = open("/Users/abhishekabhishek/downloads/l.txt").read()

text = text.replace("--", " ")

for symbol in "-'":
    text = text.replace(symbol + " ", "")
    text = text.replace(" " + symbol, "")

for symbol in ".,/'-":
    print (symbol, text.count(symbol))

回答3:

Because you don't want to import anything this will be slow and will take some time, but it should work:

file = open() # enter your file path as parameter
lines = file.readline() # enter the number of lines in your document as parameter
search_chars = [',', ';', "'", '-'] # store the values to be searched
search_values = {',':0, ';':0, "'":0, '-':0} # a dictionary saves the number of occurences
whitespaces = [' ', '--', '1', '2', ...] # you can add to this list whatever you need

for line in lines:
    for search in search_chars:
        if search in line and (search in search_chars):
            chars = line.split()
            for ch_index in chars:
                if chars [ch_index] == ',':
                    search_values [','] += 1
                elif chars [ch_index] == ';':
                    search_values [';'] += 1
                elif chars[ch_index] == "'" and not(chars[ch_index-1] in whitespaces) and not(chars[ch_index+1] in whitespaces):
                    search_values ["'"] += 1
                elif chars[ch_index] == "-" and not(chars[ch_index-1] in whitespaces) and not(chars[ch_index+1] in whitespaces):
                    search_values ["-"] += 1

for key in range(search_values.keys()):
    print(str(key) + ': ' + search_values[key])

This is obviously not optimal and it is better to use regex here, but it should work.

Feel free to ask if any questions should arise.

来源：https://stackoverflow.com/questions/55865048/counting-punctuation-in-a-given-text

标签

python

string

dictionary

substring

counting