Python - 使用字典和元组查找单词和字母的唯一计数

Question

我目前正在尝试创建一个脚本，该脚本允许我运行文件中包含的文本并计算单词数，不同单词，列出前10个最频繁的单词和计数，并将字符频率从最频繁到最不频繁进行排序。以下是我到目前为止所拥有的：import sysimport osos.getcwd()import stringpath = ""os.chdir(path)#Prompt for user to input filename:fname = input('Enter the filename: ')try:    fhand = open(fname)except IOError:    #Invalid filename error    print('\n')    print("Sorry, file can't be opened! Please check your spelling.")    sys.exit()#Initialize char counts and word counts dictionarycounts = {}worddict = {}#For character and word frequency countfor line in fhand:        #Remove leading spaces        line = line.strip()        #Convert everything in the string to lowercase        line = line.lower()        #Take into account punctuation                line = line.translate(line.maketrans('', '', string.punctuation))        #Take into account white spaces        line = line.translate(line.maketrans('', '', string.whitespace))        #Take into account digits        line = line.translate(line.maketrans('', '', string.digits))        #Splitting line into words        words = line.split(" ")        for word in words:            #Is the word already in the word dictionary?            if word in worddict:                #Increase by 1                worddict[word] += 1            else:                #Add word to dictionary with count of 1 if not there already                worddict[word] = 1        #Character count        for word in line:            #Increase count by 1 if letter            if word in counts:                counts[word] += 1            else:                counts[word] = 1#Initialize dictionarieslst = []countlst = []freqlst = []#Count up the number of lettersfor ltrs, c in counts.items():    lst.append((c,ltrs))    countlst.append(c)#Sum up the counttotalcount = sum(countlst)#Calculate the frequency in each dictionaryfor ec in countlst:    efreq = (ec/totalcount) * 100    freqlst.append(efreq)#Sort lists by count and percentage frequencyfreqlst.sort(reverse=True)lst.sort(reverse=True)

Answer 1

line = line.translate(line.maketrans('', '', string.whitespace))

您正在删除包含此代码的行中的所有空格。删除它，它应该按预期工作。

Answer 2

您的代码会删除空格以按空格拆分 - 这没有意义。由于您希望从给定的文本中提取每个单词，我建议您将所有单词彼此相邻地对齐，并在两者之间使用一个空格 - 这意味着您不仅要删除新行，不必要的空格，特殊/不需要的字符和数字，还要删除控制字符。

这应该可以解决问题：

import sys

import os

os.getcwd()

import string

path = "/your/path"

os.chdir(path)

# Prompt for user to input filename:

fname = input("Enter the filename: ")

try:

    fhand = open(fname)

except IOError:

    # Invalid filename error

    print("\n")

    print("Sorry, file can't be opened! Please check your spelling.")

    sys.exit()

# Initialize char counts and word counts dictionary

counts = {}

worddict = {}

# create one liner with undesired characters removed

text = fhand.read().replace("\n", " ").replace("\r", "")

text = text.lower()

text = text.translate(text.maketrans("", "", string.digits))

text = text.translate(text.maketrans("", "", string.punctuation))

text = " ".join(text.split())

words = text.split(" ")

for word in words:

    # Is the word already in the word dictionary?

    if word in worddict:

        # Increase by 1

        worddict[word] += 1

    else:

        # Add word to dictionary with count of 1 if not there already

        worddict[word] = 1

# Character count

for word in text:

    # Increase count by 1 if letter

    if word in counts:

        counts[word] += 1

    else:

        counts[word] = 1

# Initialize dictionaries

lst = []

countlst = []

freqlst = []

# Count up the number of letters

for ltrs, c in counts.items():

    # skip spaces

    if ltrs == " ":

        continue

    lst.append((c, ltrs))

    countlst.append(c)

# Sum up the count

totalcount = sum(countlst)

# Calculate the frequency in each dictionary

for ec in countlst:

    efreq = (ec / totalcount) * 100

    freqlst.append(efreq)

# Sort lists by count and percentage frequency

freqlst.sort(reverse=True)

lst.sort(reverse=True)

# Print out word counts sorted

for key in sorted(worddict.keys(), key=worddict.get, reverse=True)[:10]:

    print(key, ":", worddict[key])

# Print out all letters and counts:

for ltrs, c, in lst:

    print(c, "-", ltrs, "-", round(ltrs / totalcount * 100, 2), "%")