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，它允许我遍历文件中包含的文本，计算单词数、不同的单词数、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

以下是我目前掌握的情况：

import sys
import os
os.getcwd()
import string

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 = {}

#For character and word frequency count
for 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 dictionaries
lst = []
countlst = []
freqlst = []

#Count up the number of letters
for ltrs, c in counts.items():
    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
for key in list(worddict.keys()):
    print(key, ":", worddict[key])

#Print out all letters and counts:
for ltrs, c, in lst:
    print(c, '-', ltrs, '-', round(ltrs/totalcount*100, 2), '%')

当我在类似romeo.txt的东西上运行脚本时：

But soft what light through yonder window breaks
It is the east and Juliet is the sun
Arise fair sun and kill the envious moon
Who is already sick and pale with grief

我得到这个输出：

butsoftwhatlightthroughyonderwindowbreaks : 1
itistheeastandjulietisthesun : 1
arisefairsunandkilltheenviousmoon : 1
whoisalreadysickandpalewithgrief : 1
i - 14 - 10.45 %
t - 12 - 8.96 %
e - 12 - 8.96 %
s - 11 - 8.21 %
a - 11 - 8.21 %
n - 9 - 6.72 %
h - 9 - 6.72 %
o - 8 - 5.97 %
r - 7 - 5.22 %
u - 6 - 4.48 %
l - 6 - 4.48 %
d - 6 - 4.48 %
w - 5 - 3.73 %
k - 3 - 2.24 %
g - 3 - 2.24 %
f - 3 - 2.24 %
y - 2 - 1.49 %
b - 2 - 1.49 %
v - 1 - 0.75 %
p - 1 - 0.75 %
m - 1 - 0.75 %
j - 1 - 0.75 %
c - 1 - 0.75 %

在frequency.txt上运行脚本时：

I am you you you you you I I I I you you you you I am

我得到这个输出：

iamyouyouyouyouyouiiiiyouyouyouyouiam : 1
y - 9 - 24.32 %
u - 9 - 24.32 %
o - 9 - 24.32 %
i - 6 - 16.22 %
m - 2 - 5.41 %
a - 2 - 5.41 %

我可以得到一些指导，告诉我如何将每行上的单词区分开来，并以期望的方式汇总计数吗

Answer 1

您的代码删除空格以便按空格分割–这没有意义。当您想从给定文本中提取每个单词时，我建议您将相邻的所有单词对齐，中间留一个空格–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

这应该可以做到：

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), "%")