截至目前，我有一个函数可以替换 countChars 函数，

def countWords(lines):
  wordDict = {}
  for line in lines:
    wordList = lines.split()
    for word in wordList:
      if word in wordDict: wordDict[word] += 1
      else: wordDict[word] = 1
  return wordDict

但是当我运行这个程序时，它吐出这个令人厌恶的东西(这只是一个例子，大约有两页单词旁边有一个巨大的数字)

before 1478
battle-field 1478
as 1478
any 1478
altogether 1478
all 1478
ago 1478
advanced. 1478
add 1478
above 1478

虽然这显然意味着代码足够运行，但我没有从中得到我想要的东西。 需要打印每个单词在文件中出现了多少次(gb.txt，也就是葛底斯堡地址) 显然，文件中的每个单词不在那里恰好 1478 次..

我是编程新手，所以我有点难过..

from __future__ import division

inputFileName = 'gb.txt'

def readfile(fname):
  f = open(fname, 'r')
  s = f.read()
  f.close()
 return s.lower()

def countChars(t):
  charDict = {}
  for char in t:
    if char in charDict: charDict[char] += 1
    else: charDict[char] = 1
  return charDict

def findMostCommon(charDict):
  mostFreq = ''
  mostFreqCount = 0
  for k in charDict:
    if charDict[k] > mostFreqCount:
      mostFreqCount = charDict[k]
      mostFreq = k
  return mostFreq

def printCounts(charDict):
  for k in charDict:
    #First, handle some chars that don't show up very well when they print
    if k == '\n': print '\\n', charDict[k]  #newline
    elif k == ' ': print 'space', charDict[k]
    elif k == '\t': print '\\t', charDict[k] #tab
    else: print k, charDict[k]  #Normal character - print it with its count

def printAlphabetically(charDict):
  keyList = charDict.keys()
  keyList.sort()
  for k in keyList:
    #First, handle some chars that don't show up very well when they print
    if k == '\n': print '\\n', charDict[k]  #newline
    elif k == ' ': print 'space', charDict[k]
    elif k == '\t': print '\\t', charDict[k] #tab
    else: print k, charDict[k]  #Normal character - print it with its count

def printByFreq(charDict):
  aList = []
  for k in charDict:
    aList.append([charDict[k], k])
  aList.sort()     #Sort into ascending order
  aList.reverse()  #Put in descending order
  for item in aList:
    #First, handle some chars that don't show up very well when they print
    if item[1] == '\n': print '\\n', item[0]  #newline
    elif item[1] == ' ': print 'space', item[0]
    elif item[1] == '\t': print '\\t', item[0] #tab
    else: print item[1], item[0]  #Normal character - print it with its count

def main():
  text = readfile(inputFileName)
  charCounts = countChars(text)
  mostCommon = findMostCommon(charCounts)
  #print mostCommon + ':', charCounts[mostCommon]
  #printCounts(charCounts)
  #printAlphabetically(charCounts)
  printByFreq(charCounts)

main()

最佳答案

如果你需要计算一段文章中的单词数量，那么最好使用正则表达式。

让我们从一个简单的例子开始:

import re

my_string = "Wow! Is this true? Really!?!? This is crazy!"

words = re.findall(r'\w+', my_string) #This finds words in the document

结果:

>>> words
['Wow', 'Is', 'this', 'true', 'Really', 'This', 'is', 'crazy']

请注意，"is"和"is"是两个不同的词。我的猜测是您希望对它们进行相同的计数，因此我们可以将所有单词都大写，然后对它们进行计数。

from collections import Counter

cap_words = [word.upper() for word in words] #capitalizes all the words

word_counts = Counter(cap_words) #counts the number each time a word appears

结果:

>>> word_counts
Counter({'THIS': 2, 'IS': 2, 'CRAZY': 1, 'WOW': 1, 'TRUE': 1, 'REALLY': 1})

你到这里还好吗？

现在我们需要做的和上面读取文件时做的完全一样。

import re
from collections import Counter

with open('your_file.txt') as f:
    passage = f.read()

words = re.findall(r'\w+', passage)

cap_words = [word.upper() for word in words]

word_counts = Counter(cap_words)

关于在 .txt 文件中找到最常见单词的 Python 程序，必须打印单词及其计数，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10390989/