Python 如何在不使用split函数的情况下从字符串中提取单词？

如何从字符串中提取单词，用标点符号、空格、数字等分隔这些单词…而不使用任何

拆分

、

替换

，或类似

re

的库。我仍在学习python，本书建议在不使用列表和字符串方法的情况下找到解决方案

Example Input : The@Tt11end
Example Output: ["The", "Tt", "end"]

这是我迄今为止的尝试：

def extract_words(sentence):

    words_list = []
    separator = [",",".",";","'","?","/","<",">","@","!","#","$","%","^","&","*","(",")","-","_","1","2","3","4","5","6","7","8","9"]
    counter= 0
    for i in range(len(sentence)):
        i=counter
        while(is_letter(sentence[i])):
            words+= sentence[i]
            i = i+1
            counter=counter+1
        words_list.append(words)
        words=""
    return words_list

编辑：这是我的

is_letter（）

方法：

def is_letter(char):
    return ("A" <= char and char <= "Z") or \
    ("a" <= char and char <= "z")

def是字母（char）：
return（“A”最好是在那里使用正则表达式，但如果您想要一些异国情调……这里是：

str = "The@Tt11end444sooqa"
delims = [0] + [i + 1 for i, s in enumerate(str) if not s.isalpha()] + [len(str) + 1]
parts = [str[delims[i]: delims[i + 1] - 1] for i in range(len(delims) - 1) if delims[i + 1] - delims[i] != 1]


扩展版可更好地了解正在发生的事情：

str = "The@Tt11end444sooqa"

# delims will contain indexes of all non-alphabetic characters
delims = [0]  # adding 0 index as first delimiter (start of string)
for i, s in enumerate(str):  # iterating through "str"
    if not s.isalpha():  # if character is non-alphabetic store it's index
        delims.append(i + 1)  # we add 1 to not include delimiter into final string
delims += [len(str) + 1]  # adding end of string index to not miss last part

# parts will contain parts of original string stored in "str"
parts = []
for i in range(len(delims) - 1):  #iterating over "delims" using indexes
    # do not include part if delimiters goes next one to another
    if delims[i + 1] - delims[i] != 1:
        substr = str[delims[i]: delims[i + 1] - 1]  # copy substring between delimiters
        parts.append(substr)

最好是在那里使用正则表达式，但如果您想要一些异国情调…这里是：

str = "The@Tt11end444sooqa"
delims = [0] + [i + 1 for i, s in enumerate(str) if not s.isalpha()] + [len(str) + 1]
parts = [str[delims[i]: delims[i + 1] - 1] for i in range(len(delims) - 1) if delims[i + 1] - delims[i] != 1]


扩展版可更好地了解正在发生的事情：

str = "The@Tt11end444sooqa"

# delims will contain indexes of all non-alphabetic characters
delims = [0]  # adding 0 index as first delimiter (start of string)
for i, s in enumerate(str):  # iterating through "str"
    if not s.isalpha():  # if character is non-alphabetic store it's index
        delims.append(i + 1)  # we add 1 to not include delimiter into final string
delims += [len(str) + 1]  # adding end of string index to not miss last part

# parts will contain parts of original string stored in "str"
parts = []
for i in range(len(delims) - 1):  #iterating over "delims" using indexes
    # do not include part if delimiters goes next one to another
    if delims[i + 1] - delims[i] != 1:
        substr = str[delims[i]: delims[i + 1] - 1]  # copy substring between delimiters
        parts.append(substr)

此代码用于：

def extract_words(sentence):
    sentence = list(sentence)
    words_list = []
    separator = [",",".",";","'","?","/","<",">","@","!","#","$","%","^","&","*","(",")","-","_","1","2","3","4","5","6","7","8","9"]
    bufferS = []
    for i in range(len(sentence)):
        if sentence[i] not in separator:
          bufferS.append(sentence[i])

        else:
          words_list.append(''.join(bufferS))
          bufferS = []
    words_list.append(''.join(bufferS))
    words_list = [x for x in words_list if x != '']     
    return words_list

它回来了

['aaaaaaa', 'bbbbbbb', 'ccccc', 'dddd']

没有使用库。
此代码执行以下操作：

def extract_words(sentence):
    sentence = list(sentence)
    words_list = []
    separator = [",",".",";","'","?","/","<",">","@","!","#","$","%","^","&","*","(",")","-","_","1","2","3","4","5","6","7","8","9"]
    bufferS = []
    for i in range(len(sentence)):
        if sentence[i] not in separator:
          bufferS.append(sentence[i])

        else:
          words_list.append(''.join(bufferS))
          bufferS = []
    words_list.append(''.join(bufferS))
    words_list = [x for x in words_list if x != '']     
    return words_list

它回来了

['aaaaaaa', 'bbbbbbb', 'ccccc', 'dddd']

没有使用库。
您的问题是每次都将
i
设置为
计数器
，并且它不会递增超过第一个非字母

它将每次递增，直到范围（len（句子））完成，但for It的每个循环将重置回原来的is_字母故障，在这种情况下，
i=3

例如

现在变量
i
将等于4，但是变量
计数器仍然等于3，因为它在while（is_字母）块内没有递增。在这方面更合适的用法是if/else，如下所示：

def extract_words(sentence):
    words_list = []
    words = ""
    for i in range(len(sentence)):
        if is_letter(sentence[i]):
            words += sentence[i]
        else:
            if words != "":
                words_list.append(words)
                words = ""
    if words != "":
        words_list.append(words)
    return words_list


def is_letter(char):
    return ("A" <= char and char <= "Z") or \
    ("a" <= char and char <= "z")

if __name__ == '__main__':
    print(extract_words("The@Tt11end"))

在此设置中，循环将仅使用i作为递增变量，因为它已经是for循环，并且在for上下文之外更改i值可能会导致问题，如您所见

下一次，每当字符串的字符是字母时，它都会被添加到word变量中。然后，如果下一个增量是符号，它会将该单词附加到列表中，并忽略符号/数字

最后，如果两个或多个符号相邻（这导致您得到一个空字符串列表
'
），它将检查单词是否已经包含任何字符，如果没有，它将继续下一个字符。
您的问题是每次都将
i
设置为
计数器，并且它不会递增超过第一个非字母

它将每次递增，直到范围（len（句子））完成，但for It的每个循环将重置回原来的is_字母故障，在这种情况下，
i=3

例如

现在变量
i
将等于4，但是变量
计数器仍然等于3，因为它在while（is_字母）块内没有递增。在这方面更合适的用法是if/else，如下所示：

def extract_words(sentence):
    words_list = []
    words = ""
    for i in range(len(sentence)):
        if is_letter(sentence[i]):
            words += sentence[i]
        else:
            if words != "":
                words_list.append(words)
                words = ""
    if words != "":
        words_list.append(words)
    return words_list


def is_letter(char):
    return ("A" <= char and char <= "Z") or \
    ("a" <= char and char <= "z")

if __name__ == '__main__':
    print(extract_words("The@Tt11end"))

在此设置中，循环将仅使用i作为递增变量，因为它已经是for循环，并且在for上下文之外更改i值可能会导致问题，如您所见

下一次，每当字符串的字符是字母时，它都会被添加到word变量中。然后，如果下一个增量是符号，它会将该单词附加到列表中，并忽略符号/数字

最后，如果两个或多个符号相邻（这导致您得到一个空字符串列表
'
），它将检查单词是否已经包含任何字符，如果不包含，它将继续到下一个字符。
只需对当前代码进行最小的更改，您就可以一次迭代字符串一个字符，并利用您已有的分隔符列表作为O（1）的集合查找时间。这将使您不必担心递增多个计数器变量：

def extract_words(sentence):
  separator_set = set([",",".",";","'","?","/","<",">","@","!","#","$","%","^","&","*","(",")","-","_","1","2","3","4","5","6","7","8","9"])

  words_list = []
  word = []
  for c in sentence:
    if c not in separator_set:
      word.append(c)
    else:
      if len(word) > 0:
        words_list.append(''.join(word))
        word = []

  if len(word) > 0:
    words_list.append(''.join(word))

  return words_list

def is_letter(char):
  return ("A" <= char and char <= "Z") or ("a" <= char and char <= "z")

def main():
  print(extract_words("The@Tt11end"))

if __name__ == '__main__':
  main()

通过对当前代码进行最小的更改，您可以一次迭代字符串一个字符，并利用您已有的分隔符列表作为O（1）查找时间的集合。这将使您不必担心递增多个计数器变量：

def extract_words(sentence):
  separator_set = set([",",".",";","'","?","/","<",">","@","!","#","$","%","^","&","*","(",")","-","_","1","2","3","4","5","6","7","8","9"])

  words_list = []
  word = []
  for c in sentence:
    if c not in separator_set:
      word.append(c)
    else:
      if len(word) > 0:
        words_list.append(''.join(word))
        word = []

  if len(word) > 0:
    words_list.append(''.join(word))

  return words_list

def is_letter(char):
  return ("A" <= char and char <= "Z") or ("a" <= char and char <= "z")

def main():
  print(extract_words("The@Tt11end"))

if __name__ == '__main__':
  main()

您的代码陷入了混乱，没有索引到给定的句子中

你只需要反复阅读句子中的字符

def is_letter(char):
    return ("A" <= char <= "Z") or ("a" <= char <= "z")

def extract_words(sentence):
    word = ""
    words_list = []
    for ch in sentence:
        if is_letter(ch):
            word += ch
        else:
            if word:
                words_list.append(word)
                word = ""
    if word:
        words_list.append(word)
    return words_list


print(extract_words('The@,Tt11end'))

代码会遍历
语句中的每个字符。如果是字母，则会将其添加到当前单词中。如果不是，则会将当前单词（如果有）添加到输出列表中。最后，如果最后一个字符是字母，则会剩下一个单词，该单词也会添加到输出中。
您的代码陷入了一个混乱，而不是对给定句子进行索引

你只需要反复阅读句子中的字符

def is_letter(char):
    return ("A" <= char <= "Z") or ("a" <= char <= "z")

def extract_words(sentence):
    word = ""
    words_list = []
    for ch in sentence:
        if is_letter(ch):
            word += ch
        else:
            if word:
                words_list.append(word)
                word = ""
    if word:
        words_list.append(word)
    return words_list


print(extract_words('The@,Tt11end'))

代码在
语句中的每个字符中进行迭代。如果是字母，则将其添加到当前单词中。如果不是，则将当前单词（如果有）添加到输出列表中。最后，如果最后一个字符是字母，则剩余的单词也将添加到输出中。
您发布的代码不会运行。您的代码作为posted不运行。好的，这更有用，但它仍然没有解释OP代码的错误。此代码与OP中的测试数据一起工作吗？@quamrana它现在工作了好的，这更有用，但它仍然没有解释OP代码的错误。此代码与OP中的测试数据一起工作吗什么？@quamrana是的now@ggorlen“现在好点了吗？”@ggorlen，现在好点了吗？