Python 缩进错误：未缩进与else块中的任何外部缩进级别不匹配

else语句中的缩进已关闭。试试这个：

  ./babynames.py baby2008.html
  File "./babynames.py", line 51
    else:
        ^
IndentationError: unindent does not match any outer indentation level

def extract_名称（文件名）：
"""
给定baby.html的文件名，返回以年字符串开头的列表
后跟名称，按字母顺序排列字符串。
['2006'，'Aaliyah 91'，'Aaron 57'，'Abagail 895'，'…]
"""
#在这里输入您的代码+++
f=打开（文件名为'rU'）
对于f.readlines（）中的行：
match=re.search（r'（\d\d\d\d）（只需在
else:
之前删除一个额外的空格字符，并确保两者（
if
和相应的
else
）具有完全相同的缩进

def extract_names(filename):
  """
  Given a file name for baby.html, returns a list starting with the year string
  followed by the name-rank strings in alphabetical order.
  ['2006', 'Aaliyah 91', Aaron 57', 'Abagail 895', ' ...]
  """
  # +++your code here+++
  f = open(filename,'rU')
  for line in f.readlines():
      match = re.search(r'(\d\d\d\d)(</h2)',line)
      if match:
          year = match.group(1)
          print line
          print year
      else:
          match = re.search(r'<td>(\d+)</td><td)(\w+)</td><td>(\w+)</td)',line)
          if match:
              rank = match.group(1)
              boyn = match.group(2)
              girln = match.group(3)
              print rank, boyn, girln
              print year


  f.close()
  return

看起来您在其他位置前面还有一个空间。它必须与if处于相同的缩进级别。哦，谢谢！我在过去1小时内一直在打破我的头：）
def extract_names(filename):
  """
  Given a file name for baby.html, returns a list starting with the year string
  followed by the name-rank strings in alphabetical order.
  ['2006', 'Aaliyah 91', Aaron 57', 'Abagail 895', ' ...]
  """
  # +++your code here+++
  f = open(filename,'rU')
  for line in f.readlines():
      match = re.search(r'(\d\d\d\d)(</h2)',line)
      if match:
          year = match.group(1)
          print line
          print year
      else:
          match = re.search(r'<td>(\d+)</td><td)(\w+)</td><td>(\w+)</td)',line)
          if match:
              rank = match.group(1)
              boyn = match.group(2)
              girln = match.group(3)
              print rank, boyn, girln
              print year


  f.close()
  return

if match:
   #...
 else:    # This has only ONE EXTRA 'SPACE' at start!
   #...