如何迭代Python中的defaultdict（列表）？

Question

如何迭代Python中的defaultdict（list）？ 是否有更好的方式在Python中拥有一个列表字典？ 我试过正常iter(dict)，但我得到了错误：

>>> import para 
>>> para.print_doc('./sentseg_en/essentials.txt') 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
    File "para.py", line 31, in print_doc 
    for para in iter(doc): 
TypeError: iteration over non-sequence 

主类：

import para 
para.print_doc('./foo/bar/para-lines.txt') 

的para.pyc：

# -*- coding: utf-8 -*- 
## Modified paragraph into a defaultdict(list) structure 
## Original code from http://code.activestate.com/recipes/66063/ 
from collections import defaultdict 
class Paragraphs: 
    import sys 
    reload(sys) 
    sys.setdefaultencoding('utf-8') 
    # Separator here refers to the paragraph seperator, 
    # the default separator is '\n'. 
    def __init__(self, filename, separator=None): 
     # Set separator if passed into object's parameter, 
     # else set default separator as '\n' 
     if separator is None: 
      def separator(line): return line == '\n' 
     elif not callable(separator): 
      raise TypeError, "separator argument must be callable" 
     self.separator = separator 
     # Reading lines from files into a dictionary of lists 
     self.doc = defaultdict(list) 
     paraIndex = 0 
     with open(filename) as readFile: 
      for line in readFile: 
       if line == separator: 
        paraIndex+=1 
       else: 
        self.doc[paraIndex].append(line) 

# Prints out populated doc from txtfile 
def print_doc(filename): 
    text = Paragraphs(filename) 
    for para in iter(text.doc): 
     for sent in text.doc[para]: 
      print "Para#%d, Sent#%d: %s" % (
       para, text.doc[para].index(sent), sent) 

的如的./foo/bar/para-lines.txt看起来像这样：

This is a start of a paragraph. 
foo barr 
bar foo 
foo foo 
This is the end. 

This is the start of next para. 
foo boo bar bar 
this is the end. 

主类的输出应该是这样的：

Para#1,Sent#1: This is a start of a paragraph. 
Para#1,Sent#2: foo barr 
Para#1,Sent#3: bar foo 
Para#1,Sent#4: foo foo 
Para#1,Sent#5: This is the end. 

Para#2,Sent#1: This is the start of next para. 
Para#2,Sent#2: foo boo bar bar 
Para#2,Sent#3: this is the end. 

Answer 1

你链接到的食谱是相当老。它是在2001年编写的，Python有更多的现代工具，如itertools.groupby（在Python2.4中引入，released in late 2003）。这里是你的代码可能看起来像什么用groupby：

import itertools 
import sys 

with open('para-lines.txt', 'r') as f: 
    paranum = 0 
    for is_separator, paragraph in itertools.groupby(f, lambda line: line == '\n'): 
     if is_separator: 
      # we've reached paragraph separator 
      print 
     else: 
      paranum += 1 
      for n, sentence in enumerate(paragraph, start = 1): 
       sys.stdout.write(
        'Para#{i:d},Sent#{n:d}: {s}'.format(
         i = paranum, n = n, s = sentence)) 

Answer 2

的问题似乎是，你遍历你Paragraphs类，而不是字典。此外，而不是遍历键，然后访问该字典条目，可以考虑使用

for (key, value) in d.items(): 

Answer 3

它的失败，因为你没有在你的段落类中定义__iter__()，然后尝试调用iter(doc)做（其中文档是一个段落实例）。

要迭代一个类必须有__iter__()它返回迭代器。 Docs here。

Answer 4

你有行

for para in iter(doc): 

的问题是，doc是段落的一个实例，而不是一个defaultdict。您在__init__方法中使用的默认字典超出了范围并丢失。所以，你需要做两两件事：

1. 保存在__init__方法，实例变量创建doc（self.doc，例如）。

2. 要么Paragraphs本身可迭代（通过添加__iter__方法），要么允许它访问创建的doc对象。

Answer 5

我想不出为什么你在这里使用字典，更不用说defaultdict了。列表清单会简单得多。

doc = [] 
with open(filename) as readFile: 
    para = [] 
    for line in readFile: 
     if line == separator: 
      doc.append(para) 
      para = [] 
     else: 
      para.append(line) 
    doc.append(para)