Python recipe(10):逐段处理文件

代码何在？


Example Source Code [http://www.cnblogs.com/tomsheep/]

'''
Created on 2010-5-22

@author: lk
'''
class Paragraphs:
def __init__(self, fileobj, seperator = '\n'):
self.seq = fileobj.readlines()
self.line_num = 0
self.para_num = 0

if seperator[-1:]!='\n':seperator += '\n'
self.seperator = seperator

def __getitem__(self, index):
if index != self.para_num:
raise TypeError, 'only sequential access supported'
#get the first line of current paragraph
self.para_num += 1
while 1:
line = self.seq[self.line_num]
self.line_num += 1
if line != self.seperator:
break
result = [line]

#get the rest
while 1:
#            line = self.seq[self.line_num]
#tag1:

try:
line = self.seq[self.line_num]
except IndexError:
break
self.line_num += 1
if line == self.seperator:
break
result.append(line)
return ''.join(result)

if __name__ == '__main__':
text = Paragraphs(open("test.txt"))
for para in text:
print para

以上代码改写自Python Cookbook 4-9

概述：

逐段处理文件。自定义Paragraphs类，实现容器行为函数__getitem__

代码说明：

1.__getitem__函数可以使自定义类型具有容器行为,x[key]访问

2.在编写代码时，一开始在tag1处没有使用try…except，但是奇怪的是代码运行时并没有抛出IndexError,而是少打出一个para，想了一阵，觉得应该是在__getitem__外层捕捉了IndexError，自己尝试了一下，在__getitem__中手动raise IndexError，果然没有抛出（而ValueError就会抛出）可见__getitem__机制调用外层捕获了IndexError