生成函数

pre $ def group_by_heading（some_source）：
buffer = []
for some_source：
if line.startswith（Heading）：
if buffer：yield buffer
buffer = [line]
else：
buffer.append（line）
产生缓冲区

以open（some_file，r）作为源：
在group_by_heading（source）中为heading_and_lines：
heading = heading_and_lines [0]
lines = heading_and_lines [1：]
＃处理完毕。


I am an amateur using Python on and off for some time now. Sorry if this is a silly question, but I was wondering if anyone knew an easy way to grab a bunch of lines if the format in the input file is like this:

" Heading 1

Line 1

Line 2

Line 3

Heading 2

Line 1

Line 2

Line 3 "

I won't know how many lines are after each heading, but I want to grab them all. All I know is the name, or a regular expression pattern for the heading.

The only way I know to read a file is the "for line in file:" way, but I don't know how to grab the lines AFTER the line I'm currently on. Hope this makes sense, and thanks for the help!

*Thanks for all the responses! I have tried to implement some of the solutions, but my problem is that not all the headings are the same name, and I'm not sure how to work around it. I need a different regular expression for each... any suggestions?*

Generator Functions

def group_by_heading( some_source ):
    buffer= []
    for line in some_source:
        if line.startswith( "Heading" ):
            if buffer: yield buffer
            buffer= [ line ]
        else:
            buffer.append( line )
    yield buffer

with open( "some_file", "r" ) as source:
    for heading_and_lines in group_by_heading( source ):
        heading= heading_and_lines[0]
        lines= heading_and_lines[1:]
        # process away.