如何用空白填充信息，而其余部分保持不变？

Question

我正在为模拟飞行器构建场景，需要弄清楚如何在文本文件中编辑多行（其中3,579,189行）。

我有TextCrawler Pro，Node，Python SVN和Notepad ++作为工具。

原始的预编辑部分：

POLYGON_POINT -79.750000000217,42.017498354525,0
POLYGON_POINT -79.750000000217,42.016478251402,0
POLYGON_POINT -79.750598748133,42.017193264943,0
POLYGON_POINT -79.750000000217,42.017498354525,0


POLYGON_POINT -79.750000000217,42.085882815878,0
POLYGON_POINT -79.750000000217,42.082008734634,0
POLYGON_POINT -79.751045507507,42.082126409633,0
POLYGON_POINT -79.750281907508,42.083166574215,0
POLYGON_POINT -79.750781149174,42.084212672130,0
POLYGON_POINT -79.750000000217,42.085882815878,0


POLYGON_POINT -79.750000000217,42.088955814831,0
POLYGON_POINT -79.750456566883,42.087544672125,0
POLYGON_POINT -79.751642899173,42.088273325249,0
POLYGON_POINT -79.751461052298,42.088916154415,0
POLYGON_POINT -79.750000000217,42.088955814831,0

使用Notepad ++的替换功能，添加POLYGON_POINT行非常容易。 现在，我需要一些帮助使其显示为：

BEGIN_POLYGON
POLYGON_POINT -79.750000000217,42.017498354525,0
POLYGON_POINT -79.750000000217,42.016478251402,0
POLYGON_POINT -79.750598748133,42.017193264943,0
POLYGON_POINT -79.750000000217,42.017498354525,0
END_POLY
BEGIN_POLYGON
POLYGON_POINT -79.750000000217,42.085882815878,0
POLYGON_POINT -79.750000000217,42.082008734634,0
POLYGON_POINT -79.751045507507,42.082126409633,0
POLYGON_POINT -79.750281907508,42.083166574215,0
POLYGON_POINT -79.750781149174,42.084212672130,0
POLYGON_POINT -79.750000000217,42.085882815878,0
END_POLY
BEGIN_POLYGON
POLYGON_POINT -79.750000000217,42.088955814831,0
POLYGON_POINT -79.750456566883,42.087544672125,0
POLYGON_POINT -79.751642899173,42.088273325249,0
POLYGON_POINT -79.751461052298,42.088916154415,0
POLYGON_POINT -79.750000000217,42.088955814831,0

即添加BEGIN_POLYGON每个块和前END_POLY每个之后。

我怎样才能做到这一点？

Answer 1

我将使用itertools.groupby （仅使用if k条件获取非空白组）将行分为空白还是空白，并为每个组添加页眉/页脚。 然后使用itertools.chain展平组

import itertools

with open("file.txt") as f, open("fileout.txt","w") as fw:
    fw.writelines(itertools.chain.from_iterable([["BEGIN_POLYGON\n"]+list(v)+["END_POLYGON\n"] for k,v in itertools.groupby(f,key = lambda l : bool(l.strip())) if k]))

key = lambda l : bool(l.strip()))是分组键：测试空行但行终止

此方法不需要完全读取文件，因此适用于非常大的文件。 它逐行处理文件，因此不会占用内存。

Answer 2

使用sed的快速解决方案

cat -s file.txt |\
    sed -e 's/^$/END_POLY\nBEGIN_POLYGON/'\
    -e '1i BEGIN_POLYGON'\
    -e '$a END_POLY'

• cat -s将所有空白行压缩为一个
• 第一个sed用END_POLY和BEGIN_POLYGON标签替换空白行
• 第二个和最后一个sed前置并将其余标签附加到输出中