从fi中删除换行符

Question

我有一个以下格式的文本文件：

Run#1 Step#1 > Connecting to server
Run#1 Step#2 > Connected OK
Run#1 Step#3 > Sending request: {
    "path": "/testpage",
    "time": "2015-06-07T00:00:00.000Z"
}
Run#1 Step#4 > Request sent OK

我需要做的是处理这个文件。如果每个步骤都打印在单独的行上，会更容易：

Run#1 Step#1 > Connecting to server
Run#1 Step#2 > Connected OK
Run#1 Step#3 > Sending request: { "path": "/testpage", "time": "2015-06-07T00:00:00.000Z" }
Run#1 Step#4 > Request sent OK

我如何做到这一点（在bash或ruby/python/。。。脚本）？你知道吗

Answer 1

1）拆分（“\n”） 2） 替换（“运行#”，“\n运行#”） 3） 删除第一行（“\n”）

Answer 2

gnu-sed解决方案

cat file | sed ':a; N; $! ba; s/\n//g; s/Run#/\nRun#/g;' | sed '1d;' > outputfile

Answer 3

使用python根据以Run#开头的行对行进行分组，并将不以Run#开头的行的任何部分连接到上一个Run#行（无论内容如何），它还将替换原始文件，您不需要将整个文件读入内存：

from itertools import groupby
from tempfile import NamedTemporaryFile
from shutil import move

with open("file.txt") as f, NamedTemporaryFile("w",dir=".",delete=False) as  out:
    grouped = groupby(f, key=lambda x: not x.startswith("Run#"))
    for k, v in grouped:
        if not k:
            v, nxt = "".join(v), next(grouped, "  ")[1]
            out.write("{}{}\n".format(v.rstrip(), "".join(map(str.strip, nxt))))
        else:
            out.writelines(v)

move(out.name,"file.txt")

输出：

Run#1 Step#1 > Connecting to server
Run#1 Step#2 > Connected OK
Run#1 Step#3 > Sending request: {"path": "/testpage","time": "2015-06-07T00:00:00.000Z"}
Run#1 Step#4 > Request sent OK