以bash的特定顺序打印文件 [英] Print file in particular order in bash

问题描述

我的文件中包含以下内容:

I have file with content:

file.txt:

Iteration 1
RAM: +456ms
Cache: +142ms (total +417ms)

Iteration 2
Spec: +152ms
Cache: +149ms (total +413ms)

Iteration 3
RAM: +184ms
Spec: +172ms
Searchms: +131ms (total +385ms)

First launch 4
RAM: +149ms
Searchms: +188ms

在此文件中，每个First launch之间的内容可以不同，它不是固定的(例如:第一次启动3包含三个元素，而第一次启动2的内容仅包含2个元素)，因此，在First launch模式，一开始是未知的.

In this file between every First launch, content can be different, it is not fixed ( for example: First launch 3 contain three elements while First launch 2 contents only 2 elements), so any number of content can be between the First launch pattern which is not known at the beginning.

预期输出:

RAM  456 184 149
Cache 142 149  
Spec 152 172
Searchms 131 188

由于不知道确切的方法，所以我尝试了这段代码.

Due to not knowing exact approach,I have tried this code.

我的代码:

for i in {1..4}
do
awk "/First launch $i/{flag=1;next} /First launch $((i+1))/{flag=0} flag" file.txt> fl$i.txt
sed -i 's/\+//g' fl$i.txt
sed -i 's/://g' fl$i.txt
sed -i 's/(.*//g' fl$i.txt
sed -i 's/ms//g' fl$i.txt
awk '{print $1 "\t" $2}' fl$i.txt
done

我的输出有两个问题: 我正在为每个模式生成文件，这是错误的.我也想在此时间之后删除ms，但它也从模式名称中删除ms(例如:Searchms to Search)

My output has two issues: I am generating the file for each pattern which is wrong. Also I wanted to remove ms after the time but it also removes the ms from the pattern name (ex: Searchms to Search)

输出:

fl1.txt: 
    RAM     456
    Cache   142
fl2.txt : 
    Spec    152
    Cache   149
fl3.txt  :
    RAM     184
    Spec    152
    Search  131
fl4.txt : 
    RAM     149
    Search  188

请向我建议一种在不产生任何额外文件的情况下获得预期输出的方法，并且要在删除时间后删除ms.

Please suggest me an approach to get the expected output without generating any extra file with the constraint of removing ms after the time.

推荐答案

$ cat tst.awk
BEGIN { FS="[: ]+" }
/:/ { vals[$1] = vals[$1] OFS $2+0 }
END { for (key in vals) print key vals[key] }

$ awk -f tst.awk file
Cache 142 149
RAM 456 184 149
Searchms 131 188
Spec 152 172

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