Linux 如何仅使用 grep/sed 提取子字符串和数字
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how to extract substring and numbers only using grep/sed
提问by Hooloovoo
I have a text file containing both text and numbers, I want to use grep to extract only the numbers I need for example, given a file as follow:
我有一个包含文本和数字的文本文件，我想使用 grep 仅提取我需要的数字，例如，给定文件如下：
miss rate 0.21
ipc 222
stalls n shdmem 112
So say I only want to extract the data for miss ratewhich is 0.21. How do I do it with grep or sed? Plus, I need more than one number, not only the one after miss rate. That is, I may want to get both 0.21and 112. A sample output might look like this:
所以说我只想提取miss rateis的数据0.21。我如何用 grep 或 sed 做到这一点？另外，我需要多个数字，而不仅仅是miss rate. 也就是说，我可能想要同时获得0.21和112。示例输出可能如下所示：
0.21 222 112
Cause I need the data for later plot.
因为我需要稍后绘图的数据。
采纳答案by that other guy
Use awkinstead:
使用awk来代替：
awk '/^miss rate/ { print }' yourfile
To do it with just grep, you need non-standard extensions like here with GNU grep using PCRE (-P) with positive lookbehind (?<=..) and match only (-o):
要仅使用 grep 来完成此操作，您需要使用 PCRE (-P) 和正向后视 (?<=..) 并仅匹配 (-o) 的非标准扩展，例如此处与 GNU grep 一起使用：
grep -Po '(?<=miss rate ).*' yourfile
回答by kamituel
You can use:
您可以使用：
grep -P "miss rate \d+(\.\d+)?" file.txt
or:
或者：
grep -E "miss rate [0-9]+(\.[0-9]+)?"
Both of those commands will print out miss rate 0.21. If you want to extract the number only, why not use Perl, Sed or Awk?
这两个命令都会打印出来miss rate 0.21。如果您只想提取数字，为什么不使用 Perl、Sed 或 Awk？
If you really want to avoid those, maybe this will work?
如果你真的想避免这些，也许这会奏效？
grep -E "miss rate [0-9]+(\.[0-9]+)?" g | xargs basename | tail -n 1
回答by DanneJ
If you reallywant to use only grep for this, then you can try:
如果您真的只想为此使用 grep，那么您可以尝试：
grep "miss rate" file | grep -oe '\([0-9.]*\)'
It will first find the line that matches, and then only output the digits.
它将首先找到匹配的行，然后只输出数字。
Sed might be a bit more readable, though:
不过，Sed 可能更具可读性：
sed -n 's#miss rate ##p' file
回答by Gilles Quenot
回答by mariux
The grep-and-cutsolution would look like:
的grep-和-cut的解决办法是这样的：
to get the 3rd field for every successful grep use:
为每次成功使用 grep 获取第三个字段：
grep "^miss rate " yourfile | cut -d ' ' -f 3
or to get the 3rd field and the rest use:
或获取第三个字段，其余使用：
grep "^miss rate " yourfile | cut -d ' ' -f 3-
Or if you use bash and "miss rate" only occurs once in your file you can also just do:
或者，如果您使用 bash 并且“未命中率”仅在您的文件中出现一次，您也可以这样做：
a=( $(grep -m 1 "miss rate" yourfile) )
echo ${a[2]}
where ${a[2]}is your result.
${a[2]}你的结果在哪里。
If "miss rate" occurs more then once you can loop over the grep output reading only what you need. (in bash)
如果“未命中率”发生的次数更多，则您可以循环遍历 grep 输出，仅读取您需要的内容。（在 bash 中）
回答by Daniel Williams
I believe
我相信
sed 's|[^0-9]*\([0-9\.]*\)|\1 |g' fiilename
sed 's|[^0-9]*\([0-9\.]*\)|\1 |g' fiilename
will do the trick. However every entry will be on it's own line if that is ok. I am sure there is a way for sed to produce a comma or space delimited list but I am not a super master of all things sed.
会做的伎俩。但是，如果可以的话，每个条目都将在它自己的行上。我确信 sed 有一种方法可以生成逗号或空格分隔的列表，但我不是 sed 的所有方面的超级大师。
