bash sed 如何删除文件中的前 17 行和后 8 行
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sed how to delete first 17 lines and last 8 lines in a file
提问by Deano
I have a big file 150GB CSV file and I would like to remove the first 17 lines and the last 8 lines. I have tried the following but seems that's not working right
我有一个 150GB 的大文件 CSV 文件，我想删除前 17 行和最后 8 行。我已经尝试了以下但似乎不起作用
sed -i -n -e :a -e '1,8!{P;N;D;};N;ba'
and
和
sed -i '1,17d'
I wonder if someone can help with sed or awk, one liner will be great?
我想知道是否有人可以帮助使用 sed 或 awk，一个班轮会很棒吗？
回答by choroba
headand tailare better for the job than sedor awk.
head并且tail比sed或更适合这份工作awk。
tail -n+18 file | head -n-8 > newfile
回答by Ed Morton
awk -v nr="$(wc -l < file)" 'NR>17 && NR<(nr-8)' file
回答by Scrutinizer
All awk:
所有 awk：
awk 'NR>y+x{print A[NR%y]} {A[NR%y]=Try this :
sed '{[/]<n>|<string>|<regex>[/]}d' <fileName>
sed '{[/]<adr1>[,<adr2>][/]d' <fileName>
}' x=17 y=8 file
回答by Scrutinizer
LENGTH=`wc -l < file`
head -n $((LENGTH-8)) file | tail -n $((LENGTH-17)) > file
where
在哪里
/.../=delimiters
n = line number
string = string found in in line
regex = regular expression corresponding to the searched pattern
addr = address of a line (number or pattern )
d = delete
/.../=分隔符
n = 行号
字符串 = 在行中找到的字符串
regex = 搜索模式对应的正则表达式
addr = 行地址（数字或模式）
d = 删除
Refer this link
参考这个链接
回答by Adam Sznajder
LENGTH=`wc -l < file`
head -n $((LENGTH-8)) file | tail -n $((LENGTH-8-17)) > file
Edit: As mtk posted in comment this won't work. If you want to use wcand track file length you should use:
编辑：正如 mtk 在评论中发布的那样，这是行不通的。如果你想使用wc和跟踪文件长度，你应该使用：
LENGTH=`wc -l < file`
head -n $((LENGTH-8)) file > file
LENGTH=`wc -l < file`
tail -n $((LENGTH-17)) file > file
or:
或者：
sed -i '' -e $'1,17d;:a\nN;19,25ba\nP;D' file.txt
What makes this solution less elegant than that posted by choroba :)
是什么让这个解决方案不如 choroba 发布的那么优雅 :)
回答by jcsahnwaldt says GoFundMonica
Similar to Thor's answer, but a bit shorter:
类似于雷神的回答，但更短一点：
1,{front}d;:a\nN;{front+2},{front+tail}ba\nP;D
The -i ''tells sed to edit the file in place. (The syntax may be a bit different on your system. Check the man page.)
该-i ''告诉sed编辑文件到位。（语法在您的系统上可能略有不同。请查看手册页。）
If you want to delete frontlines from the front and tailfrom the end, you'd have to use the following numbers:
如果front要从前面和tail最后删除行，则必须使用以下数字：
1,17d # delete lines 1 ... 17, goto start
:a # define label a
N # add next line from file to buffer, quit if at end of file
19,25ba # if line number is 19 ... 25, goto start (label a)
P # print first line in buffer
D # delete first line from buffer, go back to start
(I put them in curly braces here, but that's just pseudocode. You'll have to replace them by the actual numbers. Also, it should work with {front+1}, but it doesn't on my machine (macOS 10.12.4). I think that's a bug.)
（我在这里将它们放在花括号中，但这只是伪代码。您必须用实际数字替换它们。此外，它应该可以使用{front+1}，但它在我的机器（macOS 10.12.4）上不起作用。我认为这是一个错误。）
I'll try to explain how the command works. Here's a human-readable version:
我将尝试解释该命令的工作原理。这是一个人类可读的版本：
{
ghead -17 > /dev/null
sed -n -e :a -e '1,8!{P;N;D;};N;ba'
} < my-bigfile > subset-of
First we skip 17 lines. That's easy. The rest is tricky, but basically we keep a buffer of eight lines. We only start printing lines when the buffer is full, but we stop printing when we reach the end of the file, so at the end, there are still eight lines left in the buffer that we didn't print - in other words, we deleted them.
首先我们跳过 17 行。这很容易。其余的很棘手，但基本上我们保留了 8 行的缓冲区。我们只在缓冲区满时开始打印行，但是当我们到达文件末尾时我们停止打印，所以最后，缓冲区中还有八行我们没有打印 - 换句话说，我们删除了它们。
回答by sotapme
I learnt this today for the shell.
我今天为 shell 学习了这个。
##代码##One has to use a non consuming head, hence the use of gheadfrom the GNU coreutils.
必须使用非消耗的head，因此使用ghead来自 GNU coreutils 的 。
