如何仅打印 BASH 中的唯一行？
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How to print only the unique lines in BASH?
提问by Village
How can I print only those lines that appear exactly once in a file? E.g., given this file:
如何仅打印在文件中仅出现一次的那些行？例如，给定这个文件：
mountain
forest
mountain
eagle
The output would be this, because the line mountainappears twice:
输出是这样的，因为该行mountain出现了两次：
forest
eagle
- The lines can be sorted, if necessary.
- 如有必要，可以对行进行排序。
采纳答案by anubhava
Using awk:
使用 awk：
awk '{!seen[sort inputfile | uniq -u
]++};END{for(i in seen) if(seen[i]==1)print i}' file
eagle
forest
回答by devnull
Use sortand uniq:
使用sort和uniq：
-u, --unique
only print unique lines
The -uoption would cause uniqto print only unique lines. Quoting from man uniq:
该-u选项将导致uniq仅打印唯一的行。引自man uniq：
eagle
forest
For your input, it'd produce:
对于您的输入，它会产生：
##代码##Obs:Remember to sortbefore uniq -ubecause uniqoperates on adjacentlines. So what uniq -uactually does is to print lines that don't have identical neighbor lines, but that doesn't mean they are really unique. When you sort, all the identical lines get grouped together and only the lines that are reallyunique in the file will remain after uniq -u.
Obs：记住sort之前，uniq -u因为uniq在相邻的行上操作。所以uniq -u实际上做的是打印没有相同相邻线的线，但这并不意味着它们真的是独一无二的。当你sort，所有相同的行被组合在一起，只有在文件中真正独特的行会保留在uniq -u.
回答by Oliver Matthews
You almost had the answer in your question:
你的问题几乎有了答案：
sort filename | uniq -u
sort filename | uniq -u
