如何在不进行预排序(或类似操作)的情况下执行uniq -d [英] How to do uniq -d without presorting (or something similar)

问题描述

我知道我可以不进行预排序就删除重复的行，例如:

I am aware that I can remove duplicated lines without presorting, for example:

awk '!x[$0]++' file

但是，我的目标是仅打印重复且仅一次的行.如果不是因为预分类问题

However, my goal is to only print lines which are duplicated and only once. If it were not for the presorting problem

sort | uniq -d

将是完美的.但是订单对我来说非常重要.有没有办法用awk，grep或类似的方法做到这一点?

would be perfect. BUT the order is of great importance to me. Is there a way to do this with awk, grep or something similar?

我正在寻找一种衬纸，如果可能的话，它不需要编写脚本.

I am looking for a one liner which does not require writing a script if possible.

推荐答案

只需检查 x [$ 0] 的值:

awk 'x[$0]++ == 1' file.txt

以上内容将在第二次显示时打印一行.

The above will print a line when it's seen for the second time.

或带前缀 ++ :

awk '++x[$0] == 2' file.txt

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