按列消除部分重复的行并保留最后一行 [英] Eliminate partially duplicate lines by column and keep the last one

问题描述

我有一个看起来像这样的文件:

I have a file that looks like this:

2011-03-21 name001 line1
2011-03-21 name002 line2
2011-03-21 name003 line3
2011-03-22 name002 line4
2011-03-22 name001 line5

对于每个名字，我只想要它的最后一次出现.所以，我希望结果是:

for each name, I only want its last appearance. So, I expect the result to be:

2011-03-21 name003 line3
2011-03-22 name002 line4
2011-03-22 name001 line5

有人可以用 bash/awk/sed 给我一个解决方案吗?

Could someone give me a solution with bash/awk/sed?

推荐答案

此代码通过第二个字段获取 uniq 行，但从文件或文本的末尾(如您的结果示例中)

This code get uniq lines by second field but from the end of file or text (like in your result example)

tac temp.txt | sort -k2,2 -r -u

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