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
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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
对于每个名称,我只希望其最后一次出现.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
有人可以给我bash/awk/sed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(但从文件或文本的末尾开始,仅在第二个字段中获得uniq行)(如您的结果示例中一样)
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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