保留具有公共字段的记录组中的最后记录（awk）

Question

输入文件 ：

0,,M00,100,,SEK,20190411T10:02:12+0530,20190425T23:59:59+0530,,vsuser,,5774126050,12345685865,,,test,,,,,,,,,,,,,,,C
0,,M00,100,,SEK,20190411T10:02:12+0530,20190425T23:59:59+0530,20190411T10:02:12+0530,,vsuser,5774126050,12345685865,,,test,,,,,,,,,,,,,,,H
5,,M00,100,,SEK,20190411T10:02:12+0530,20190425T23:59:59+0530,20190411T15:42:17+0530,,vsuser,5774126050,12345685865,,,test,,,,,,,,,,,,,,,H
0,,M00,100,,SEK,20190411T10:02:12+0530,20190425T23:59:59+0530,20190411T15:42:37+0530,,vsuser,5774126050,12345685865,,,test,,,,,,,,,,,,,,,H
0,,M00,100,,SEK,20190411T10:03:16+0530,20190425T23:59:59+0530,,vsuser,,7318713283,12345729902,,,test,,,,,,,,,,,,,,,C
0,,M00,100,,SEK,20190411T10:03:16+0530,20190425T23:59:59+0530,20190410T10:03:16+0530,,vsuser,7318713283,12345729902,,,test,,,,,,,,,,,,,,,H

输出要求：

0,,M00,100,,SEK,20190411T10:02:12+0530,20190425T23:59:59+0530,20190411T15:42:37+0530,,vsuser,5774126050,12345685865,,,test,,,,,,,,,,,,,,,H
0,,M00,100,,SEK,20190411T10:03:16+0530,20190425T23:59:59+0530,20190410T10:03:16+0530,,vsuser,7318713283,12345729902,,,test,,,,,,,,,,,,,,,H

选择公共值为第13列的行。

Answer 1

awk下面可能会有所帮助：

awk -F, '{found[$13]=$0}END{for(i in found){print found[i]}}' file

产量

0,,M00,100,,SEK,20190411T10:02:12+0530,20190425T23:59:59+0530,20190411T15:42:37+0530,,vsuser,5774126050,12345685865,,,test,,,,,,,,,,,,,,,H
0,,M00,100,,SEK,20190411T10:03:16+0530,20190425T23:59:59+0530,20190410T10:03:16+0530,,vsuser,7318713283,12345729902,,,test,,,,,,,,,,,,,,,H

Answer 2

$ tac file | awk -F, '!seen[$13]++' | tac
0,,M00,100,,SEK,20190411T10:02:12+0530,20190425T23:59:59+0530,20190411T15:42:37+0530,,vsuser,5774126050,12345685865,,,test,,,,,,,,,,,,,,,H
0,,M00,100,,SEK,20190411T10:03:16+0530,20190425T23:59:59+0530,20190410T10:03:16+0530,,vsuser,7318713283,12345729902,,,test,,,,,,,,,,,,,,,H