{"id":2397,"date":"2017-12-08T09:22:58","date_gmt":"2017-12-08T09:22:58","guid":{"rendered":"http:\/\/www.scmgalaxy.com\/tutorials\/?p=2397"},"modified":"2020-01-09T10:00:18","modified_gmt":"2020-01-09T10:00:18","slug":"shell-script-merge-two-list-and-remove-duplicates","status":"publish","type":"post","link":"https:\/\/www.devopsschool.com\/blog\/shell-script-merge-two-list-and-remove-duplicates\/","title":{"rendered":"Shell script merge two list and remove duplicates"},"content":{"rendered":"

rajeshkumar created the topic: shell script merge two list and remove duplicates<\/strong><\/p>\n

You want all the records from list_A supplemented by all the records from list_B for which there is not already a matching name in list A. Mathematically this is:<\/p>\n

A + B - {w in B | (w,value) in A }<\/code><\/p>\n

\\<\/p>\n

There are many ways of accomplishing this, depending on access and needed efficiencies.<\/p>\n

* If you can modify DB1 (with A), then download table B from DB2, upload it to DB1, then extract your data with the appropriate join
\n* If you can’t modify DB1, then download both A and B and concatenate them to the same stream, with A followed by B. Then sort by the first field. Then process the stream one record at time. Duplicate names will be side-by-side. If the same name appears more than one time, print the first and ignore subsequent records with the same name.<\/p>\n

Here is a sample solution to your problem (starting with two lists of names\/values)<\/p>\n

#!\/bin\/bash<\/p>\n

A=\"Smith value1
\nJones value2
\nWilson value3\"<\/p>\n

B=\"Smith value10
\nWilson value11
\nFox value12
\nBrown value13\"<\/p>\n

PrevName=\"Not a valid name\"
\necho \"$A
\n$B\" | sort -k1 |
\nwhile read Name Value
\ndo
\n if [ \"$Name\" != \"$PrevName\" ]; then
\n echo $Name $Value
\n fi
\n PrevName=\"$Name\"
\ndone > outfile<\/code><\/p>\n

You want all the records from list_A supplemented by all the records from list_B for which there is not already a matching name in list A. Mathematically this is:<\/p>\n

A + B – {w in B | (w,value) in A }<\/p>\n

There are many ways of accomplishing this, depending on access and needed efficiencies.<\/p>\n

* If you can modify DB1 (with A), then download table B from DB2, upload it to DB1, then extract your data with the appropriate join
\n* If you can’t modify DB1, then download both A and B and concatenate them to the same stream, with A followed by B. Then sort by the first field. Then process the stream one record at time. Duplicate names will be side-by-side. If the same name appears more than one time, print the first and ignore subsequent records with the same name.<\/p>\n

Here is a sample solution to your problem (starting with two lists of names\/values):<\/p>\n

#!\/bin\/bash<\/p>\n

A=”Smith value1
\nJones value2
\nWilson value3″<\/p>\n

B=”Smith value10
\nWilson value11
\nFox value12
\nBrown value13″<\/p>\n

PrevName=”Not a valid name”
\necho “$A
\n$B” | sort -k1 |
\nwhile read Name Value
\ndo
\nif [ “$Name” != “$PrevName” ]; then
\necho $Name $Value
\nfi
\nPrevName=”$Name”
\ndone > outfile<\/p>\n

Here is the output:
\nBrown value13
\nFox value12
\nJones value2
\nSmith value1
\nWilson value11
\nRegards,
\nRajesh Kumar
\nTwitt me @ twitter.com\/RajeshKumarIn<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

rajeshkumar created the topic: shell script merge two list and remove duplicates You want all the records from list_A supplemented by all the records from list_B for which there is not already a matching name in list A. Mathematically this is: A + B – {w in B | (w,value) in A } \\ There…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_kad_post_classname":"","_joinchat":[],"footnotes":""},"categories":[454],"tags":[138],"class_list":["post-2397","post","type-post","status-publish","format-standard","hentry","category-shell-script","tag-shell"],"_links":{"self":[{"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/posts\/2397","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/comments?post=2397"}],"version-history":[{"count":1,"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/posts\/2397\/revisions"}],"predecessor-version":[{"id":2398,"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/posts\/2397\/revisions\/2398"}],"wp:attachment":[{"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/media?parent=2397"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/categories?post=2397"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.devopsschool.com\/blog\/wp-json\/wp\/v2\/tags?post=2397"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}