I am trying to find out the decrements in a column and if found then print the last highest value.
For example:
From 111 to 445 there is a continous increment in the column.But 333 is less then the number before it.
111 aaa
112 aaa
112 aaa
113 sdf
115 aaa
222 ddd
333 sss
333 sss
444 sss
445 sss
333 aaa<<<<<<this is less then the number above it (445)
If any such scenario is found then print 445 sss
Like this, for example:
$ awk '{if (before>$1) {print before_line}} {before=$1; before_line=$0}' a
445 sss
What is it doing? Check the variable before and compare its value with the current. In case it is bigger, print the line.
It works for many cases as well:
$ cat a
111 aaa
112 aaa
112 aaa
113 sdf
115 aaa <--- this
15 aaa
222 ddd
333 sss
333 sss
444 sss
445 sss <--- this
333 aaa
$ awk '{if (before>$1) {print before_line}} {before=$1; before_line=$0}' a
115 aaa
445 sss
Store each number in a single variable called prevNumber then when you come to print the next one do a check e.g. if (newNumber < prevNumber) print prevNumber;
dont really know what language you are using
You can say:
awk '$1 > max {max=$1; maxline=$0}; END{ print maxline}' inputfile
For your input, it'd print:
445 sss
