how to get max number from a file with Linux bash shell scripts

Question

How to get the maximum "rate" and the corresponding "log2c" value from a file as follows? e.g: the max rate is 89.5039 , and log2c 3.0 . thanks a lot.

log2c=5.0 rate=88.7619
log2c=-1.0 rate=86.5412
log2c=11.0 rate=86.1482
log2c=3.0 rate=89.5039
log2c=-3.0 rate=85.5614
log2c=9.0 rate=81.4302

Answer 1

Use sort:

sort -t= -nr -k3 inputfile | head -1

For the given input, it'd return:

log2c=3.0 rate=89.5039

If you want to read the values into variables, you can use the builtin read:

$ IFS=$' =' read -a var <<< $(sort -t= -nr -k3 inputfile | head -1)
$ echo ${var[1]}
3.0
$ echo ${var[3]}
89.5039

Answer 2

For very large files, using sort will be quite slow. In this case, it's better to use something like awk, which needs only one pass:

$ awk -F= 'BEGIN { max = -inf } { if ($3 > max) { max = $3; line = $0 } } END { print line }' test.txt
log2c=3.0 rate=89.5039

The time complexity of this operation is linear, while the space complexity is constant (and small). Explanation:

1. awk -F= '...' test.txt: Invoke awk on test.txt, using = as the field separator
2. BEGIN { max = -inf }: Initialise max to something that will always be smaller than whatever you're reading.
3. { if ($3 > max) { max = $3; line = $0; } }: For each input line, if max is less than the value of the third field ($3), then update it and remember the value of the current line ($0)
4. END { print line }: Finally, print the line we remembered while reading the input.