bash, find nearest next value, forward and backward

Question

I have a data.txt file

1    2     3    4      5       6        7   
cat data.txt
13 245 1323 10.1111 10.2222 60.1111 60.22222
13 133 2325 11.2222 11.333  61.2222 61.3333
13 245 1323 12.3333 12.4444 62.3333 62.44444444
13 245 1323 13.4444 13.5555 63.4444 63.5555

Find next nearest: My target value is 11.6667 and it should find the nearest next value in column 4 as 12.3333

Find previous nearest: My target value is 62.9997 and it should find the nearest previous value in column 6 as 62.3333

I am able to find the next nearest (case 1) by

awk -v c=4 -v t=11.6667 '{a[NR]=$c}END{
        asort(a);d=a[NR]-t;d=d<0?-d:d;v = a[NR]
        for(i=NR-1;i>=1;i--){
                m=a[i]-t;m=m<0?-m:m
                if(m<d){
                    d=m;v=a[i]
                }
        }
        print v
}' f
12.3333

Any bash solution? for finding the previous nearest (case 2)?

Answer 1

Try this:

$ cat tst.awk
{
    if ($fld > tgt) {
        del = $fld - tgt
        if ( (del < minGtDel) || (++gtHit == 1) ) {
            minGtDel = del
            minGtVal = $fld
        }
    }
    else if ($fld < tgt) {
        del = tgt - $fld
        if ( (del < minLtDel) || (++ltHit == 1) ) {
            minLtDel = del
            minLtVal = $fld
        }
    }
    else {
        minEqVal = $fld
    }
}
END {
    print (minGtVal == "" ? "NaN" : minGtVal)
    print (minLtVal == "" ? "NaN" : minLtVal)
    print (minEqVal == "" ? "NaN" : minEqVal)
}

.

$ awk -v fld=4 -v tgt=11.6667 -f tst.awk file
12.3333
11.2222
NaN

$ awk -v fld=6 -v tgt=62.9997 -f tst.awk file
63.4444
62.3333
NaN

$ awk -v fld=6 -v tgt=62.3333 -f tst.awk file
63.4444
61.2222
62.3333

Answer 2

For the first part:

awk -v v1="11.6667" '$4>v1 {print $4;exit}' file
12.3333

And second part:

awk -v v2="62.9997" '$6>v2 {print p;exit} {p=$6}' file
62.3333

---

Both in one go:

awk -v v1="11.6667" -v v2="62.9997" '$4>v1 && !p1 {p1=$4} $6>v2 && !p2 {p2=p} {p=$6} END {print p1,p2}' file
12.3333 62.3333

Answer 3

I don't know if this is what you're looking for, but this is what I came up with, not knowing awk:

#!/bin/sh

IFSBAK=$IFS
IFS=$'\n'

best=

for line in `cat $1`; do
    IFS=$' \t'
    arr=($line)


    num=${arr[5]}
    [[ -z $best ]] && best=$num

    if [ $(bc <<< "$num < 62.997") -eq 1 ]; then 
        if [  $(bc <<< "$best < $num") -eq 1 ]; then
            best=$num
        fi
    fi

    IFS=$'\n'
done

IFS=$IFSBAK
echo $best

If you want, you can add the column and the input value 62.997 as paramters, I didn't to demonstrate that it would look for specifically what you want.

Edited to remove assumption that file is sorted.

Answer 4

You solution looks unnecessarily complicated (storing a whole array and sorting it) and I think you would see the bash solution if you re-thought your awk.

In awk you can detect the first line with

FNR==1 {do something}

so on the first line, set a variable BestYet to the value in the column you are searching.

On subsequent lines, simply test if the value in the column you are checking is

a) less than your target AND 
b) greater than `BestYet`

if it is, update BestYet. At the end, print BestYet.

In bash, apply the same logic, but read each line into a bash array and use ${a[n]} to get the n'th element.