CODEWITHSUNDEEP
arraysfor-loopawkif-statement
MortenM
MortenM

Reputation: 33

awk for loop with if else conditions to array

I seek your help in storing awk returning values in an array for my awk for loop using if else conditions.

If  $3 == $7
then print $9 multiplied by $4
else print $4 multiplied by (2 minus $9)

I got this working so far by:

awk '{if ($3 == $7) print $9*$4; else print $4*(2-$9);}' file >outfile

the above code works for the first data column ($9). However, I want to loop through all columns from 9 to 1547 and return an array containing the returning values. This should be simple enough but I cant seem to understand some basic concepts here.

So far I understand the need to declare the number of loops, before the actual function, by:

awk ' {for(i=9;i<=NF;i++)} END {if ($3 == $7) print $i*$4; else print $4*(2-$i);}'

However, how and when to declare the array is beyond me (biologist). Any help would be highly appreciated.

Example:

input (big file.. here column 1-10):

rs2070501 22 A 0.0206 0.337855 rs2070501 G A 0.977 0.066

output:

0.0210738

here the else statement kicks in ($3 * (2-$9)

How to get awk to print out the array 9-Nth, and not just column 9

Upvotes: 3

Views: 44908

Answers (1)

tripleee
tripleee

Reputation: 189417

Try with this.

awk '{
    for(i=9; i<=NF; ++i)
        printf "%s%f",
            (i==9 ? "" : " "),
            ($3 == $7 ? $i*$4 : $4*(2-$i));
    printf "\n"
}' filename

The ( test ? when : else ) is just a shorthand; the stuff after ? gets evaluated if the test is true, and the stuff after : otherwise. So it prints an empty separator for the first field, and a space otherwise; and chooses how to calculate the value of the field depending on whether $3 == $7 is true.

Upvotes: 5

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