Using for-loop variable in awk print

Question

I'm trying to print out

for i in 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40; do
awk -v a="$i" '{printf "%10.2f %10.2f\n", a, ($8*627.509)}' e1.txt > e2.txt
done

But when I open this file,

  1.40 -12111939.85
  1.40 -12112479.17
  1.40 -12112817.98
  1.40 -12112997.55
  1.40 -12113047.39
  1.40 -12112998.93
  1.40 -12112873.57
  1.40 -12112695.74
  1.40 -12112504.02
  1.40 -12112346.74
  1.40 -12112316.49
  1.40 -12112204.51
  1.40 -12112149.56

Ignore the second column as it reads the value and operates from other txt file, e1.txt.

As it is shown, only the last for-loop index variable is used in this case. But I wish to print the for-loop values of 0.80 ~ 1.40 accordingly to each line.

Your output redirection overwrites the output file for every iteration of the loop. Remove the output redirection from the "awk" line and add it to the "done" line: done > e2.txt — glenn jackman
– glenn jackman, Commented Jan 2, 2020 at 22:04

glenn jackman · Accepted Answer · 2020-01-02 22:16:02Z

3

For efficiency, I would avoid processing the same file 13 times.

The BEGIN block looks awkward because awk can't declare an array literal.

awk '
  BEGIN {
    a = "0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40"
    n = split(a, as)
  }
  {
    for (i=1; i <= n; i++)
      printf "%10.2f %10.2f\n", as[i], ($8 * 627.509)
  }
' e1.txt > e2.txt

If you want all the 0.80 first and all the 1.40 last, you can:

awk '...' e1.txt | sort -g > e2.txt

answered Jan 2, 2020 at 22:16

glenn jackman

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1 Comment

Ed Morton Over a year ago

you could just do split("0.80 0.85 ...", as) though, no need for a.

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