Using awk to right justify file

Shell Programming and Scripting
1

I have a csv file that contain 3 fields of data. I need to reformat it so that the output, particularly in the 3rd field is right justified.

Input

-159.4031,22.2341,96.52
-159.4009,22.2341,95.85
-159.4774,22.2319,114.09
-159.4054,22.2319,97.65

This is the awk command that I am working with but 3rd field is left justified.

awk '{ printf " %.4f %.4f %.2f\n", $1, $2, $3 }'

 -159.4031  22.2341  96.52
 -159.4009  22.2341  95.85
 -159.4774  22.2319  114.09
 -159.4054  22.2319  97.65

My desired output is:

 -159.4031  22.2341    96.52
 -159.4009  22.2341    95.85
 -159.4774  22.2319   114.09
 -159.4054  22.2319    97.65

What am I doing wrong?

2

@ncwxpanther , welcome

something along ...

awk -F, '{ printf " %.4f %.4f %6.2f\n", $1, $2, $3 }' /tmp/input 
 -159.4031 22.2341  96.52
 -159.4009 22.2341  95.85
 -159.4774 22.2319 114.09
 -159.4054 22.2319  97.65

I edited mine ...

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3

I have edited my desired output.

4

After the dot there is the number of digits after the decimal point.

Before the dot there is the total field width, right justified and space-padded. A - before that number means left justified.

5

It seems possible that the first two fields might also vary in length at some time, so it might be wise to apply the (maximum possible ) total width to all columns.

Note that awk (or any printf in C or Bash, for example) will never truncate a field to the total field width. It will always show all the data, in preference to retaining the formatting.

The part after the decimal point is treated differently: that part of the value is rounded to fit in the specified length given after the dot.

6

%.4f is %0.4f, the field width is zero, all values exceed the field width, so no adjustment within the field is possible.
You might want
printf "%10.4f %9.4f %7.2f\n", $1, $2, $3
so the given values fit in the fields and are adjusted.

Assume the longest possible numbers (values) for each column, and make a test with them!

1 Like