first character goes to last when formatting XYZ columns with printf awk

Question

I have a file with 7 columns and I want to get the first 4 and format them with awk. When I use awk without formatting it works however when I put it with printf it happens that the first character of the first line goes to the last line.

Example when I use sed -n "323,384p" file.xyz | awk '{print $1, $2, $3, $4}'

H 1.159798 1.491313 -0.946869
O 0.854926 2.591706 0.311355
O 1.225042 0.465961 2.017823
O 3.035434 -1.589327 2.087115
H 2.323008 -0.909968 6.144897
H 0.166459 -2.029119 3.043016
H 1.551022 -2.693028 4.563796
H 0.353505 -2.316287 5.790394

However, when I use sed -n "323,384p" file.xyz | awk '{printf "%15s %15s %15s \n" $1, $2, $3, $4}'

       1.159798        1.491313       -0.946869  
H       0.854926        2.591706        0.311355  
O       1.225042        0.465961        2.017823  
O       3.035434       -1.589327        2.087115  
O       0.588302       -2.894040        1.612656  
O       2.323008       -0.909968        6.144897  
H       0.166459       -2.029119        3.043016  
H       1.551022       -2.693028        4.563796  
H       0.353505       -2.316287        5.790394  
H

The letter H that should appear first for some reason goes on a new line after the last line. Would anyone please know how to fix this problem?

For simplicity's sake I didn't put all lines from the xyz file

Answer 1

There are three placeholders in the printf statement. $1 is appended to the format string after the newline since there is no comma before $1. Fields $2, $3 and $4 are formatted with printf. The result is that $1 is always printed after the newline.

You probably want to format all four fields with printf:

... | awk '{printf "%s %15s %15s %15s\n", $1, $2, $3, $4}'