0

I have some data in this form:

0.123456 11.234567 -22.345678
-3.456789 4.567891 -5.678912 etc

I would like to convert them in a matrix with 8columns in a scientific format with +- sign, one number before and 6 numbers after the decimal point, such as:

+0.123456E+00 +1.123456E+01  -2.234567E+01 etc

I've tried something like this but it doesn't work:

sed -n '31,46p' $filename1|xargs -n8|printf "%13.6e" >>file2.txt

Any ideas?

eva_thess
  • 101
  • 1
  • 1
  • 7

1 Answers1

2

I would use this:

xargs -n 8 < file | awk '{for (i=1;i<NF;i++) printf ("%13.6e%s", $i, FS); printf ("%13.6e\n", $NF)}'
  • xargs -n X < file comes from your previous question: How to format the data of a file in Unix?
  • awk '{for (i=1;i<NF;i++) printf ("%13.6e%s", $i, FS); printf ("%13.6e\n", $NF)}' loops through the items in a line and prints them with the specific format %13.6e, together with a space. Then, it prints the last field and a new line.

See a test

$ cat a
0.123456 11.234567 -22.345678
-3.456789 4.567891 -5.678912
2.342 23.111 123.23 22
$ xargs -n 8 < a | awk '{for (i=1;i<NF;i++) printf ("%13.6e%s", $i, FS); printf ("%13.6e\n", $NF)}'
1.234560e-01  1.123457e+01 -2.234568e+01 -3.456789e+00  4.567891e+00 -5.678912e+00  2.342000e+00  2.311100e+01
1.232300e+02  2.200000e+01
Community
  • 1
  • 1
fedorqui
  • 275,237
  • 103
  • 548
  • 598