CODEWITHSUNDEEP
linuxshellunixawk
Marjer
Marjer

Reputation: 1403

awk to print more than 2 decimal values with trailing zero

Below is my input file,

input.txt

17.0
17.12
17.123
17.1234
17.120
17.1230
17.12340
17.00
17.000
17.0000

And i want the out put as below

output.txt

17.120
17.1230
17.12340
17.000
17.0000

i.e. i need the values with more than 2 decimal points with trailing zero. i Have used the below awk but the output file contains 17.00 also. How to ignore this.

awk '/^[0-9]+[.][0-9][1-9]*0/' input.txt

Output of awk

17.120
17.1230
17.12340
17.00
17.000
17.0000

Upvotes: 0

Views: 869

Answers (3)

Adrian Frühwirth
Adrian Frühwirth

Reputation: 45556

I'm not exactly sure why you want to do this using awk, grep suffices. Also, if you know that your input only contains valid numbers matching anything in front of the dot is quite unnecessary.

$ grep '[.]...*0$' input.txt
17.120
17.1230
17.12340
17.000
17.0000

$ grep -P '[.][0-9]{2,}0$' input.txt
17.120
17.1230
17.12340
17.000
17.0000

To fix your pattern just add one more [0-9] class; matching "at least two" means matching exactly two ([0-9][0-9]) plus zero or more [0-9]*):

$ awk '/[.][0-9][0-9][0-9]*0$/' input.txt
17.120
17.1230
17.12340
17.000
17.0000

or use +:

$ awk '/[.][0-9][0-9]+0$/' input.txt
17.120
17.1230
17.12340
17.000
17.0000

Upvotes: 3

fedorqui
fedorqui

Reputation: 289535

Count the number of fields after the .:

$ awk -F. '/0$/ && (split($2,a,"")>2)' file
17.120
17.1230
17.12340
17.000
17.0000
  • check if the line ends with 0.
  • count the number of digits after the dot with split(). If there are more than 2, the check is True.
  • when both conditions are accomplished, the line will be printed.

Upvotes: 1

Avinash Raj
Avinash Raj

Reputation: 174696

Through sed,

$ sed -n '/^[0-9][0-9]\.[0-9][0-9]\+0$/p' file
17.120
17.1230
17.12340
17.000
17.0000

Through awk,

$ awk '/^[0-9][0-9]\.[0-9][0-9]+0$/' file
17.120
17.1230
17.12340
17.000
17.0000

Upvotes: 1

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