How to extract only values greater than a threshold from a file?

I have this file:

names average
john:15.02
Mark:09.63
James:12.58

I want to extract only the averages greater than 10 from it, so the output in this example should be:

15.02
12.58

3 Answers

With awk

awk -F: '{if($2>10)print$2}' <filename

Explanations

• -F: – sets the Field separator to :
• {if($2>10)print$2} – for each line, test whether the 2nd field is >10, if so print it
• <filename – let the shell open file filename, that's better than letting awk do that, see Stéphane Chazelas' answer on the topic

Example run

$ <filename awk -F: '{if($2>10)print$2}'
15.02
12.58

It's also possible to add spaces and put the pattern outside the brackets, so these are equal – thanks to Stefan for pointing that out:

awk -F: '{if($2>10)print$2}' <filename
awk -F: '{ if ( $2 > 10 ) print $2 }' <filename
awk -F: '$2>10{print$2}' <filename
awk -F: '$2 > 10 { print $2 }' <filename

4

With grep you'd have to work with regular expressions; e.g.

grep -E ':[^0-9]*[1-9][0-9][0-9]*\.' file | cut -d':' -f2

as with sed:

sed -n 's/.*:[^0-9]*\([1-9][0-9][0-9]*\..*\)/\1/p' file

But using RegEx on ordered data is error prone (in my experience) and difficult to read ;-).

2

Im posting also this solution

sed '1d' File|cut -d: -f2 |awk '$0>10' 

sed '1d' = removes the header

cut -d: -f2 = keeps values after the delimiter (: in this case)

awk '$0>10' = prints all values greater than 10

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