CODEWITHSUNDEEP
regexlinuxbashunixgrep
arnpry
arnpry

Reputation: 1141

Only Print Lines that End with Certain Numerical Values

I have a CSV file made up of a space delimiter and two columns.

I need to grep any lines that end with either 06, 12, 18, or 00 in the first column.

file.txt

2017121106 22.9
2017121109 19.4
2017121112 17.2
2017121115 9999.0
2017121118 9999.0
2017121121 9999.0
2017121200 9999.0
2017121203 9999.0
2017121206 16.3
2017121209 13.1
2017121212 8.8
2017121215 8.1
2017121218 10.5
2017121221 8.6

Attempted Code:

egrep '(00|06|12|18)$' file.txt

Expected Output:

2017121106 22.9
2017121112 17.2
2017121118 9999.0
2017121200 9999.0
2017121206 16.3
2017121212 8.8
2017121218 10.5

I am getting an empty return when running this code in the terminal.

What am I doing wrong?

Upvotes: 3

Views: 75

Answers (4)

karakfa
karakfa

Reputation: 67557

awk to the rescue!

awk '!(substr($1,length($1)-1)%6)' file

will give

2017121106 22.9
2017121112 17.2
2017121118 9999.0
2017121200 9999.0
2017121206 16.3
2017121212 8.8
2017121218 10.5

you're looking the the multiples of 6 in the last two digits, translates into awk as print line when remainder by dividing to 6 is zero

This solution works due to the domain of the data its 24h representation, so it won't have false positives due to other multiples of 6.

Upvotes: 3

123
123

Reputation: 11236

Using basic grep

grep '0[06] \|1[28] ' file

2017121106 22.9
2017121112 17.2
2017121118 9999.0
2017121200 9999.0
2017121206 16.3
2017121212 8.8
2017121218 10.5

Upvotes: 2

ghoti
ghoti

Reputation: 46886

So... The grep command doesn't understand "fields". It only understands patterns. Since your "first column" is a pattern of numbers followed by a space, you'd match that instead of using $:

$ egrep '^[0-9]+(00|06|12|18) ' file.txt

Note the space character at the end of the parenthesized expression. The initial [0-9]+ is in place so that we can anchor this regex to the beginning of the line, which makes sure we're matching the first "field".

A better solution might be to use awk, which does understand fields:

$ awk '$1~/(00|06|12|18)$/' file.txt

Upvotes: 4

RavinderSingh13
RavinderSingh13

Reputation: 133750

Following awk code may help you in same.

awk '(substr($1,length($1)-1)+0== 06 || substr($1,length($1)-1)+0 == 12 || substr($1,length($1)-1)+0 == 18  || substr($1,length($1)-1)+0 == 00 )'  Input_file

Adding a non-one liner form of solution too now.

awk '
(substr($1,length($1)-1)+0== 06 ||\
substr($1,length($1)-1)+0 == 12 ||\
substr($1,length($1)-1)+0 == 18 ||\
substr($1,length($1)-1)+0 == 00 )
'   Input_file

Upvotes: 1

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