如何从awk或sed的行中提取单引号中的数字？

Question

I have this line, tab delimited:

chr1    11460   11462   '16/38' 421     +       chr1    11460   11462   '21/29' 724     +       2
chr1    11479   11481   '11/29' 379     +       chr1    11479   11481   '20/5' 667     +       2

What I want to do is to test if all the second number inside ' ' are greater or equal to 10. If so, I'll output this line. So the result should be to print the first line

chr1    11460   11462   '16/38' 421     +       chr1    11460   11462   '21/29' 724     +       2

I can write a perl code to do it. But this seems to be something awk can do easily.. anyone has a solution?

Thanks.

Answer 1

如果设置正确的字段分隔符，则非常简单：

awk -F "['/]" '{for (i=3; i<=NF; i+=3) if ($i<10) next; print}' file

Answer 2

Easiest way fetch the content inside single quotes might be just to strip off everything from both ends of each line, up to and including the single quote:

$ sed "s/^[^']*'//;s/'.*//" file
16/38
11/29

This sed expression consists of two commands:

• s/^[^']*'// -- strips off all text to the first single quote,
• s/'.*// -- strips off all text from the first (remaining) single quote to EOL.

To wrap this in a shell script that does something with that data requires .. well, a shell script...

You can parse this stuff using bash's read command. For example:

#!/bin/bash
IFS=/
sed "s/^[^']*'//;s/'.*//" file \
| while read left right; do
  echo "$left / $right"
done

To implement something that grabs contents of multiple single-quoted numbers, you can expand the sed script appropriately, and implement if statements for the conditions you want. For example, a sed expression to grab the TWO single-quoted strings might be:

sed "s/^[^']*'\([^']*\)'[^']*'\([^']*\)'.*/\1 \2/"

This is a single large regex that uses two sets of brackets \\( and \\) , to mark patterns that will be placed in the output, \\1 and \\2 .

But you might be better off parsing things according to column position:

$ while read _ _ _ A _ _ _ _ _ B _; do echo "$A .. $B"; done < file
'16/38' .. '21/29'
'11/29' .. '20/5'

Actually implementing your programming logic is left as an exercise to the reader. If you'd like us to help you with your script, please include your work so far.

Answer 3

As long as those are the only ' characters in the string and the numbers won't have leading zeros you could use the regular expression:

\d\d+'.*\d\d+'

If either of those preconditions isn't true there are changes that could be made, but it would depend on the situation.

You should be able to use grep to get the lines you want using that regex. The following puts just the first line to stdout:

grep \d\d+'.*\d\d+' "chr1    11460   11462   '16/38' 421     +       chr1    11460   11462   '21/29' 724     +       2
chr1    11479   11481   '11/29' 379     +       chr1    11479   11481   '20/5' 667     +       2"

Answer 4

My version, serious overkill but should work with any amount of 'xx/xx' per line:

awk -F'\t' "{
    found=1;
    for(i=0;i<NF;i++){
        if(match(\$i, /'[[:digit:]]+\/([[:digit:]]+)'/, capts)){
            if(capts[1] < 10){
                found=0;
                break;
            }
        }
    }
    if(found){
        print;
    }
}" file.txt

Explanation:

This will loop through each field of the line and apply a regex against the field to find the last digits of 'xx/xx'. If the last digits are less than 10 it will break out of the loop and go to the next line. If all fields have been processed by the if loop and no last digits were less than 10, it will print the line.

Note:

Seeing that i'm using the match function to capture regex groups this will only work with GNU awk.