Bash - 仅使用awk打印矩阵的某些部分

Question

I want to read a matrix of numbers

1 3 4 5 
2 4 9 0

And only want my awk statement to print out the first and last, so 1 and 0. I have this so far, but nothing will print. What is wrong with my logic?

    awk 'BEGIN {for(i=1;i<NF;i++)
        if(i==1)printf("%d ", $i);
        else if(i==NF && i==NR)printf("%d ", $i);}'

Answer 1

$ awk '{ if (NR==1) { print $1}} END{print $NF}' matrix
1
0

The above awk program has two parts. The first is:

{ if (NR==1) { print $1}}

This prints the first field (column) of the first record (line) of the file.

The second part is:

END{print $NF}

This parts runs only at the end after the last record (line) has been read. It prints the last field (column) of that line.

Answer 2

awk 'NR==1{print $1;} END{print $NF;}'

Answer 3

Borrowing from unix.com , you can use the following:

awk 'NR == 1 {print $1} END { print $NF }'

This will print the first column of the first line (NR == 1) and end input has finished (END), print the final column of the last line.

If I understand the output format you're looking for, this code should capture those values and print them:

awk 'NR == 1 {F = $1} END { L = $NF ; printf("%d %d", F, L) }'

Answer 4

awk is line based, NR is the current record (line) number. and awk is essentially match => action,

echo "1 3 4 5
2 4 9 0" |
awk 'NR == 1 {print $1;}
    END {print $NF;}'

for the first record print the first field; for the last record print the last field.

Answer 5

由于awk有这么多解决方案，这里是sed的另一种方式。

sed -r ':a;$!{N;ba};s/\s+.*\s+/ /' file

Answer 6

另一种sed变种：

$ echo $'1 3 4 5\n2 4 9 0' | sed -n '1s/ .*//p;$s/.* //p'