结合awk和sed以匹配行并替换字符

Question

I'm trying to match first letter, in this example "B", and second column "2". When match is found replace characters [38-41] with white spaces.

Here is the data I'm trying to modify:

A1234A123 1 2 12345.12345 1234.1234.112341234

B1234A123 2 2 12345.12345 1234.1234.112341234

A1234A123 2 2 12345.12345 1234.1234.112341234

I can match the conditions with awk using:

awk '/^B/ && $2=="2" {print}' 

and I can modify the lines with sed using:

sed -r 's/^(.{37})(.{4})/\1    /' 

I'm trying to find the lines in the file which contains the two conditions and then modify the characters, while still printing the entire line of lines that don't match. Can you combine the two commands in order to introduce some sort of if/then statement?

I've tried to combine the commands, but it edited all of the lines:

awk '/^B/ && $2=="2" {print}' ¦ sed -r 's/^(.{37})(.{4})/\1    /' data

Resulting data should look like this:

A1234A123 1 2 12345.12345 1234.1234.112341234

B1234A123 2 2 12345.12345 1234.1234.1    1234

A1234A123 2 2 12345.12345 1234.1234.112341234

Thanks in advance.

Answer 1

You can use single awk to combine both commands:

awk '/^B/ && $2=="2"{$0=substr($0, 1, 37) "    " substr($0, 38, 4)} 1' file
A1234A123 1 2 12345.12345 1234.1234.112341234
B1234A123 2 2 12345.12345 1234.1234.1    1234
A1234A123 2 2 12345.12345 1234.1234.112341234

Answer 2

您可以通过在正则表达式前添加sed来指示仅替换匹配行（ /^B[^ ]* 2/ ）：

sed -r '/^B[^\s]*\s2\s/s/^(.{37}).{4}/\1    /' data

Answer 3

使用GNU awk：

gawk '/^B/ && $2=="2" {print gensub(/(.{37}).{4}/,"\\1    ","")}' data

Answer 4

In Gnu Awk version 4 you could try:

gawk 'BEGIN { FIELDWIDTHS = "1 9 1 26 4 20"; OFS="" }
$1=="B" && $3=="2" {
    $5="    "
} 1' file

with output:

A1234A123 1 2 12345.12345 1234.1234.112341234
B1234A123 2 2 12345.12345 1234.1234.1    1234
A1234A123 2 2 12345.12345 1234.1234.112341234