So the text is the following:

1a fost odata
2un balaur
care fura
mere de aur

and after using this command:

sed 's/\([a-z]*\)\(.*\)\( [a-z]*\)/\1 ... \2/' filename

the result is this:

... 1a fost
 ... 2un balaur
care ...
mere ...  de

I know that \1 is for the first [a-z]* subexpression and so on, but I just can't figure this out.. also, what's the difference between the first subexpression and the last one? why is there a space before [a-z]?

CodePudding user response：

The first [a-z]* matches the first sequence of letters on the line. The * quantifier matches 0 or more repetitions, so this can also match an empty string.

On the first line it matches the empty string before 1a. On the second line it matches the empty string before 2un. On the third line it matches care, and on the fourth line it matches mere. These matches will go into capture group 1.

.* matches zero or more of any characters, so this will skip over everything in the middle of the line. These matches go into capture group 2.

[a-z]* matches a space followed by zero or more letters. The space is needed to make .* stop matching when it gets to the last space on the line. These matches go into capture group 3.

The replacement is capture groups 1 and 2 with ... between them. This is the letters at the beginning of the line, ..., then everything after that except the last word.