How to match a decimal letter and blank in vim?

Question

Hi Guys,

I need to change

1  A
2  B
3  C
4  D

to

A
B
C
D

which means the decimal letter at the begining of every line and the following one or more blank should be deleted .

I'm only familiar with Reqex in Perl, so I try to use :%s/^\d\s+// to solve my problem, but it does not work. so does anyone of you can tell me how to get the work done using vim ?

thanks.

Answer 1

Use the following

:%s/^[0-9]  *//

Answer 2

Vim needs a backslash for +, so try

:%s/^\d\s\+//

Answer 3

One way is to use the global search and replace command:

:g/^[0-9]  */s//

It searches for the sequence:

• start of line ^
• a digit [0-9]
• a space <space>
• zero or more spaces <space>*

and then substitutes it for nothing (s//).

Answer 4

Escape the plus sign:

:%s/^\d\s\+//

Answer 5

You should use instead

:%s/^\d\s\+//

Being a text editor, vim tends to treat more characters literally‒as text‒when matching a pattern than perl. In the default mode, + matches literal +.

Of course, this is configurable. Try

:%s/\v^\d\s+//

and read the help file.

:help magic

Answer 6

You can also use the Visual Block mode (Ctrl+V), then move down and to the right to highlight a block of characters and use 'x' to remove them. Depending on the layout of the line, that may in fact be quicker (and easier to remember).

Answer 7

If you still want to use Perl for this, you can:

:%!perl -pe 's/^\d\s+//'

Vim will write the file to a temporary file, run the given Perl script on it, and reload the file into the edit buffer.

Answer 8

If it's in a column like that you could go into the column visual mode by pressing:

esc ctrl+q

then you can highlight what you want to delete