How can I generate a list of sequential numbers, one per line?

Question

Starting from a blank slate, how can I obtain a document that contains

1
2
3
4
5
6
7
8
9
10
…
100

To be clear, I don't want these numbers displayed in the margin; I want them inserted into the document itself.

Answer 1

Use :put and range():

:put =range(1,100)

To avoid the blank line at the top (kudos to romainl), use :0put:

:0put =range(1,100)

Answer 2

In addition to Undo's pure-vim :put =range(1,100) (which actually leaves you with a blank line up top), you can, depending on your OS, use one of its commands. E.g., on a Unix/Linux box:

%!seq 1 100

The above works by piping the entire (empty) buffer to seq, which ignores its input and just outputs the numbers 1 to 100. Vim then replaces the entire buffer with seq's output.

That's useful when you're already familiar with some command-line way to get what you want.

Answer 3

For the record, and definitely not the shortest way (see @Undo's awesome solution), but sequence of keystrokes will do it too:

i1EscqaYpCtrl+aq98@a

Let me break that down for you:

1. i1<Esc> -- insert the number 1, then get back out to command mode
2. qa -- start recording a macro in register "a"
3. Y -- copy the current line
4. p -- paste the current line (cursor will also move to the pasted line)
5. <Ctrl>a -- increment the next number on the line, making it 2
6. q -- stop recording the macro (was register "a")
7. 98@a -- replay the macro in register "a" 98 times

Answer 4

Here is a different approach, that needs a newer Vim (something like 7.4.800)

This assumes an empty buffer and '1' in register a. First enter 100 1 into your buffer, "a100P.

Then visually select lines 2 till 100 :2EnterVG.

Now press gCtrl+A.

Read the help at :h v_g_CTRL-A

Answer 5

I know this question is really old, but there's another way you can do it also. Try this:

99o<esc>:%s/^/\=line('.')<cr>

99o<esc> will just open up 100 blank lines. Then, we run a substitute command:

:%                  " On every line
  s/                " substitute
    ^/              " the start of the line. (This will always match no matter what)
      \=            " Evaluate:
        line('.')   " The current line.