问题
I have a file which looks like below.
1 1 0 # 1
6 1 0 # 2
8 1 0 # 3
10 1 0 # 4
12 1 0 # 6
How can I add .0 to all numbers, except the numbers behind the #. I think this should not be too difficult to do with a regular expression, but my regex knowledge is too rusty..
回答1:
If your numbers after the # don't have spaces after them, you can use:
:g/\([0-9]\+\) /s//\1.0 /g
The use of () creates groups which you can refer to as \D in the replacement text, where D is the position of the group within the search string. This will give you:
1.0 1.0 0.0 # 1
6.0 1.0 0.0 # 2
8.0 1.0 0.0 # 3
10.0 1.0 0.0 # 4
12.0 1.0 0.0 # 6
If they do have spaces after them (which yours seem to have), you'll get:
1.0 1.0 0.0 # 1.0
6.0 1.0 0.0 # 2.0
8.0 1.0 0.0 # 3.0
10.0 1.0 0.0 # 4.0
12.0 1.0 0.0 # 6.0
in which case you can then do:
:g/\.0\( *\)$/s//\1/g
to fix it up.
回答2:
With VIM:
:%s/\v(#.*)@<!\d+/&.0/g
Explanation: \v = very magic (see help \v), @<! Matches with zero width if the preceding atom does NOT match just before what follows (see help \@<!). The rest of the pattern replaces strings of 1 or more digits with the same string followed by .0.
回答3:
Assuming it's well formed, using sed this should work.
EDIT: Just to clarify, vim was not attached as a tag to this questions when i saw it.
sed 's/\([0-9]\+\) \+\([0-9]\+\) \+\([0-9]\+\)/\1.0 \2.0 \3.0/' file
回答4:
you can try this also sed 's/[0-9]/&.0/g' | sed 's/.0[ ]*$//g'
First sed add .0 all numbers and second one removes the trailing .0
来源：https://stackoverflow.com/questions/3332973/regex-find-and-replace-in-vim-adding-0-to-numbers