UNIX for Dummies Questions & Answers

I have a file with numbers like:

2235900
2234500
2220000
22300
235700
2200
0

and I need a search and replace to
change the number to a 3 place decimal...
like this :

2235.900
2234.500
2220.000
22.300
235.700
2.200
.000

Can someone please help me with the command
I need to use in the VI editor to get these results
of a file that has 3724 lines of numbers I need
to make into 3 place decimal numbers?

Thanks so much.

I dont think you can do this in VI. Unless someone who knows vi macros can knock something together .... if you only have 1 number per line and you dont mind 0.000 instead of .000 then the following will work:

perl -e 'while($line=<>) { printf ("%.3f\n",($line/1000));}' inputfile.txt > outputfile.txt

Or if you had multiple numbers on each line then maybe:

perl -e 'while($line=<>){$line =~ s{(\d+)}{sprintf("%.3f",$1/1000)}eg; print "$line";}' inputfile.txt > outputfile.txt

Unbeliever, Thanks for the reply. I will need to
get Perl loaded and give this a try.... running
windoz ... would really like to get back to the REAL OS but have too many apps that only run under windoz ....

Thanks,

Code:

sed -e 's/\(.*\)\([0-9]\{3\}\)/\1.\2/;ta' -e 's/\(.*\)\([0-9]\{2\}\)/\1.0\2/;ta' -e 's/\(.*\)\([0-9]\{1\}\)/\1.00\2/;ta' -e ':a' fileName

Unbeliever, I am getting a
" > was unexpected at this time. " when I try this command at a command prompt.

Any suggestions?

Unbeliever's code works perfectly for me. *shrug*

... although his SECOND sample has a missing single-quote....
lemme add it in.