VI - how to replace arbitary column(s)

I've used vi and unix for years but I've never found a good way to do this.

I want to search and replace based on column/character number

the search part I can do by

/^......<correct number of .>...X/

given its an 'X' in the column I want but I can't do

:1,$s/^.............................X/.....................Y/

unless I want all .'s in the file
LVL 19
Nick UpsonPrincipal Operations EngineerAsked:
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tfewsterCommented:
The way I understand this is:
Look for X in the 5th column;  "Remember" the characters in the columns 1-4 (4 .'s) in a sed variable.
Subsitute the "pattern" found with \1Y, where \1 is the contents of the variable
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tfewsterCommented:
:1,$s/^\(....\)X/\1Y/
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