How to use sed to delete last several character of a pattern - bash

Question

I've gone through all of the threads but still cannot find the answer.
For example.
I have a timestamp of format: yyyy-mm-dd hh:mm:ss.xxx
where xxx indicates the milliseconds.
I want to get rid of the xxx part, notice that this timestamp is not in certain position so we cannot take it as a part in end of line or start of line.(in unix command or in bash script)
The method I can think of is to use sed, but all i can do is to get the pattern, but don't know what to do to process the pattern, it seems that all pattern does is to locate the lines instead of the pattern itself. So generally we can think of the question like: how to use sed to delete last several letters of a certain pattern.
Thanks for reading.
Note that xxx can be 0-999, so it can be 1,2,3 digits, sample is like:
asfd,asasfsf,afas,2017-10-20 13:22:22.0,333,222,0.002
nyh,nyhny,nhy,2 23 4 23 32:23:14.czxv,2017-10-20 13:22:22.234,12.0,234.22
nyh,nyhny,nhy,2017-10-20 13:22:22.234,12.0
wn,rrwn,daff,2017-10-20 13:22:32.543,12,32
What I expect is:
asfd,asasfsf,afas,2017-10-20 13:22:22,333,222,0.002
nyh,nyhny,nhy,2 23 4 23 32:23:14.czxv,2017-10-20 13:22:22,12.0,234.22
nyh,nyhny,nhy,2017-10-20 13:22:22,12.0
wn,rrwn,daff,2017-10-20 13:22:32,12,32

Answer 1

As per OP's shown Input_file proposing the new following solution.
awk '{sub(/\.[^,]*/,"",$2)} 1' Input_file
Explanation: Adding explanation of awk code also here.
awk '{
sub(/\.[^,]*/,"",$2) ##sub is awk in-built utility, which will substitute on basis of sub(text/regex which we need to replace,"new_text"/variable_value,For a current line/variable/field), so in this case I am using a REGEX which will look from a DOT to first occurrence of comma(,) which I am substituting with NULL in 2nd field(your 2nd field is the one which is having timing details because awk has space as delimiter by default).
}
1 ##awk works on method of condition then action. So Here I am making condition TRUE by mentioning 1 and no action is mentioned so be default print action will happen.
' Input_file

Answer 2

This might work for you (GNU sed):
sed 's/\(....-..-.. ..:..:..\)\..../\1/g' file
This is very lazy but most likely will work 99% of the time. It matches on the time stamp separators and then removes the .xxx at the end. If you want, you can be more specific i.e.
sed 's/\([0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\} [0-9]\{2\}:[0-9]\{2\}:[0-9]\{2\}\)\.[0-9]\{3\}/\1/g' file
Using the -r option, removes the toothpick mess:
sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2})\.[0-9]{3}/\1/g' file