The following sed command assumes that there is exactly one == in the line and extracts the parts before and after it as groups 1 and 2 which can be used in the substitution.
sed 's/\(.*\)==\(.*\)/\1>=\2,<\2+1.0/'
With the input
Something==x.y.z
argcomplete==1.12.3
youtube-dl==2021.6.6
systemd-python==234
the output is
Something>=x.y.z,<x.y.z+1.0
argcomplete>=1.12.3,<1.12.3+1.0
youtube-dl>=2021.6.6,<2021.6.6+1.0
systemd-python>=234,<234+1.0
To edit the file in-place, add option -i and the name of the input file:
sed -i 's/\(.*\)==\(.*\)/\1>=\2,<\2+1.0/' example.txt
Explanation:
Pattern:
. = any character* = the preceding pattern repeated from 0 to any number- -->
.* = any number of any characters
\(...\) = capture the text matching the enclosed pattern== = literal text\(.*\)==\(.*\) = any text captured as group 1 followed by == and any text captured as group 2
Replacement:
\1, \2 = text from capture group 1 or 2- other parts are literal text here
\1>=\2,<\2+1.0 = group 1 >= group2 ,< group2 +1.0
As mentioned in they's comment, the first pattern before the literal == can be omitted, resulting in
sed 's/==\(.*\)/>=\1,<\1+1.0/'
The explanation is similar with the difference that sed will only modify the matching part of the line. So the part before == will be preserved, and only one capture group for the part after the == is necessary.
A difference in the behavior of the two patterns is that .*==... will match the last == while ==... will match the first one because the .* part matches the longest possible text that is followed by ==.
