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    <p>One solution is to first split the intervals, so that they are sometimes equal but never partially overlap, and them remove the duplicates. The problem is that we are left with many small abutting intervals, and merging them does not look straightforward.</p> <pre><code>library(reshape2) library(sqldf) d$machine &lt;- as.character( d$machine ) # Duplicated levels... ddply( d, c("username", "machine"), function (u) { # For each username and machine, # compute all the possible non-overlapping intervals intervals &lt;- sort(unique( c(u$start, u$end) )) intervals &lt;- data.frame( start = intervals[-length(intervals)], end = intervals[-1] ) # Only retain those actually in the data u &lt;- sqldf( " SELECT DISTINCT u.username, u.machine, intervals.start, intervals.end FROM u, intervals WHERE u.start &lt;= intervals.start AND intervals.end &lt;= u.end " ) # We have non-overlapping, but potentially abutting intervals: # ideally, we should merge them, but I do not see an easy # way to do so. u } ) </code></pre> <p><strong>EDIT:</strong> Another, conceptually cleaner solution, that fixes the non-merged abutting intervals problem, is to count the number of open sessions for each user and machine: when it stops being zero, the user has logged in (with one or more session), when it drops to zero, the user has closed all his/her sessions.</p> <pre><code>ddply( d, c("username", "machine"), function (u) { a &lt;- rbind( data.frame( time = min(u$start) - 1, sessions = 0 ), data.frame( time = u$start, sessions = 1 ), data.frame( time = u$end, sessions = -1 ) ) a &lt;- a[ order(a$time), ] a$sessions &lt;- cumsum(a$sessions) a$previous &lt;- c( 0, a$sessions[ - nrow(a) ] ) a &lt;- a[ a$previous == 0 &amp; a$sessions &gt; 0 | a$previous &gt; 0 &amp; a$sessions == 0, ] a$previous_time &lt;- a$time a$previous_time[-1] &lt;- a$time[ -nrow(a) ] a &lt;- a[ a$previous &gt; 0 &amp; a$sessions == 0, ] a &lt;- data.frame( username = u$username[1], machine = u$machine[1], start = a$previous_time, end = a$time ) a } ) </code></pre>
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    1. PORemove rows from data: overlapping time intervals?
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    1. USVincent Zoonekynd
    1. USVincent Zoonekynd
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    1. COThanks for your contribution. This seems to add rows, not remove them. This seems to be quite difficult task to solve.
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      1. USjrara
    2. COIf a user has two sessions on the same machine, say from 08:00 to 10:00 and from 09:00 to 12:00, those overlapping intervals are split into 08:00 to 09:00, 09:00 to 10:00 and 10:00 to 12:00: there are no more overlapping intervals, there are no duplicates, but since the intervals have been cut into smaller pieces, there are more of them. Merging them when they abut (here, into a single interval 09:00 to 12:00) looks harder.
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      1. USVincent Zoonekynd
    3. COI'm not sure if I understand what you mean. My idea was to remove thos rows (except one of them) for user from the same computer which have overlapping interval. E.g if I have intervals 8.30 to 9:30 and 8:45 to 10:00, the 8:45 to 10:00 should be removed because it has one overlapping interval row in the data.
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