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On February 20, 2014 6:30:37 PM PST, umair durrani <umairdurrani at outlook.com> wrote:
>I have a data frame of vehicle trajectories. Here's a snapshot:
>>dput(head(df))structure(list(vehicle = c(2L, 2L, 2L, 2L, 2L, 2L),
>frame = 43:48, globalx = c(6451214.156, 6451216.824, 6451219.616,
>6451222.548, 6451225.462, 6451228.376), class = c(2L, 2L, 2L, 2L,
>2L, 2L), velocity = c(37.76, 37.9, 38.05, 38.18, 38.32, 38.44 ),
>lane = c(2L, 2L, 2L, 2L, 2L, 2L)), .Names = c("vehicle", "frame",
>"globalx", "class", "velocity", "lane"), row.names = c(NA, 6L), class =
>"data.frame")
>where, vehicle= vehicle id, frame= frame id of time frames in which it
>was observed, globalx = x coordinate of the front center of the
>vehicle, class=type of vehicle (1=motorcycle, 2=car, 3=truck),
>velocity=speed of vehicles in feet per second, lane= lane number (there
>are 6 lanes).The 'frame' represents one tenth of a second i.e. one
>frame is 0.1 seconds long. At frame 't' the vehicle has globalx
>coordinate x(t) and at frame 't-1' (0.1 seconds before) it was x(t-1).
>If the reference location has globalx coordinate=6451179.1116 then I
>simply want a new column in df called 'u' which has 'yes' in the row
>where globalx of the vehicle was greater than reference coordinate at
>'U' AND the previous consecutive globalx coordinate of this vehicle was
>less than reference coordinate at 'U'(i.e. reference coordinate is
>between the 2 locations of vehicle in two consecutive frames). This
>means that if df has 100 vehicles then there will be 100 'yes' in 'u'
>column because every vehicle wil!
>
>
>l meet the above criteria only once. I have tried to do this by running
>the function with ifelse and also tried to do the same using a for loop
>but it doesn't work for me.
>
>
>
> [[alternative HTML version deleted]]
>
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