根据列条件连接数据框行

Question

For subsequent discussion, I will refer to the example data frame below:

Now, what I wish to achieve is to group all the packet times that are similar - ie all the 7s, 12s, etc. Furthermore, the PacketTime field should contain the difference in min and max ( max(PacketTime) - min(PacketTime) ), and the FrameLen , IPLen and TCPLen fields should be lists of all the values that correspond to the grouped time. For example for the 7s group, FrameLen would contain c(304, 276, 276) .

My solution for the above is as follows:

df <- packets %>%
  group_by(round(PacketTime)) %>%
  summarise(
    PTime=max(PacketTime)-min(PacketTime),
    FLen=list(FrameLen),
    ILen=list(IPLen),
    Movement=0
  ) %>%
  rename(PacketTime=PTime) %>%
  rename(FrameLen=FLen) %>%
  rename(IPLen=ILen)
df$"round(PacketTime)" <- NULL # Remove the group_by

However, some of these crossover (ie 1480s also includes part of 1481s). The part here, which makes this a little easier (in some regard) is that each of the groups are separated by 5s timing window (via Python time.sleep(5) ).

How can I achieve the previous result, but only relying on the 5s difference between the groups that also takes into account the crossover ?

EDIT: As suggested by Ben, here is the dput() of my dataframe df[1:20,] :

structure(list(PacketTime = c(7.083779, 7.147268, 7.147462, 12.084768, 
12.153246, 12.153951, 17.095972, 17.159268, 17.159876, 22.11384, 
22.176926, 22.177467, 27.134427, 27.199108, 27.200064, 32.144442, 
32.208648, 32.20922, 37.144255, 37.205622), FrameLen = c(304L, 
276L, 276L, 304L, 276L, 276L, 304L, 276L, 276L, 304L, 276L, 276L, 
304L, 276L, 276L, 304L, 276L, 276L, 304L, 276L), IPLen = c(300L, 
272L, 272L, 300L, 272L, 272L, 300L, 272L, 272L, 300L, 272L, 272L, 
300L, 272L, 272L, 300L, 272L, 272L, 300L, 272L), TCPLen = c(260L, 
232L, 232L, 260L, 232L, 232L, 260L, 232L, 232L, 260L, 232L, 232L, 
260L, 232L, 232L, 260L, 232L, 232L, 260L, 232L), Movement = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), row.names = c(NA, 
20L), class = "data.frame")

Answer 1

Here is a base R solution using aggregate + transform

u <- aggregate(
    . ~ PacketTime,
    transform(df,
        PTime = ave(PacketTime, trunc(PacketTime), 
        FUN = function(x) diff(range(x))), PacketTime = trunc(PacketTime)
    ),
    c
)
dfout <- transform(u, PTime = sapply(PTime, unique))

which gives

> dfout
  PacketTime      FrameLen         IPLen        TCPLen Movement    PTime
1          7 304, 276, 276 300, 272, 272 260, 232, 232  0, 0, 0 0.063683
2         12 304, 276, 276 300, 272, 272 260, 232, 232  0, 0, 0 0.069183
3         17 304, 276, 276 300, 272, 272 260, 232, 232  0, 0, 0 0.063904
4         22 304, 276, 276 300, 272, 272 260, 232, 232  0, 0, 0 0.063627
5         27 304, 276, 276 300, 272, 272 260, 232, 232  0, 0, 0 0.065637
6         32 304, 276, 276 300, 272, 272 260, 232, 232  0, 0, 0 0.064778
7         37      304, 276      300, 272      260, 232     0, 0 0.061367

Answer 2

One approach is to use seq and cut . Create a sequence from your minimum to maximum times, every 5 seconds. Then, use cut to put your times in intervals. You can use the interval for the labels, for example: (7-12 sec) by omitting the labels argument. Or just use the lower time of the interval (7 sec) as done below.

library(tidyverse)

my_breaks <- seq(trunc(min(packets$PacketTime)), max(packets$PacketTime) + 5, 5)
packets$Interval <- cut(packets$PacketTime, breaks = my_breaks, labels = my_breaks[-length(my_breaks)], right = FALSE)

packets %>%
  group_by(Interval) %>%
  summarise(
    PTime=max(PacketTime)-min(PacketTime),
    FLen=list(FrameLen),
    ILen=list(IPLen),
    Movement=0
  ) %>%
  rename(PacketTime=PTime) %>%
  rename(FrameLen=FLen) %>%
  rename(IPLen=ILen)

Output

# A tibble: 7 x 5
  Interval PacketTime FrameLen  IPLen     Movement
  <fct>         <dbl> <list>    <list>       <dbl>
1 7            0.0637 <int [3]> <int [3]>        0
2 12           0.0692 <int [3]> <int [3]>        0
3 17           0.0639 <int [3]> <int [3]>        0
4 22           0.0636 <int [3]> <int [3]>        0
5 27           0.0656 <int [3]> <int [3]>        0
6 32           0.0648 <int [3]> <int [3]>        0
7 37           0.0614 <int [2]> <int [2]>        0