Group Data in R for consecutive rows

Question

If there\'s not a quick 1-3 liner for this in R, I\'ll definitely just use linux sort and a short python program using groupby, so don\'t bend over

Answer 1

In dplyr, I would do this by creating another grouping variable for the consecutive rows. This is what the code cumsum(c(1, diff(weight) != 0) is doing in the code chunk below. An example of this is also here.

The group creation can be done within group_by, and then you can proceed accordingly with making any summaries by group.

library(dplyr)

df_in %>%
    group_by(ID, group_weight = cumsum(c(1, diff(weight) != 0)), weight) %>%
    summarise(start_day = min(start_day), end_day = max(end_day))

Source: local data frame [5 x 5]
Groups: ID, group_weight [?]

     ID group_weight weight start_day end_day
  (dbl)        (dbl)  (dbl)     (dbl)   (dbl)
1     1            1    150         1       7
2     1            2    151         7      10
3     1            3    150        10      30
4     2            4    170         5      20
5     2            5    171        20      30

This approach does leave you with the extra grouping variable in the dataset, which can be removed, if needed, with select(-group_weight) after ungrouping.

Answer 2

First we combine ID and weight. The quick-and-dirty way is using paste:

df_in$id_weight <- paste(df_in$id, df_in$weight, sep='_')
df_in
   ID weight start_day end_day id_weight
1   1    150         1       4     1_150
2   1    150         4       7     1_150
3   1    151         7      10     1_151
4   1    150        10      11     1_150
5   1    150        11      30     1_150
6   2    170         5      10     2_170
7   2    170        10      15     2_170
8   2    170        15      20     2_170
9   2    171        20      25     2_171
10  2    171        25      30     2_171

Safer way is to use interaction or group_indices: Combine values in 4 columns to a single unique value

We can group consecutively using rle.

rlel <- rle(df_in$id_weight)$lengths
df_in$group <- unlist(lapply(1:length(rlel), function(i) rep(i, rlel[i])))
df_in
   ID weight start_day end_day id_weight group
1   1    150         1       4     1_150     1
2   1    150         4       7     1_150     1
3   1    151         7      10     1_151     2
4   1    150        10      11     1_150     3
5   1    150        11      30     1_150     3
6   2    170         5      10     2_170     4
7   2    170        10      15     2_170     4
8   2    170        15      20     2_170     4
9   2    171        20      25     2_171     5
10  2    171        25      30     2_171     5

Now with the convenient group number we can summarize by group.

df_in %>% 
  group_by(group) %>% 
  summarize(id_weight = id_weight[1], 
            start_day = min(start_day), 
            end_day = max(end_day))
# A tibble: 5 x 4
  group id_weight start_day end_day
  <int> <chr>         <dbl>   <dbl>
1     1 1_150             1       7
2     2 1_151             7      10
3     3 1_150            10      30
4     4 2_170             5      20
5     5 2_171            20      30

Answer 3

with(df_in, {
  aggregate(day, list('ID'=ID, 'weight'=weight), 
            function(x) c('min_day' = min(x), 'max_day' = max(x)))
})

Produces:

  ID weight x.min_day x.max_day
1  1    150         1         5
2  1    151         3         3
3  2    170         1         3
4  2    171         4         5