r - Fill missing sequence values with dplyr -

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i have data frame missing values "snap_id". i'd fill in missing values floating point values based on sequence previous non-missing value (lag()?). achieve using dplyr if possible.

assumptions:

  1. there never missing data first or last row i'm generating missing dates based on missing days between min , max of data set
  2. there can multiple gaps in data set

current data:

                  end snap_id 1 2015-06-26 12:59:00     365 2 2015-06-26 13:59:00     366 3 2015-06-27 00:01:00      na 4 2015-06-27 23:00:00      na 5 2015-06-28 00:01:00      na 6 2015-06-28 23:00:00      na 7 2015-06-29 09:00:00     367 8 2015-06-29 09:59:00     368

what want achieve:

                  end snap_id 1 2015-06-26 12:59:00     365.0 2 2015-06-26 13:59:00     366.0 3 2015-06-27 00:01:00     366.1 4 2015-06-27 23:00:00     366.2 5 2015-06-28 00:01:00     366.3 6 2015-06-28 23:00:00     366.4 7 2015-06-29 09:00:00     367.0 8 2015-06-29 09:59:00     368.0

as data frame:

df <- structure(list(end = structure(c(1435323540, 1435327140, 1435363260,      1435446000, 1435449660, 1435532400, 1435568400, 1435571940), tzone = "utc", class = c("posixct",      "posixt")), snap_id = c(365, 366, na, na, na, na, 367, 368)), .names = c("end",      "snap_id"), row.names = c(na, -8l), class = "data.frame")

this attempt @ achieving goal, works first missing value:

df %>%    arrange(end) %>%   mutate(snap_id=ifelse(is.na(snap_id),lag(snap_id)+0.1,snap_id))                    end snap_id 1 2015-06-26 12:59:00   365.0 2 2015-06-26 13:59:00   366.0 3 2015-06-27 00:01:00   366.1 4 2015-06-27 23:00:00      na 5 2015-06-28 00:01:00      na 6 2015-06-28 23:00:00      na 7 2015-06-29 09:00:00   367.0 8 2015-06-29 09:59:00   368.0

the outstanding answer @mathematical.coffee below:

df %>%    arrange(end) %>%   group_by(tmp=cumsum(!is.na(snap_id))) %>%   mutate(snap_id=snap_id[1] + 0.1*(0:(length(snap_id)-1))) %>%   ungroup() %>%   select(-tmp)

edit: new version works number of na runs. 1 doesn't need zoo, either.

first, notice tmp=cumsum(!is.na(snap_id)) groups snap_ids such groups of same tmp consist of 1 non-na value followed run of na values.

then group variable , add .1 first snap_id fill out nas:

df %>%    arrange(end) %>%   group_by(tmp=cumsum(!is.na(snap_id))) %>%   mutate(snap_id=snap_id[1] + 0.1*(0:(length(snap_id)-1)))                    end snap_id tmp 1 2015-06-26 12:59:00   365.0   1 2 2015-06-26 13:59:00   366.0   2 3 2015-06-27 00:01:00   366.1   2 4 2015-06-27 23:00:00   366.2   2 5 2015-06-28 00:01:00   366.3   2 6 2015-06-28 23:00:00   366.4   2 7 2015-06-29 09:00:00   367.0   3 8 2015-06-29 09:59:00   368.0   4

then can drop tmp column afterwards (add %>% select(-tmp) end).

edit: old version doesn't work subsequent runs of nas.

if aim fill each na previous value + 0.1, can use zoo's na.locf (which fills each na previous value), along cumsum(is.na(snap_id))*0.1 add 0.1.

library(zoo) df %>%    arrange(end) %>%   mutate(snap_id=ifelse(is.na(snap_id),                        na.locf(snap_id) + cumsum(is.na(snap_id))*0.1,                        snap_id))

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