После группировки по «Дате», «Типу», «Идентификатору» получите sum
значения «Сумма» и затем получите distinct
строк
library(dplyr)
df1 %>%
group_by(Date, Type, ID) %>%
mutate(Amount = sum(Amount)) %>%
select(-Hour) %>%
ungroup %>%
distinct(Date, Type, ID, .keep_all = TRUE)
# A tibble: 5 x 8
# Date Type ID Amount Remaining Price1 Price2 Sell
# <chr> <chr> <chr> <int> <chr> <chr> <chr> <chr>
#1 01/01/2019 Devolution D 16 - 2,8 2,8 3,19
#2 01/01/2019 Correction D 354 0 2,8 2,8 3,19
#3 01/01/2019 Client L 1 - 2,8 2,8 3,19
#4 02/01/2019 Enter D 46 - 2,2 - 4,29
#5 02/01/2019 Selling L 5 -99 2,0 2,5 3,59
данных
df1 <- structure(list(Date = c("01/01/2019", "01/01/2019", "01/01/2019",
"01/01/2019", "02/01/2019", "02/01/2019", "02/01/2019", "02/01/2019",
"02/01/2019"), Hour = c("12:03:50", "12:03:50", "13:00:01", "18:30:20",
"02:00:30", "08:38:18", "19:00:20", "22:12:00", "23:47:52"),
Type = c("Devolution", "Devolution", "Correction", "Client",
"Enter", "Enter", "Selling", "Selling", "Selling"), ID = c("D",
"D", "D", "L", "D", "D", "L", "L", "L"), Amount = c(6L, 10L,
354L, 1L, 36L, 10L, 1L, 3L, 1L), Remaining = c("-", "-",
"0", "-", "-", "-", "-99", "-", "-"), Price1 = c("2,8", "2,8",
"2,8", "2,8", "2,2", "2,2", "2,0", "2,0", "2,0"), Price2 = c("2,8",
"2,8", "2,8", "2,8", "-", "-", "2,5", "2,5", "2,5"), Sell = c("3,19",
"3,19", "3,19", "3,19", "4,29", "4,29", "3,59", "3,59", "3,59"
)), class = "data.frame", row.names = c(NA, -9L))