Here is a data frame
# 5 companies observed each day for 10 days
df <- tibble(
company = rep(LETTERS[1:5], 10),
value = rep(sample(100, 5), 10),
date = rep(seq(as.Date("2020-01-01"), as.Date("2020-01-10"), 1), each = 5)
)
df
Now something happens to the data and some of the company E rows are removed.
df_error <- df[-c(5, 10, 15, 20), ]
df_error
What is the simplest Tidyverse way to add back the E rows. Value doesn't matter. The date of the E row is the same as the D row above it.
I started with the following and wasn't sure how to proceed:
# Find all D occurrences
e_idx <- which(df_error$company == "D")
e_idx
# If there is not an E in the next row, get the index. These need E rows below each index value.
rows_need_e_below <- ifelse(df_error[e_idx + 1, 1] != "E", e_idx, NA)
rows_need_e_below
If you know that your data should have "A" to "E" companies you can use complete :
tidyr::complete(df_error, date, company = LETTERS[1:5])
Or more generally :
unique_company <- c('A', 'B', 'C', 'D', 'E')
tidyr::complete(df_error, date, company = unique_company)
# A tibble: 50 x 3
# date company value
# <date> <chr> <int>
# 1 2020-01-01 A 87
# 2 2020-01-01 B 5
# 3 2020-01-01 C 40
# 4 2020-01-01 D 67
# 5 2020-01-01 E NA
# 6 2020-01-02 A 87
# 7 2020-01-02 B 5
# 8 2020-01-02 C 40
# 9 2020-01-02 D 67
#10 2020-01-02 E NA
# … with 40 more rows
The value column is by default given NA value. If you want to fill it with specific value you can use fill parameter of complete. For example, to fill with 0's you can do :
tidyr::complete(df_error, date, company = unique_company, fill = list(value = 0))