如何在 data.frame 中创建一个新列，以便该列计算该 data.frame 中不同行的数量？

Question

I have a huge data.frame like this.

First, how can I add a new column "date1" into this data.frame so that column counts the number of the UNIQUE different day in this data.frame then arranged in ascending order in that newly created column.

Second, how can I add another column "date2" into this data.frame so that column counts total different id in a day?

    year  month day id
    2011    1   5   31
    2011    1   14  22
    2011    2   6   28
    2011    2   17  41
    2011    3   9   55
    2011    1   5   34
    2011    1   14  25
    2011    2   6   36
    2011    2   17  11
    2011    3   12  10

The result I expect looks like this. Please help!

    year month day  id date1 date2
    2011    1   5   31  1     2
    2011    1   14  22  2     2
    2011    2   6   28  3     2
    2011    2   17  41  4     2
    2011    3   9   55  5     1
    2011    1   5   34  1     2
    2011    1   14  25  2     2
    2011    2   6   36  3     2
    2011    2   17  11  4     2
    2011    3   12  10  6     1

Answer 1

We can first combine year , month and day into one column using unite and give a unique number to each group of that combination, then group_by same combination and count the unique id for each combination using n_distinct .

library(dplyr)
library(tidyr)

df %>%
  unite(date, year, month, day, sep = "-", remove = FALSE) %>%
  mutate(date1 = as.integer(factor(date,level = unique(date)))) %>%
  group_by(date) %>%
  mutate(date2 = n_distinct(id)) %>%
  ungroup() %>%
  select(-date)


#    year month   day    id date1 date2
#   <int> <int> <int> <int> <int> <int>
# 1  2011     1     5    31     1     2
# 2  2011     1    14    22     2     2
# 3  2011     2     6    28     3     2
# 4  2011     2    17    41     4     2
# 5  2011     3     9    55     5     1
# 6  2011     1     5    34     1     2
# 7  2011     1    14    25     2     2
# 8  2011     2     6    36     3     2
# 9  2011     2    17    11     4     2
#10  2011     3    12    10     6     1

Answer 2

We can do this more compactly in tidyverse by getting the group_indices of 'year', 'month', 'day' in the group_by and then create the 'date2' as the number of distinct elements of 'id' ( n_distinct )

librarytidyverse)
df1 %>% 
     group_by(date1 = group_indices(., year, month, day)) %>% 
     mutate(date2 = n_distinct(id))
# A tibble: 10 x 6
# Groups:   date1 [6]
#    year month   day    id date1 date2
#   <int> <int> <int> <int> <int> <int>
# 1  2011     1     5    31     1     2
# 2  2011     1    14    22     2     2
# 3  2011     2     6    28     3     2
# 4  2011     2    17    41     4     2
# 5  2011     3     9    55     5     1
# 6  2011     1     5    34     1     2
# 7  2011     1    14    25     2     2
# 8  2011     2     6    36     3     2
# 9  2011     2    17    11     4     2
#10  2011     3    12    10     6     1

---

Or another compact option with data.table (using the same logic)

library(data.table)
setDT(df1)[, date1 := .GRP, .(year, month, day)][, date2 := uniqueN(id), date1][]
#     year month day id date1 date2
# 1: 2011     1   5 31     1     2
# 2: 2011     1  14 22     2     2
# 3: 2011     2   6 28     3     2
# 4: 2011     2  17 41     4     2
# 5: 2011     3   9 55     5     1
# 6: 2011     1   5 34     1     2
# 7: 2011     1  14 25     2     2
# 8: 2011     2   6 36     3     2
# 9: 2011     2  17 11     4     2
#10: 2011     3  12 10     6     1

---

Or this can be done with interaction and ave from base R

df1$date1 <- with(df1, as.integer(interaction(year, month, day, 
         drop = TRUE, lex.order = TRUE)))
df1$date2 <- with(df1, ave(id, date1, FUN = function(x) length(unique(x))))

data

df1 <- structure(list(year = c(2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 
2011L, 2011L, 2011L, 2011L), month = c(1L, 1L, 2L, 2L, 3L, 1L, 
1L, 2L, 2L, 3L), day = c(5L, 14L, 6L, 17L, 9L, 5L, 14L, 6L, 17L, 
12L), id = c(31L, 22L, 28L, 41L, 55L, 34L, 25L, 36L, 11L, 10L
)), class = "data.frame", row.names = c(NA, -10L))