Loop through rows and count number of rows that matches multiple criteria in R

Question

I have a dataset that looks like this:

        city period_day       date 
1  barcelona    morning 2017-01-15         
2  sao_paulo  afternoon 2016-12-07         
3  sao_paulo    morning 2016-11-16         
4  barcelona    morning 2016-11-06         
5  barcelona  afternoon 2016-12-31         
6  sao_paulo  afternoon 2016-11-30         
7  barcelona    morning 2016-10-15         
8  barcelona  afternoon 2016-11-30         
9  sao_paulo  afternoon 2016-12-24         
10 sao_paulo  afternoon 2017-02-02

For every row, I want to count how many rows have an older date than the date of the row, both for city and period_day. In this case, I want this result:

        city period_day       date row_count
1  barcelona    morning 2017-01-15         2
2  sao_paulo  afternoon 2016-12-07         1
3  sao_paulo    morning 2016-11-16         0
4  barcelona    morning 2016-11-06         1
5  barcelona  afternoon 2016-12-31         1
6  sao_paulo  afternoon 2016-11-30         0
7  barcelona    morning 2016-10-15         0
8  barcelona  afternoon 2016-11-30         0
9  sao_paulo  afternoon 2016-12-24         2
10 sao_paulo  afternoon 2017-02-02         3

When row_count equals to 0, it means that it's the older date.

I came up with a solution, but it took too long with more data. That's the code:

get_count_function <- function(df) {
  idx <- 1:nrow(df)

  count <- sapply(idx, function(x) {
    name_city <-
      df %>% select(city) %>% filter(row_number() == x) %>% pull()
    name_period <-
      df %>% select(period_day) %>% filter(row_number() == x) %>% pull()

    date_row <- df %>%
      select(date) %>%
      filter(row_number() == x) %>%
      pull()

    date_any_row <- df %>%
      filter(dplyr::row_number() != x,
             city == name_city,
             period_day == name_period) %>%
      select(date) %>%
      pull()

    how_many <- sum(date_row > date_any_row)

    return(how_many)

  })

  return(count)

}

How could I turn this function more efficient?

Try out this one to see if that works - https://stackoverflow.com/questions/23528862/summarize-with-conditions-in-dplyr — Joni Hoppen, May 09 '19 at 18:27

score 1 · Accepted Answer · answered May 09 '19 at 19:51

Try this one:

library(tidyverse)

dat %>%
  group_by(city, period_day) %>%
  mutate(row_count = order(date) - 1) %>%
  ungroup()

When you call order it returns indices, pointing to the order of the value in a selected group of values (date). Subtracting 1 from the indices, you obtain the count of values preceding current value, in a particular group. E.g. if it is the min. value in a group, it has index 1, so nothing preceding it (1 - 1 = 0), if the index is 2 - only one value is preceding it (one older date before it) etc.

Data:

dat <- read.table(
  text = "        city period_day       date
  barcelona    morning 2017-01-15
  sao_paulo  afternoon 2016-12-07
  sao_paulo    morning 2016-11-16
  barcelona    morning 2016-11-06
  barcelona  afternoon 2016-12-31
  sao_paulo  afternoon 2016-11-30
  barcelona    morning 2016-10-15
  barcelona  afternoon 2016-11-30
  sao_paulo  afternoon 2016-12-24
  sao_paulo  afternoon 2017-02-02",
  header = T,
  colClasses = c("character", "character", "Date")
)

it worked! unfortunately i cant +1 to you because of reputation, but thank you! — Gabriel Dias, May 09 '19 at 20:10

score 1 · Answer 2 · answered May 09 '19 at 19:52

This should work if you are willing to use the data.table package:

library(data.table)

dat <- read.table(header=T, row.names=1, text="
        city period_day       date 
1  barcelona    morning 2017-01-15         
2  sao_paulo  afternoon 2016-12-07         
3  sao_paulo    morning 2016-11-16         
4  barcelona    morning 2016-11-06         
5  barcelona  afternoon 2016-12-31         
6  sao_paulo  afternoon 2016-11-30         
7  barcelona    morning 2016-10-15         
8  barcelona  afternoon 2016-11-30         
9  sao_paulo  afternoon 2016-12-24         
10 sao_paulo  afternoon 2017-02-02   
")

dat <- as.data.table(dat)

dat[, row_count := (order(as.Date(date)) - 1), by=.(city, period_day)]

# Check
dat

##          city period_day       date row_count
##  1: barcelona    morning 2017-01-15         2
##  2: sao_paulo  afternoon 2016-12-07         1
##  3: sao_paulo    morning 2016-11-16         0
##  4: barcelona    morning 2016-11-06         1
##  5: barcelona  afternoon 2016-12-31         1
##  6: sao_paulo  afternoon 2016-11-30         0
##  7: barcelona    morning 2016-10-15         0
##  8: barcelona  afternoon 2016-11-30         0
##  9: sao_paulo  afternoon 2016-12-24         2
## 10: sao_paulo  afternoon 2017-02-02         3

it worked! unfortunately i cant +1 to you because of reputation, but thank you! — Gabriel Dias, May 09 '19 at 20:10

Loop through rows and count number of rows that matches multiple criteria in R

2 Answers2