遍历行并计算与 R 中的多个条件匹配的行数

Question

我有一个如下所示的数据集：

        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         

对于每一行，我想计算有多少行的日期早于该行的日期，包括 city 和 period_day。在这种情况下，我想要这个结果：

        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

row_count 为 0 时，表示为旧日期。

我想出了一个解决方案，但是需要更多数据才需要很长时间。代码如下：

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)

}

我怎样才能让这个功能更有效率？

Answer 1

如果您愿意使用 data.table 包，这应该可以：

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

Answer 2

试试这个：

library(tidyverse)

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

当您调用 order 时，它会返回索引，指向选定值组 (date) 中值的顺序。从索引中减去 1，您将获得特定组中 preceding 当前值的 count 个值。例如。如果是最小值。一个组中的值，它有索引1，所以它前面没有任何东西（1 - 1 = 0），如果索引是2 - 只有一个值在它前面（一个旧的date在它前面）等等。

数据：

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")
)