根据属性组件匹配列表元素

Question

我有一个数据集，我将其分成两个列表 int1 和 int2。

library(lubridate)
library(tidyverse)
library(purrr)

date <- rep_len(seq(dmy("01-01-2011"), dmy("01-01-2013"), by = "days"), 300)
ID <- rep(c("A","B", "C"), 300)
df <- data.frame(date = date,
                 x = runif(length(date), min = 60000, max = 80000),
                 y = runif(length(date), min = 800000, max = 900000),
                 ID)

df$month <- month(df$date)
df$year <- year(df$date)

# Create first list
int1 <- df %>%
  mutate(new = floor_date(date, '10 day')) %>%
  mutate(new = if_else(day(new) == 31, new - days(10), new)) %>% 
  group_by(ID, new) %>%
  filter(month == "1") %>% 
  group_split()

# Create second list
int2 <- df %>%
  mutate(new = floor_date(date, '10 day')) %>%
  mutate(new = if_else(day(new) == 31, new - days(10), new)) %>% 
  group_by(ID, new) %>%
  filter(month == "2") %>% 
  group_split()


names(int1) <- sapply(int1, function(x) paste(x$ID[1],
                                                   x$year[1], sep = '_'))

names(int2) <- sapply(int2, function(x) paste(x$ID[1],
                                                    x$year[1], sep = '_'))

然后我为每个列表分配一个属性 (match)。我创建了一个函数 check 来更轻松地获取此属性。我为这个例子从一个列表中删除了一些元素。


int1 <- int1[-c(3,6)]

# Convenience function to grab the attributes for you
check <- function(x) {
  return(attr(x, "match"))
}

# Add an attribute to hold the attributes of each list element
attr(int1, "match") <- data.frame(id = sapply(int1, function(x) paste(x$ID[1])),
                                     interval_start_date = sapply(int1, function(x) paste(x$new[1]))
)

# Check the attributes
check(int1)

# Add an attribute "tab" to hold the attributes of each list element
attr(int2, "match") <- data.frame(id = sapply(int2, function(x) paste(x$ID[1])),
                                     interval_start_date = sapply(int2, function(x) paste(x$new[1]))
) 

# Check the attributes
check(int2)

我想根据我添加的属性删除不在另一个元素中的元素。具体来说，我想删除任何不具有相同 interval_start_date 和 ID 关联的内容。对于 interval_start_date，只有年份和日期必须匹配，因为两个列表之间的月份很可能不同。在这种情况下，我希望 int2 匹配 int1。关于如何做到这一点的任何想法？如果可能，首选基本 r 方法。

# Expected results
expected_int2 <- list(int2[[1]], int2[[2]], int2[[3]], int2[[4]], int2[[5]], 
                      int2[[6]], int2[[7]])

names(expected_int2) <- sapply(int1, function(x) paste(x$ID[1],
                                                       x$year[1], sep = "_"))

Answer 1

我们可以在 pasteing 'id' 和 formatted 'interval_start_date' 之后创建一个 %in% 的索引，即删除 'month' 部分

i1 <-  with(check(int2), paste(id, format(as.Date(interval_start_date), 
     "%Y-%d"))) %in%  with(check(int1), paste(id, 
      format(as.Date(interval_start_date), "%Y-%d")))
> which(i1)
[1] 1 2 4 5 7 8 9
out <- int2[i1]

根据属性组件匹配列表元素

Match list elements based on attribute component

r

list

attributes