如何检查嵌套列表的所有元素是否是 R 中另一个列表的子集

Question

我为此尝试了许多不同的方法，包括此堆栈，但没有任何方法可以正常工作。

我的数据框“SiteVisits”（一个小的子集 dput 位于底部）由Date （类 = 日期）、 TagID （类 = 数字）、 SiteVisits （字符列表）和NumSites （类 = 数字）列组成。 每一行都列出了在每个日期找到单个生物体 ( TagID ) 的所有站点。

我想根据标签访问的站点来分配标签是在“内部”、“外部”还是“传输”中度过了一整天。 如果它从不访问外部站点，则只能是“内部”，如果它从不访问内部站点，则只能是“外部”

首先，我想确定此列表中是否包含某个日期的 TagID 的所有站点：

inside <- list(c("Release","IC1", "IC2", "IC3","RGD1"))

如果 TRUE SiteVisit$Location = "INSIDE" ELSE 测试某个日期的 TagID 的所有站点是否包含在此列表中：

outside <- list(c("ORS1","WC1","WC2","WC3","RGU1","ORN1","ORN2","ORS3","GL1","CVP1","CLRS"))

如果 TRUE SiteVisit$Location = "OUTSIDE" ELSE SiteVisit$Location = "TRANSITING"

我已经尝试了许多不同的dplyr和base版本来实现这一点，但似乎没有一个是正确的。 我认为这是因为我没有正确检查SiteVisit$SiteVisits每个元素

我目前的尝试是：

SiteVisit <- SiteVisit %>%
  mutate(Location = ifelse(all(SiteVisits[[]] %in% inside), "INSIDE",
                           ifelse(all(SiteVisits[[]] %in% outside),"OUTSIDE","TRANSITING")))

产生所有“内部”

和

SiteVisit <- SiteVisit %>%
  mutate(Location = ifelse(all(SiteVisits[] %in% inside), "INSIDE",
                           ifelse(all(SiteVisits[] %in% outside),"OUTSIDE","TRANSITING")))

产生所有“TRANSITING”

此外，尝试在 for 循环中执行此操作并不完全有效

for (i in 1: nrow(SiteVisit)) {SiteVisit$Inside <- 
all(SiteVisit$SiteVisits[[i]] %in% inside)}

产生所有 FALSE 而

all(SiteVisit$SiteVisits[[2]] %in% inside)

是真的

这是我的数据框“SiteVisit”dput 的一小部分：

structure(list(Date = structure(c(15828, 15828, 15847, 15847, 
15847, 15847, 15847, 15847, 15848, 15848, 15848, 15848, 15848, 
15848, 15848, 15848, 15849, 15849, 15849, 15849, 15849, 15849, 
15849, 15850, 15850, 15850, 15850, 15850, 15850, 15850, 15851, 
15851, 15851, 15851, 15851, 15851, 15851, 15851, 15852, 15852, 
15852, 15852, 15852, 15852, 15852, 15853, 15853, 15853, 15853, 
15853, 15853, 15853, 15853, 15853, 15854, 15854, 15854, 15854, 
15854, 15854, 15854, 15854, 15855, 15855, 15855, 15855, 15855, 
15855, 15855, 15855, 15855, 15855, 15855, 15855, 15855, 15855, 
15856, 15856, 15856, 15856, 15856, 15856, 15856, 15856, 15856, 
15856, 15856, 15856, 15856, 15857, 15857, 15857, 15857, 15857, 
15857, 15857, 15857, 15857, 15857, 15857), class = "Date"), TagID = c(5717.06, 
6277.06, 5073.06, 5717.06, 11121.1, 11191.1, 11387.1, 11415.1, 
5717.06, 6277.06, 11121.1, 11191.1, 11219.1, 11289.1, 11387.1, 
11415.1, 5717.06, 11121.1, 11191.1, 11219.1, 11289.1, 11387.1, 
11415.1, 5717.06, 11121.1, 11191.1, 11219.1, 11289.1, 11387.1, 
11415.1, 5717.06, 11121.1, 11191.1, 11219.1, 11289.1, 11317.1, 
11387.1, 11415.1, 5717.06, 6277.06, 11191.1, 11219.1, 11289.1, 
11387.1, 11415.1, 5717.06, 6277.06, 9015.01, 9833.06, 11191.1, 
11219.1, 11289.1, 11387.1, 11415.1, 5717.06, 6277.06, 9015.01, 
11191.1, 11219.1, 11289.1, 11387.1, 11415.1, 5641.22, 5717.06, 
6221.06, 6277.06, 7909.22, 9015.01, 9833.06, 11121.1, 11191.1, 
11219.1, 11289.1, 11317.1, 11387.1, 11415.1, 5717.06, 6277.06, 
6529.06, 8119.01, 8545.06, 9015.01, 9497.06, 9833.06, 11191.1, 
11219.1, 11289.1, 11387.1, 11415.1, 5717.06, 6277.06, 6529.06, 
9015.01, 9497.06, 9833.06, 11191.1, 11219.1, 11289.1, 11387.1, 
11415.1), SiteVisits = list("Release", "Release", c("IC2", "IC1", 
"Release"), "IC3", "WC2", "RGD1", c("WC1", "WC3"), "WC3", "IC3", 
    "IC3", "WC2", "RGD1", "IC2", "IC1", "WC1", "WC3", "IC3", 
    "WC2", "RGD1", c("IC2", "IC1"), "IC1", "WC1", "WC3", "IC3", 
    "WC2", "RGD1", "IC2", "IC1", "WC1", "WC3", "IC3", "WC2", 
    "RGD1", "IC2", "IC1", "WC1", "WC1", "WC3", "IC3", "IC3", 
    "RGD1", "IC2", "IC1", "WC1", "WC3", "IC3", "IC3", c("IC3", 
    "Release"), c("IC3", "IC2", "IC1", "Release"), "RGD1", "IC2", 
    "IC1", "WC1", "WC3", "IC3", "IC3", c("IC3", "IC2"), "RGD1", 
    "IC2", "IC1", "WC1", "WC3", "Release", "IC3", "Release", 
    "IC3", c("RGD1", "Release"), c("IC3", "IC2"), c("IC3", "IC1"
    ), "WC2", "RGD1", "IC2", "IC1", "WC1", "WC1", "WC3", "IC3", 
    "IC3", c("RGD1", "Release"), c("RGD1", "Release"), "Release", 
    c("IC3", "IC2", "IC1"), "Release", c("IC3", "IC2", "IC1", 
    "RGD1"), "RGD1", "IC2", "IC1", "WC1", "WC3", "IC3", "IC3", 
    "RGD1", c("IC3", "IC2", "IC1"), "RGD1", c("IC3", "IC1", "RGD1"
    ), "RGD1", "IC2", c("IC2", "IC1"), "WC1", "WC3"), NumSites = c(1L, 
1L, 3L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 4L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 
3L, 1L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 3L, 1L, 1L, 
2L, 1L, 1L)), row.names = c(NA, -100L), groups = structure(list(
    Date = structure(c(15828, 15847, 15848, 15849, 15850, 15851, 
    15852, 15853, 15854, 15855, 15856, 15857), class = "Date"), 
    .rows = list(1:2, 3:8, 9:16, 17:23, 24:30, 31:38, 39:45, 
        46:54, 55:62, 63:76, 77:89, 90:100)), row.names = c(NA, 
-12L), class = c("tbl_df", "tbl", "data.frame"), .drop = TRUE), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"))

Answer 1

以下工作一次inside和outside存储为array而不是list

inside <- c("Release", "IC1", "IC2", "IC3", "RGD1")
outside <- c("ORS1", "WC1", "WC2", "WC3", "RGU1", "ORN1", "ORN2", "ORS3", "GL1", "CVP1", "CLRS")

df1$Location <- lapply(df1$SiteVisits, function(x) ifelse(all(x %in% inside), "INSIDE", ifelse(all(x %in% outside), "OUTSIDE", "TRANSIT")))

Answer 2

想要一个快 1/100 的答案？ （不是打字错误*，这比 manotheshark 的答案更糟糕，但它适用于您的结构化数据）。 *这是一个错字！ 1/100 不是 1/10

for (i in 1:nrow(SiteVisit)) {
  SiteVisit_test$Location[i] <- if (all(unlist(SiteVisit[i, ]$SiteVisits) %in% unlist(inside))) {
    "INSIDE"
  } else if (all(unlist(SiteVisit[i, ]$SiteVisits) %in% unlist(outside))) {
    "OUTSIDE"
  } else {"TRANSITIONING"}
}

两种方法的基准：

microbenchmark(
  for_statement = for (i in 1:nrow(SiteVisit)) {
    SiteVisit_test$Location[i] <- if (all(unlist(SiteVisit[i, ]$SiteVisits) %in% unlist(inside))) {
      "INSIDE"
    } else if (all(unlist(SiteVisit[i, ]$SiteVisits) %in% unlist(outside))) {
      "OUTSIDE"
    } else {"TRANSITIONING"}
  },
  lapply_statemnt = lapply(SiteVisit$SiteVisits, function(x) ifelse(all(x %in% inside2), "INSIDE", ifelse(all(x %in% outside2), "OUTSIDE", "TRANSIT")))
)

Unit: microseconds
            expr     min      lq      mean  median       uq     max neval
   for_statement 28874.4 30082.0 32411.968 31008.3 33108.90 48878.1   100
 lapply_statemnt   268.4   284.2   346.201   295.5   310.85  4114.9   100

我真的不明白为什么 lapply 方法在这里要快得多......可能是因为我正在为循环中的每个 i 取消列出。