如何檢查嵌套列表的所有元素是否是 R 中另一個列表的子集

Question

我為此嘗試了許多不同的方法，包括此堆棧，但沒有任何方法可以正常工作。

我的數據框“SiteVisits”（一個小的子集 dput 位於底部）由Date （類 = 日期）、 TagID （類 = 數字）、 SiteVisits （字符列表）和NumSites （類 = 數字）列組成。 每一行都列出了在每個日期找到單個生物體 ( TagID ) 的所有站點。

我想根據標簽訪問的站點來分配標簽是在“內部”、“外部”還是“傳輸”中度過了一整天。 如果它從不訪問外部站點，則只能是“內部”，如果它從不訪問內部站點，則只能是“外部”

首先，我想確定此列表中是否包含某個日期的 TagID 的所有站點：

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

如果 TRUE SiteVisit$Location = "INSIDE" ELSE 測試某個日期的 TagID 的所有站點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"

我已經嘗試了許多不同的dplyr和base版本來實現這一點，但似乎沒有一個是正確的。 我認為這是因為我沒有正確檢查SiteVisit$SiteVisits每個元素

我目前的嘗試是：

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

產生所有“內部”

和

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

產生所有“TRANSITING”

此外，嘗試在 for 循環中執行此操作並不完全有效

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

產生所有 FALSE 而

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

是真的

這是我的數據框“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存儲為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

想要一個快 1/100 的答案？ （不是打字錯誤*，這比 manotheshark 的答案更糟糕，但它適用於您的結構化數據）。 *這是一個錯字！ 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"}
}

兩種方法的基准：

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

我真的不明白為什么 lapply 方法在這里要快得多......可能是因為我正在為循環中的每個 i 取消列出。