我有一个;; -----------------------------------------
;; begin Point class
;; -----------------------------------------
(defrecord Point [x y methods] )
(def someMethods
{
:getX (fn [this] (:x this) )
:getY (fn [this] (:y this) )
:distance (fn [this other]
(def dx (- (:x this) (:x other)))
(def dy (- (:y this) (:y other)))
(Math/sqrt (+ (* dx dx) (* dy dy) ))
)
}
)
;;
;; Point constructor
;;
(defn newPoint [x y]
(Point. x y someMethods)
)
;; -----------------------------------------
;; end Point class
;; -----------------------------------------
;; -----------------------------------------
;; helper to call methods
;; -----------------------------------------
(defn call
([obj meth] ((meth (:methods obj)) obj))
([obj meth param1] ((meth (:methods obj)) obj param1))
([obj meth param1 param2] ((meth (:methods obj)) obj param1 param2))
)
;; -----------------------------------------
;; main()
;; -----------------------------------------
(def p1 (newPoint 3 4))
(def p2 (newPoint 0 0))
(call p1 :getY) ;; == ((:getX (:methods p1)) p1)
(call p1 :distance p2) ;; == ((:distance (:methods p1)) p1 p2)
，其中每行表示是否在某个特定位置发现了一只动物。
我想在此示例data.frame中创建一个标记为data.frame的新列。此值将为1或0，具体取决于是否在同一位置找到捕食者的猎物（每个位置都有唯一的"prey"）。
问题是每只动物都有一个单独的行，所以关于猎物存在于与捕食者不同的行中的信息。这两个掠食者是狮子和猎豹。
对于这个例子，狮子的猎物是羚羊和斑马，所以：
ID 1，由于在该位置发现了羚羊和狮子，猎物列的狮子座应该有1个。ID 2，未发现羚羊或斑马，所以狮子行的猎物列为0。下面是示例ID，我提出的解决方案非常低效，而且我正在寻找更快/更整洁的东西。
data.frame答案 0 :(得分：0)
考虑使用tidyr::spread来简化第一个数据框的结构。
df <- df %>% spread(species, present)
#> ID antelope cheetah gazelles impala lion zebra
#>1 1 1 1 0 1 1 0
#>2 2 0 1 1 1 1 0
然后继续dplyr。
df %>%
spread(species, present) %>%
mutate(lion_prey = case_when(antelope == 1 | zebra == 1 ~ 1,
TRUE ~ 0),
cheetah_prey = case_when(antelope == 1 | gazelles == 1 | impala == 1 ~ 1,
TRUE ~ 0)) %>%
gather(species, present, -ID, -lion_prey, -cheetah_prey) %>%
mutate(prey = case_when(species == "lion" ~ lion_prey,
species == "cheetah" ~ cheetah_prey,
TRUE ~ 0)) %>%
select(-lion_prey, -cheetah_prey)
#> ID species present prey
#> 1 1 antelope 1 0
#> 2 2 antelope 0 0
#> 3 1 cheetah 1 1
#> 4 2 cheetah 1 1
#> 5 1 gazelles 0 0
#> 6 2 gazelles 1 0
#> 7 1 impala 1 0
#> 8 2 impala 1 0
#> 9 1 lion 1 1
#> 10 2 lion 1 0
#> 11 1 zebra 0 0
#> 12 2 zebra 0 0
答案 1 :(得分：0)
由于您描述的原因，这涉及一些混乱的逻辑表达式，但这是一种方法。这具有可推广的优点。如果您想添加捕食者，只需将它们添加到predators并相应地将它们的猎物添加到predators_prey。 predators_prey是一个容纳（在此处发生）具有不同猎物数量的捕食者的列表：
# define the predators
predators <- c("lion", "cheetah")
# create a list of their prey from which to programmatically extract
predators_prey <- list(lion = c("antelope", "zebra"), cheetah = c("antelope", "gazelles", "impala"))
# initialize the $prey column
df$prey <- 0
# use for loop because we're assigning a value in global env
for (predator in predators ){
for (ID in unique(df$ID)){
# is the predator here?
predator_here = df[df$ID == ID & df$species == predator,]$present
# is that predator's prey here?
prey_here = any(df[df$ID == ID & df$present == 1,]$species %in% predators_prey[[predator]])
# if both, then set $prey to 1
if(predator_here & prey_here){
df[df$ID == ID & df$species == predator,]$prey <- 1
}
}
}
# lets look at the result
df
# ID species present prey
# 1 1 lion 1 1
# 2 1 antelope 1 0
# 3 1 zebra 0 0
# 4 1 cheetah 1 1
# 5 1 impala 1 0
# 6 1 gazelles 0 0
# 7 2 lion 1 0
# 8 2 antelope 0 0
# 9 2 zebra 0 0
# 10 2 cheetah 1 1
# 11 2 impala 1 0
# 12 2 gazelles 1 0
数据：
df <- data.frame(ID=c(1,1,1,1,1,1, 2, 2, 2, 2, 2, 2),
species=c("lion", "antelope", "zebra", "cheetah", "impala", "gazelles", "lion", "antelope", "zebra", "cheetah", "impala", "gazelles"),
present=c(1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1),
stringsAsFactors=FALSE)