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r - Purr::map2 过滤多个条件

问题描述

我已经生成了一系列随机游走，每个人都在有界的舞台上移动（data.frame fishmoves）。在这个竞技场内，有一些感兴趣的网格方块（data.frame 蚂蚁，盒子的 x 和 y 坐标）。我想过滤 fishmoves 以仅包含感兴趣的网格正方形内的点。然后我想计算每个人在每个框中花费的个人数量和时间步长。

解决了我在过滤上遇到了障碍。我正在使用 purr::map2 从蚂蚁传递过滤条件，但是当我调用所有四个参数时它不起作用——而是返回 null。我确保（data.frame tester）有实际值可以找到，所以一定是我在调用中做错了什么。结束 解决

这使我无法进行下一步，即按蚂蚁拆分并按个人和时间步数进行汇总。任何帮助将不胜感激！

library(tidyverse)

n.times<-1000
OUT <-data.frame(time=vector("numeric", n.times), x.a = vector("numeric", n.times),y.a = vector("numeric", n.times))

walker <- function(n.times,
                   xlim=c(0,40),
                   ylim=c(0,20),
                   start=c(0,0),
                   stepsize=c(1,1)) {
  ## extract starting point
  x <- start[1]
  y <- start[2]

  for (i in 1:n.times) {
    repeat {
      ## pick jump sizes
      xi <- stepsize[1]*sample(rnorm(n = n.times, mean = 0, sd = .5),1)
      yi <- stepsize[2]*sample(rnorm(n = n.times, mean = 0, sd = .5),1)
      ## new candidate locations
      newx <- x+xi
      newy <- y+yi
      ## IF new locations are within bounds, then
      ##    break out of the repeat{} loop (otherwise
      ##    try again)
      if (newx>xlim[1] && newx<xlim[2] &&
          newy>ylim[1] && newy<ylim[2]) break
    }
    ## set new location to candidate location
    x <- newx
    y <- newy
    OUT[i,"time"]<-i
    OUT[i,"x.a"] <-x
    OUT[i, "y.a"] <-y
  }
  return(OUT)
}


#generate fake fish
fish<-data.frame(fish=as.character(letters[1:10]))

#apply walker to fake fish
fishmoves <- fish %>% 
  mutate(data= map(fish,~walker(n.times))) %>% 
  unnest(data)


#ants <- data.frame(ant=c("a", "b"),x.min=seq(from=2, to = 38, by= 4)),x.max=c(1,4),y.min=c(0,2), y.max=c(1,3)) 
ants <- data.frame(ant=LETTERS[1:16]) %>% 
  bind_cols(x.min=c(seq(from=4, to = 32, by= 4),seq(from=4, to = 32, by= 4)),
            y.min=c(rep(4,each=8),rep(12,each=8))) %>% 
  mutate(x.max=x.min+2,
         y.max=y.min+2) %>% 
  group_by(ant) 


#filter fishmoves based on the filter parameters - works separately for both x and y

ant_fish1 <- map2(ants$x.min, ants$x.max, ~ fishmoves %>%
         filter(between(x.a, ..1[1], ..2[1])) )

ant_fish2 <- map2(ants$y.min, ants$y.max, ~ fishmoves %>%
                    filter(between(y.a, ..1[1], ..2[1])) )  

#test to demonstrate that there are individuals that meet the joint criteria
tester <- fishmoves %>% filter (between(x.a, ants$x.min[1], ants$x.max[1]) & between(y.a, ants$y.min[1], ants$y.max[1]))

#### switched map2 to pmap

ant_fish <- pmap(list(ants$x.min, ants$x.max, ants$y.min, ants$y.max), ~ fishmoves %>%
                    filter(between(x.a, ..1[1], ..2[1]) & between(y.a, ..3[1], ..4[1])) )  

#conceptual approach? does not work...
ant_fish <- ants %>% nest(ant_loc = c(x.min, x.max, y.min, y.max)) %>% 
 pmap(list(ants$x.min, ants$x.max, ants$y.min, ants$y.max), ~ fishmoves %>%
                    filter(between(x.a, ..1[1], ..2[1]) & between(y.a, ..3[1], ..4[1])) ) %>% 
  group_by(fish) %>% 
  summarise(counts=n())

标签： rpurrr

解决方案

问题是你不能像那样使用 map2 。map2 中的 2 表示您用于映射的函数必须接受 2 个（并且只有 2 个）参数，但您想使用一个接受 4 个参数的函数。执行您想要的操作的映射函数称为pmap。

而不是这个...

#filter returns null
ant_fish3 <- map2(ants$x.min, ants$x.max, ants$y.min, ants$y.max, ~ fishmoves %>%
                    filter(between(x.a, ..1[1], ..2[1]) & between(y.a, ..3[1], ..4[1])) )

你会用这个...

ant_fish3 <- pmap(list(ants$x.min, ants$x.max, ants$y.min, ants$y.max), function(one, two, three, four) { 
    fishmoves %>%
        filter(between(x.a, one[1], two[1]) & between(y.a, three[1], four[1])) })

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