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我刚刚在我的模型中发现了一个错误，我创建了一个具有家庭规模分布和受抚养人年龄分布的人口。我的问题是，在确定年龄时，我的程序似乎没有正确分配每个结果的概率。我应该有 16% 的老年人，但我的模型给了我 3%（占总人口）。

我认为这与程序的内部顺序有关，因为老年人排在最后。请参见下面的代码。

分配家庭规模的程序：

to hatch-family ; hatching the rest of the household using percentual distribution of household size in berlin population
  ifelse random-float 1 <= 0.01 [set household-size 6  set dwelling patch-here hatch 5]
  [ifelse random-float 1 <= 0.02 [set household-size 5 set dwelling patch-here hatch 4]
    [ifelse random-float 1 <= 0.06 [set household-size 4  set dwelling patch-here hatch 3]
      [ifelse random-float 1 <= 0.11 [set household-size 3 set dwelling patch-here hatch 2]
        [ifelse random-float 1 <= 0.31 [set household-size 2 set dwelling patch-here hatch 1]
          [ifelse random-float 1 <= 0.49 [set household-size 1 set dwelling patch-here]
          [hatch-family]
  ]]]]]
end

分配年龄的程序：

to assign-age; assign age depending on household size according to distribution on csv file,
             ;i = the different rows representing different household sizes
  get-age-data
  if household-size = 1 [get-age-breed 1]
  if household-size = 2 [get-age-breed 2]
  if household-size = 3 [get-age-breed 3]
  if household-size = 4 [get-age-breed 4]
  if household-size = 5 [get-age-breed 5]
  if household-size = 6 [get-age-breed 6]
end

to get-age-data ; reading csv file and creating dataset with age distribution of the population
  file-close-all                          ; close all open files
    if not file-exists? "Data/age-dist-breeds.csv"  [
    user-message "No file 'age-dist-breeds' exists!"
    stop
  ]

  file-open "Data/age-dist-breeds.csv"  ; open the file with the turtle data
  set age-data csv:from-file "Data/age-dist-breeds.csv"
  file-close                              ; making sure to close the file
end

to get-age-breed [i]; process for assigning age after distribution for household size i, will repeat itself if no age is assigned in previous run
  let child-var item 1 item i age-data
ifelse (random-float 1 <= child-var)
   [set age random 14 set breed children set age-susceptibility 0.0025]
   [let teen-var item 2 (item i age-data)
     ifelse (random-float 1 <= teen-var)
    [set age 15 + random 4 set breed teens set age-susceptibility 0.005]
    [let adult-var item 3 (item i age-data)
        ifelse ( random-float 1 <= adult-var)
           [let age-susc-var random 49
            set age 20 + age-susc-var set breed adults set age-susceptibility (age-susc-var / 49) * 0.075 + 0.015]
           [let elder-var item 4 (item i age-data)
  ifelse ( random-float 1 <= elder-var )
        [ set age 70 + random 30 set breed elderly set age-susceptibility 0.08]
        [get-age-breed i]
    ]]
  ]
end

我试图在开始时制作一个随机变量，而不是为每一步重新计算它，但最终我没有任何老人。

有没有比随机浮点程序更好的方法来准确分配非正态概率分布？

4

2 回答 2

1

您的问题是您继续绘制一个新的随机数。这是您的代码结构，缩进以便您可以看到您的逻辑：

to hatch-family
  ifelse random-float 1 <= 0.01
  []
  [ ifelse random-float 1 <= 0.02
    []
    [ ifelse random-float 1 <= 0.06
      []
      [ ifelse random-float 1 <= 0.11
        []
        [ ifelse random-float 1 <= 0.31
          []
          [ ifelse random-float 1 <= 0.49
            []
            [ hatch-family ]
          ]
        ]
      ]
    ]
   ]
end

所以它会抽取一个数字，满足第一个条件的概率为 1%，否则它会抽取另一个数字，依此类推，直到结束。更重要的是，它在大约一半的时间内再次完成了整个事情，因为该过程将自己称为最终平局的假分支。

所以，第一步是只绘制一次随机数，然后简单地比较那个值，看起来像这样：

to hatch-family
  let my-draw random-float 1
  ifelse my-draw <= 0.01 []
  [ ifelse my-draw <= 0.02 []
    [ ifelse my-draw <= 0.06 []

但是您也可以使用多项选择ifelse来清理代码（假设您使用的是当前版本的 NetLogo），它看起来像：

to hatch-family
  let my-draw random-float 1
  ( ifelse
     my-draw <= 0.01 []
     my-draw <= 0.02 []
     my-draw <= 0.06 []

注意开头 ( 让 NetLogo 知道需要两个以上的子句。

或者您可以使用rnd史蒂夫已经指出的扩展名。

于 2021-04-02T08:00:13.353 回答
1

我终于通过@JenB 的一些输入解决了这个问题。

代码现在如下所示：

其余户按照-累积-户数分布孵化的程序：

; hatching the rest of the household using percentual distribution of household size of Berlin population (Zensus Datenbank)
; (nr of households with specific size / total number of households in Berlin, 
; hence the percentage of the total population will be accurate as the population increases with each hatching)
to hatch-family
ask turtles
  [ 
  let my-draw random-float 1
  (ifelse
   my-draw <= 0.01 [set household-size 6 set dwelling patch-here hatch 5]
   my-draw <= 0.03 [set household-size 5 set dwelling patch-here hatch 4]
   my-draw <= 0.09 [set household-size 4 set dwelling patch-here hatch 3]
   my-draw <= 0.20 [set household-size 3 set dwelling patch-here hatch 2]
   my-draw <= 0.51 [set household-size 2 set dwelling patch-here hatch 1]
      my-draw <= 1 [set household-size 1 set dwelling patch-here]
    )
  ]
end

根据每个家庭规模的年龄分布分配品种和年龄的程序：

（我用up-to-n-of命令而不是随机浮点数解决了这个问题，以解决概率并不总是按最低->最高顺序的问题）

to assign-age; assign age depending on household size according to distribution on csv file,
             ;second item [i] = the different rows representing different household sizes
  ask turtles [set age 0]
  get-age-data ; reading csv file and naming it age-data
  get-age-breed 1
  get-age-breed 2
  get-age-breed 3
  get-age-breed 4
  get-age-breed 5
  get-age-breed 6
end

to get-age-breed [i]; process for assigning age after distribution for household size i
  let child-var up-to-n-of ((count turtles with [household-size = i]) * item 1 (item i age-data)) (turtles with [household-size = i and age = 0])
  ask child-var [set age random 14 set breed children set age-susceptibility 0.0025]
  let teen-var up-to-n-of (count turtles with [household-size = i] * item 2 item i age-data) (turtles with [household-size = i and age = 0])
  ask teen-var [set age 15 + random 4 set breed teens set age-susceptibility 0.005]
  let elder-var up-to-n-of (count turtles with [household-size = i] * item 4 item i age-data) (turtles with [household-size = i and age = 0])
  ask elder-var [set age 70 + random 30 set breed elderly set age-susceptibility 0.08]
  let adult-var up-to-n-of (count turtles with [household-size = i] * item 3 item i age-data) (turtles with [household-size = i and age = 0])
  ask adult-var [let age-susc-var random 49 set age 20 + age-susc-var set breed adults set age-susceptibility (age-susc-var / 49) * 0.075 + 0.015]
 
end
于 2021-04-03T13:59:00.990 回答