如何決議Netlogo中的字串？

語境

對于我的模型，我希望有一個輸入，用戶可以在其中輸入一系列值。

例如

我想從上面顯示的輸入中得到一個包含五個數字的串列，例如[0.5 0.2 0 0.2 0.5]，這樣我就可以使用他們輸入的數字進行一些計算。

問題

不幸的是，"0.5 0.2 0 0.2 0.5"如果我將型別設定為字串，那么像上面那樣設定輸入會吐出來。如果我將型別設定為數字，它將只允許輸入一個數字。

那么，我怎樣才能決議字串作為空格的基礎（即“”）？我也對替代方案持開放態度，盡管我更愿意將它保留在 Netlogo 中（例如，不讀取值的文本檔案）以使其更容易更改，因為我懷疑它會被大量使用。

我試過的

我試過使用read-from-string，但它也不喜歡上面輸入的一系列數字。我還嘗試使用explode字串擴展 ( https://github.com/NetLogo/String-Extension ) 中的函式，但我的 Netlogo (6.2.0) 版本不喜歡該擴展中的 API 并且不允許我用它。

我對 NetLogo 很陌生，如果我的問題很愚蠢或者我沒有說清楚，很抱歉！

uj5u.com熱心網友回復：

根據它的檔案，read-from-string可以決議文字值串列。根據編程指南的常量串列部分，您遇到的問題是 NetLogo 串列文字必須有方括號才能打開和關閉。因此，您需要做的就是將[和添加]到用戶的輸入中。

to test
  let s "0.5 0.2 0 0.2 0.5"
  let l read-from-string (word "[" s "]")
  show l
  show item 2 l
end

輸出：

observer> test
observer: [0.5 0.2 0 0.2 0.5]
observer: 0

不過，我要提醒的是，用戶很容易輸入不同格式的數字，例如0, 2, 3, 5.0，使用逗號分隔值。檢查轉換是否有效是明智的，因為您從失敗中獲得的錯誤訊息read-from-string可能對模型用戶沒有幫助。

uj5u.com熱心網友回復：

你可以組合做position，substring，read-from-string和fput。

這是作業流程：

1. 創建一個回圈，只要字串包含多個數字（= 只要它包含至少一個空格，使用 進行檢查position " " string）；
2. 提取從第一個字符到排除的第一個空格的子字串（用 完成substring）；
3. 將該子字串作為數值（with read-from-string）讀取并將其添加到list-of-numbers（with fput）中；
4. 洗掉字串中的第一個數字（使用position " " string,repeat和but-first）并再次開始回圈；
5. 當回圈條件計算為 asFALSE這意味著字串中只剩下一個數字。將最后一個數字（即整個剩余的字串）添加到list-of-numbers回圈外，就完成了。

下面的程序是一個報告器程序，它執行此作業流并報告從字串中讀取的值串列（它只需要user-string界面中的輸入框）：

to-report convert-user-string [str]
  let temp-string user-string
  let list-of-numbers (list)
  
  while [position " " temp-string != FALSE] [
   let next-number-as-string  (substring temp-string 0 position " " temp-string)
   set list-of-numbers lput (read-from-string next-number-as-string) (list-of-numbers)
   
   repeat (position " " temp-string   1) [
     set temp-string (but-first temp-string)
   ]
  ]
  
  set list-of-numbers lput (read-from-string temp-string) (list-of-numbers)
  
  report list-of-numbers
end

例如：

observer> set user-string "0.5 0.2 0 0.2 0.5"
observer> show user-string
observer: "0.5 0.2 0 0.2 0.5"
observer> show convert-user-string user-string
observer: [0.5 0.2 0 0.2 0.5]

我在上面發布的程序是我制作的初始代碼的精簡版本，我將在下面留下大量評論：

globals [
  list-of-numbers   ; The list where values from the input string will be stored.
  
  temp-string       ; A temporary variable being the alter-ego of 'user-list'. This is needed because
                    ; the 'trim-string-to-next-nonspace' procedure won't let me change the value of
                    ; 'user-string' directly (I am not sure why, anyone please feel free to say if I'm
                    ; missing something here) but also because you might want to keep the value of the
                    ; user input intact - hence we use this 'temp-string' to trim the string without worries.
]


to convert-user-string [str]
; As long as there are at least two numbers in the string (identified by the presence of at least one
; space), the while loop extracts the first number with 'substring' and then assigns it as a numeric
; value to 'list-of-numbers' by using 'read-from-string' and 'lput'. At that point, it trims the
; string up to the next non-space character.
; When there is only one number left in the string (identified by the absence of spaces in the string),
; the 'more-than-one-number-in-string? temp-string'condition evaluates as 'FALSE' and the while loop
; stops. At that point, the last line of code adds what is left of the string (i.e. the last number)
; to the 'list-of-numbers' list.
  
  set list-of-numbers (list)   ; Initiating this variable as a list in order to be able to use 'lput'.
  
  set temp-string user-string
  
  while [more-than-one-number-in-string? temp-string] [
   let next-number-as-string  (substring temp-string 0 position-of-next-space temp-string)
   set list-of-numbers lput (read-from-string next-number-as-string) (list-of-numbers)
   
   trim-string-to-next-nonspace temp-string
  ]
  
  set list-of-numbers lput (read-from-string temp-string) (list-of-numbers)
end


to-report more-than-one-number-in-string? [str]
; This reporter is needed as a condition for the while loop in 'convert-user-string'. The reason is that
; the 'position' command (used by the 'position-of-next-space' procedure) reports either a number (i.e.
; the position of the character in the given string) or 'FALSE' (in case the item is not present in the
; string). Therefore, this procedure is needed in order to get either TRUE or FALSE to be used in the
; while condition.
  
  ifelse (position-of-next-space str = FALSE)
   [report FALSE]
   [report TRUE]
end


to-report position-of-next-space [str]
; Simply reporting the position of the next space in the string. Note that positions (indexes) in NetLogo
; are numbered starting from 0.
  
  report position " " str
end


to trim-string-to-next-nonspace [str]
; By using 'but-first' repeatedly, this procedure gets rid of the first number (which has already been stored
; in 'list-of-numbers' by the 'convert-user-string' procedure) and the following space in the string.
; Note that the '  1' bit is needed because the count of positions in NetLogo starts from 0 for the first item.

  let x temp-string
  
  repeat (position-of-next-space temp-string   1) [
   set x (but-first x) 
  ]
  
  set temp-string x
end

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