どのように相互排他的な比較を事前に決定するのですか？

Question

人間の目には値xが条件

x<1 & x>2 

を満たしていないことがわかりますが、どのように私はRがそれを見ることができます。私は渡された比較（文字列など）を取得し、必ずしもデータではない関数でこれを使用したいと思います。のは、私は私は1つが行うことができます意味この

areTherePossibleValues <- function(someString){ 
    someCode 
} 

areTherePossibleValues("x<1 & x>2") 
[1] FALSE 

などのように、比較の兆候であり、部分文字列を解釈することによって、比較の組み合わせが今までとにかく満たすことができるかどうかをチェックする機能を書きたいとしましょう、私もっと良い方法があるように感じる。 Rの比較関数（ '<'、 '>'、 '='など）自体が実際これに対する答えかもしれません。

Answer 1

関数は組み合わせまたはそれらの組み合わせ<又は>、>=、<=、または=（又は==）を含む比較ののリストを処理する関数

。

areTherePossibleValues = function(condition = "x<1 & x>2", tolerance = 1e-10){ 

    #Attach all comparison into one 
    condition = paste(condition, collapse = "&") 

    #PARSE 
    condition = tolower(condition) #make everything lowercase just in case  
    condition = gsub("[ ,x]","",condition) #Remove whitespace and 'x'  
    condition = gsub(">=","g",condition) # >= to g(reater than or equal to) 
    condition = gsub("<=","s",condition) # <= to s(maller than or equal to) 
    condition = gsub("[==,=]","e",condition) # == or = to e(qual) 

    #Separate conditions into a list 
    condition = unlist(strsplit(condition,"&")) 

    #Initiate vector of upper and lower bounds with NA 
    Upper = rep(x = NA, times = length(condition)) 
    Lower = rep(x = NA, times = length(condition)) 

    #Fill the vector of upper and lower bounds based on comparators and numbers 
    for (i in 1:length(condition)){ 
     number = as.numeric(gsub(pattern = "[<,>,e,g,s]", replacement = "", condition[i]))  
     comparator = substr(condition[i], start = 1, stop = 1) 
     if (comparator == ">"){ 
      Lower[i] = number + tolerance #just to the right of the number so as to exclude it 
     } else if (comparator == "<"){ 
      Upper[i] = number - tolerance #just to the left of the number so as to exclude it 
     } else if (comparator == "g"){ 
      Lower[i] = number   #Include the number 
     } else if (comparator == "s"){ 
      Upper[i] = number   #Include the number 
     } else if (comparator == "e"){ 
      Upper[i] = number   #For =, make upper and lower bounds same 
      Lower[i] = number 
     } 
    } 

    Upper = as.numeric(Upper[which(is.na(Upper) == FALSE)]) #Remove NAs 
    Lower = as.numeric(Lower[which(is.na(Lower) == FALSE)]) #Remove NAs 

    if (length(Upper) == 0 & length(Lower) > 0){ 
     #1. If Upper has 0 length and Lower has more than 0, it means 
     # x is constrained only by lower bounds. x will always be fulfilled 
     ans = TRUE 
    } else if (length(Lower) == 0 & length(Upper) > 0){ 
     #2. If Lower has 0 length and Upper has more than 0, it means 
     # x is constrained only by upper bounds. x will always be fulfilled 
     ans = TRUE 
    } else { 
     # If the smallest upper bound is bigger than the largest lower bound, 
     #x will be fulfilled. 
     ans = (min(Upper) - max(Lower)) >= 0 
    } 

    if (ans == FALSE){ 
    return(ans) 
    } else { 
    return(paste(ans," for (",max(Lower)," < x < ",min(Upper),")",sep = "")) 
    } 
} 

USAGE

areTherePossibleValues(">=5 & <50 & >30 & >45") 
#[1] "TRUE for (45.0000000001 < x < 49.9999999999)" 

areTherePossibleValues("x>5 & x<3") 
#[1] FALSE 

areTherePossibleValues(c("<5",">=2 & =4")) 
#[1] "TRUE for (4 < x < 4)"