使用列表推导中的函数,Haskell?

时间:2012-01-10 18:43:26

标签: haskell list-comprehension

我们可以放置函数来确定列表推导的条件。这是我试图实现的代码:

mQsort :: [String] -> [F.Record] -> [F.Record]
mQsort [] _ = []
mQsort c@(col:cond:cs) l@(x:xs) = (mQsort c small) ++ mid ++ (mQsort c large)
   where
    small = [y | y<-xs, (qGetStr col y) (qGetCond cond) (qGetStr col x)]
    mid   = mQsort cs [y | y<-l, (qGetStr col y) == (qGetStr col x)]
    large = [y | y<-xs, (qGetStr col y) (qGetCond' cond) (qGetStr col x)]

qGetStr :: String -> F.Record -> String
qGetStr col r | U.isClub col = F.club r
            | U.isMap col = F.mapName r
            | U.isTown col = F.nearestTown r
            | U.isTerrain col = F.terrain r
            | U.isGrade col =F.mapGrade r
            | U.isSW col = F.gridRefOfSWCorner r
            | U.isNE col = F.gridRefOfNECorner r
            | U.isCompleted col = F.expectedCompletionDate r
            | U.isSize col = F.sizeSqKm r
            | otherwise = ""

qGetCond "ascending" = (<)
qGetCond "decending" = (>)
qGetCond' "ascending" = (>)
qGetCond' "decending" = (<)

我收到一条错误,指出函数qGetStr应用于4个参数而不是2 此外,qGetCond - 这是返回运算符的正确语法。由于编译错误,我不得不将操作符放在括号中,但我感觉这是不正确的

1 个答案:

答案 0 :(得分:8)

替换

small = [y | y<-xs, (qGetStr col y) (qGetCond cond) (qGetStr col x)]

small = [y | y<-xs, (qGetCond cond) (qGetStr col y) (qGetStr col x)]

同样适用于large

原因与您在qGetCond中返回运算符的语法的原因相同。操作员真的只是一个功能。

  • foo < bar(<) foo bar
  • 相同
  • foo `fire` barfire foo bar
  • 相同

所以你必须移动&#34;运算符&#34;到列表理解中的表达式的开头。 (n.b.这一般是正确的,特别是与列表理解无关。)

编辑:扩展Chris Kuklewicz的观点:

  • 反引号仅适用于单个标识符。因此foo `fire` bar是有效的语法,但更复杂的内容如foo `fire forcefully` barfoo `(fire forcefully)` bar是语法错误。

  • 括号轮操作符更灵活。这些表达式都评估为相同的值:

    • foo < bar
    • (<) foo bar
    • (foo <) bar
    • (< bar) foo

    最后两个表单称为运算符部分。它们在将函数传递给另一个函数时很有用,例如map (+1) listOfIntegers。语法有几个细微之处:

    (quux foo <)      -- function calls are fine
    (baz + foo <)     -- syntax error because of the extra operator...
    ((baz + foo) <)   -- ...but this is ok
    (-3)              -- `-` is a special case: this is a numeric literal,
                      --                        not a function that subtracts three
    (subtract 3)      -- is a function that subtracts three