我必須撰寫elemFreqByFirstOcc :: Eq a => [a] -> [(a, Int)]應該計算串列中每個元素的出現次數，并將元素和結果回傳到元組串列中。
例如：
elemFreqByFirstOcc "adbcba" == [('a',2),('d',1),('b',2),('c',1)]
elemFreqByFirstOcc [1,2,1,3,3,2,1,4,3,2,1,1,1,4,6,5] == [(1,6),(2,3),( 3,3),(4,2),(6,1),(5,1)]
到目前為止，我有這段代碼，如果串列中的所有元素只顯示一次，它就可以正常作業，但是當一個元素出現更多次時，它們每次都會被計算在內。（所以對于第一個例子，它回傳[('a',2),('d',1),('b',2),('c',1),('b',1),('a',1)]）
elemFreqByFirstOcc :: Eq a => [a] -> [(a, Int)]
elemFreqByFirstOcc [] = []
elemFreqByFirstOcc [x] = [(x, 1)]
elemFreqByFirstOcc (x:xs) = zip [x] [(length $ filter (==x) (x:xs))] elemFreqByFirstOcc xs
uj5u.com熱心網友回復：
您需要過濾掉已經計算過的專案，因此：
elemFreqByFirstOcc :: Eq a => [a] -> [(a, Int)]
elemFreqByFirstOcc [] = []
elemFreqByFirstOcc (x:xs) = (x, length (filter (x ==) xs) 1) : elemFreqByFirstOcc (filter (x /=) xs)轉載請註明出處，本文鏈接：https://www.uj5u.com/ruanti/411545.html
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