OAML 使用开关计算 char 的频率

Question

我试图计算每个字符在字符串中出现的时间，我使用开关和 for 循环，但是，它们没有正确增加。这是我的代码

let countChar x = 
  match x with
    'A' -> countA := !countA + 1;
  | 'C' -> countC := !countC + 1;
  | 'T' -> countT := !countT + 1;
  | 'G' -> countG := !countG + 1;   
;;

let demoStri = "ACGTACGT" in 
for j = 0 to 7 do
  countChar demoStri.[j];
  let tempA = !countA in
  print_int tempA;
  print_string "\n";
  let tempC = !countC in
  print_int tempC;
  print_string "\n";
  let tempG = !countG in
  print_int tempG;
  print_string "\n";
  let tempT = !countT in 
  print_int tempT;
  print_string "\n";
done

但由于某种原因，它只递增 1，并且返回 1 0 0 0、2 0 0 0、3 0 0 0 等等......我想知道这个过程中是否出了问题？

Answer 1

我认为这段代码当前的形式没有问题，它对我有用。您没有显示

countA

、

countC

、

countT

和

countG

的初始化，但如果我按如下方式初始化：

let countA = ref 0
let countC = ref 0
let countT = ref 0
let countG = ref 0

然后运行你的代码，我得到这一系列数字（将四个折叠成一行以节省空间）：

1 0 0 0
1 1 0 0
1 1 1 0
1 1 1 1
2 1 1 1
2 2 1 1
2 2 2 1
2 2 2 2

Answer 2

这里最大的问题是您使用

tempH

变量来索引字符串，而不是您应该使用的

j

。

let () = 
    let demoStri = "ACGTACGT" in 
    let countA = ref 0 in
    let countC = ref 0 in
    let countT = ref 0 in
    let countG = ref 0 in
    for j = 0 to String.length demoStri - 1 do
        match demoStri.[j] with
        | 'A'-> countA := !countA +1
        | 'C'-> countC := !countC +1
        | 'T'-> countT := !countT +1
        | 'G'-> countG := !countG +1
        | _ -> assert false
    done;
    print_int !countA; print_string "\n";
    print_int !countC; print_string "\n";
    print_int !countT; print_string "\n";
    print_int !countG; print_string "\n"