如何建立数组或集合

Question

我想建立多个最高点的集合，并像在数组中一样访问所有这些最高点。我怎么做？我的代码如下：

var float lastHigh = 0
if (highfound)
    lastHigh := high

现在我尝试这个：

x := lastHigh[3]

...但是x只是最后的lastHigh值（就像它将是lastHigh [1]）。所以lastHigh只是一个“扁平”变量，对吧？我如何收集多于最后的高点？

Answer 1

假设您的highfound函数未在原始问题中演示，但结果为false或true，则您的代码应通过从:变量中删除x来工作：

x = lastHigh[3]

以下内容对我来说很好：

//@version=4
study("My Script")

var float lastHigh = 0
if (close>open)
    lastHigh := high

x = lastHigh[3]
plot(x, color=color.red)

Answer 2

我想出了下一种方法：

//@version=4
study("Three last highs", overlay=true)

// here goes the logic for finding the high
// NOTE: if you are updating this function,
// then it should return either new value for the array or n/a. 
// Because the non-n/a value will be added to the array and the oldest removed
getNewHighOrNa() =>
    pivothigh(high, 3, 3)

newHigh = getNewHighOrNa()




// ======= Array lifecicle =========
ARRAY_LENGTH = 3

arrayCell = label(na)
if bar_index < ARRAY_LENGTH
    arrayCell := label.new(0, 0)
    label.set_y(arrayCell, 0)
else
    if na(newHigh)
        val = label.get_y(arrayCell[1])
        for i = 2 to ARRAY_LENGTH
            v = label.get_y(arrayCell[i])
            label.set_y(arrayCell[i-1], v)
        arrayCell := arrayCell[ARRAY_LENGTH]
        label.set_y(arrayCell, val)
    else
        arrayCell := arrayCell[ARRAY_LENGTH]
        label.set_y(arrayCell, newHigh)
// ==================================

// Array getter by index. Note, that numeration is right to left
// so 0 is the last bar (current) and 1 it's a left to current bar
get(index) => label.get_y(arrayCell[index])

// example of using of the array for calculation average of all elements of the array
mean() =>
    sum = 0.0
    for i = 0 to ARRAY_LENGTH - 1
        sum := sum + get(i) / ARRAY_LENGTH
    sum
plot(mean(), color=color.green)




// the rest is just checking that the code works:
plot(get(0))
plot(get(1))
plot(get(2))

// mark the bars where we found new highs
plotshape(not na(newHigh), style=shape.flag)

我试图使它变得更容易，但是由于松木的限制，没有对其进行管理。但是此代码有效，您可以获得最后的3个局部高点，并且可以对其进行迭代。希望我能正确回答你的问题。