Pine Script需求：4小时级别MA50与MA100交叉点绘射线并自动删除

我是Pine Script新手，希望实现以下功能：

• 在4小时分辨率下，捕捉MA50（50周期简单移动平均线）与MA100（100周期简单移动平均线）的交叉点
• 在交叉点位置绘制水平射线（向右延伸）
• 当价格穿越射线后，自动删除所有已创建的射线

最初代码（无法实现需求） #

//@version=5
indicator("CU",shorttitle = "CU", overlay=true)

ma50 = ta.sma(close, 50)
ma100 = ta.sma(close, 100)
bcu = ma50 == ma100

bcu4h = input.timeframe("4H", "BCU4H")
bcu_sec = request.security(syminfo.tickerid, bcu4h, close)

line_bcu1d = line.new(na, na, na, na, style=line.style_solid,extend=extend.right, color=color.green)

//PLOT
plot(ma50, "MA50", color.rgb(255, 255, 255, 0), 2)
plot(ma100, "MA100", color.rgb(255, 235, 59, 0), 2)

调整后代码（射线绘制成功，但无法自动删除） #

//@version=5
indicator("CU2",shorttitle = "CU2", overlay=true)

var float   last_intersect_x    = na
var float   last_intersect_y    = na

// Gets intersection point of two line segments [(x1, y1), (x2, y2)] and [(u1, v1), (u2, v2)].
f_get_intersect_coordinates(x1, y1, x2, y2, u1, v1, u2, v2) =>
    x = -1 * ((x1 - x2) * (u1 * v2 - u2 * v1) - (u2 - u1) * (x2 * y1 - x1 * y2)) / ((v1 - v2) * (x1 - x2) - (u2 - u1) * (y2 - y1))
    y = -1 * (u1 * v2 * y1 - u1 * v2 * y2 - u2 * v1 * y1 + u2 * v1 * y2 - v1 * x1 * y2 + v1 * x2 * y1 + v2 * x1 * y2 - v2 * x2 * y1) / (-1 * u1 * y1 + u1 * y2 + u2 * y1 - u2 * y2 + v1 * x1 - v1 * x2 - v2 * x1 + v2 * x2)
    [x,y]

_ma50 = ta.sma(close, 50)
_ma100 = ta.sma(close, 100)
line line_bcu = na

if ta.crossover(_ma50, _ma100)
    x1 = bar_index[1]
    y1 = _ma50[1]
    x2 = bar_index
    y2 = _ma50
    u1 = bar_index[1]
    v1 = _ma100[1]
    u2 = bar_index
    v2 = _ma100
    
    [x,y] = f_get_intersect_coordinates(x1, y1, x2, y2, u1, v1, u2, v2)
    
    last_intersect_x := x
    last_intersect_y := y

    line_bcu := line.new(bar_index-1, last_intersect_y, bar_index, last_intersect_y, extend=extend.right, color=color.rgb(255, 255, 255), style=line.style_solid, width = 2)
//=====================================================
//I tried with this loop, but it is not working, only some of the lines are deleted

for i = 0 to 100
        if line.get_y1(line_bcu[i]) > low
            line.delete(line_bcu[i])
//=====================================================
plot(_ma50, "MA50", color.rgb(255, 255, 255, 0), 2)
plot(_ma100, "MA100", color.rgb(255, 235, 59, 0), 2)

正确解决方案代码 #

//@version=5
indicator("MA Cross Ray with Auto-Delete", shorttitle="MA Ray", overlay=true)

// 输入参数
ma_len1 = input.int(50, "MA50 周期")
ma_len2 = input.int(100, "MA100 周期")
target_tf = input.timeframe("4H", "目标周期")
ray_color = input.color(color.new(color.white, 0), "射线颜色")
ray_width = input.int(2, "射线宽度")

// 用数组存储所有创建的射线对象
var array<line> ray_array = array.new_line()

// 在目标周期（4H）获取MA交叉信号与交叉价格
[ma50_4h, ma100_4h, cross_over, cross_under] = request.security(syminfo.tickerid, target_tf, 
     [ta.sma(close, ma_len1), ta.sma(close, ma_len2), ta.crossover(ta.sma(close, ma_len1), ta.sma(close, ma_len2)), ta.crossunder(ta.sma(close, ma_len1), ta.sma(close, ma_len2))])

// 当目标周期出现交叉时，绘制水平射线
if cross_over or cross_under
    // 获取交叉点的价格（取两个MA在交叉K线的平均值）
    cross_price = (ma50_4h + ma100_4h) / 2
    // 创建射线，从当前K线左边缘延伸到右边缘并向右扩展
    new_ray = line.new(bar_index - 1, cross_price, bar_index, cross_price, extend=extend.right, color=ray_color, width=ray_width)
    // 将射线对象存入数组
    array.push(ray_array, new_ray)

// 遍历数组，检查价格是否穿越射线，符合条件则删除
if barstate.islast
    // 倒序遍历避免数组索引混乱
    for i = array.size(ray_array) - 1 downto 0
        current_ray = array.get(ray_array, i)
        ray_price = line.get_y1(current_ray)
        // 判断价格是否穿越射线（当前K线的low低于射线价格 或 high高于射线价格）
        if low < ray_price or high > ray_price
            line.delete(current_ray)
            array.remove(ray_array, i)

// 绘制当前周期的MA（可选，用于参考）
plot(ta.sma(close, ma_len1), "MA50", color=color.white, linewidth=2)
plot(ta.sma(close, ma_len2), "MA100", color=color.yellow, linewidth=2)

关键改进点说明 #

1. 数组管理射线对象：用array<line>类型数组保存所有射线，解决单个变量无法批量管理的问题，方便遍历和删除操作。
2. 目标周期信号获取：通过request.security()在4H周期内判断MA交叉，确保交叉信号来自指定的4小时分辨率，而非当前图表周期。
3. 精准穿越判断：判断当前K线的low是否低于射线价格，或high是否高于射线价格，满足任一条件即删除射线。
4. 倒序遍历数组：删除数组元素时采用倒序遍历，避免因数组长度变化导致的索引错位问题。

内容的提问来源于stack exchange，提问作者xlameee