如何判断3个移动平均线在4根蜡烛或以下的持续时间内何时穿过

问题描述

我一直在研究2个移动平均线交叉，这很简单。我想添加三分之一，然后尝试找出是否在4支蜡烛或更少的蜡烛内发生这种情况。

对于两个移动平均线，我使用以下内容：

  // if shift5MA > shift0MA 
   if (shift5MAArray[1] > shift0MAArray[1]) {
      
      //if shift5MA < shift0MA VALID SELL
      if (shift5MAArray[2] < shift0MAArray[2]) {
         signal = "sell";
      }
      }
   
   if (shift5MAArray[1] < shift0MAArray[1]) {
      
      //if shift5MA > shift0MA
      if (shift5MAArray[2] > shift0MAArray[2]) {
         signal = "buy";
      }
      }

如何检查3个移动均线何时在4根蜡烛或更少的蜡烛内彼此交叉，如图像中的三根交叉：

解决方法

此代码并未针对速度进行优化，但总体上说明了如何解决此问题。该答案仅回答您的问题，即发现三个移动平均线是否交叉。它不会给您卖出或买入信号，但是您可以通过检查diff阵列中的方向变化来轻松实现该信号。

注意：由于在4个烛台的范围内看，该代码当前可以给出“三个MA均已交叉”的重复。

import numpy as np

MA1 = np.asarray([0,1,4,3,5,6,7,8,9,10])
MA2 = np.asarray([0,2,10,11])
MA3 = np.asarray([0,11,12,13])

haveCrossed = False
for i in range(len(MA1)-3):
  # These are the differences between the moving averages,if one MA
  # crosses another the sign of the difference changes from positive to 
  # negative or vice versa.
  diff1 = MA1[i:i+4] - MA2[i:i+4]
  diff2 = MA1[i:i+4] - MA3[i:i+4]
  diff3 = MA2[i:i+4] - MA3[i:i+4]

  # Check if all signs are equal. If the signs are equal,the moving averages
  # did not intersect.

  # Check if MA1 and MA2 crossed.
  if np.all(diff1 > 0) if diff1[0] > 0 else np.all(diff1 < 0):
    cross1Flag = False
  else:
    cross1Flag = True

  # Check if MA1 and MA3 crossed.
  if np.all(diff2 > 0) if diff2[0] > 0 else np.all(diff2 < 0):
    cross2Flag = False
  else:
    cross2Flag = True

  # Check if MA2 and MA3 crossed.
  if np.all(diff3 > 0) if diff3[0] > 0 else np.all(diff3 < 0):
    cross3Flag = False
  else:
    cross3Flag = True

  if cross1Flag and cross2Flag and cross3Flag:
    haveCrossed = True
    print(f"The three Moving Averages crossed at time: [{i},{i+3}]")

if not haveCrossed:
  print("The three Moving Averages have not crossed.")

输出：

The three Moving Averages crossed at time: [0,3]
The three Moving Averages crossed at time: [1,4]