如何解决如何缩短此代码以避免冗余?
这是我的代码片段:
string trend()
{
double emaTrend0 = iMA (NULL,200,MODE_EMA,PRICE_CLOSE,0);
double emaTrend1 = iMA (NULL,1);
double emaTrend2 = iMA (NULL,2);
double emaTrend3 = iMA (NULL,3);
double emaTrend4 = iMA (NULL,4);
double emaTrend5 = iMA (NULL,5);
double emaTrend6 = iMA (NULL,6);
double emaTrend7 = iMA (NULL,7);
double emaTrend8 = iMA (NULL,8);
double emaTrend9 = iMA (NULL,9);
string signal;
double HighCandle0 = High[0];
double HighCandle1 = High[1];
double HighCandle2 = High[2];
double HighCandle3 = High[3];
double HighCandle4 = High[4];
double HighCandle5 = High[5];
double HighCandle6 = High[6];
double HighCandle7 = High[7];
double HighCandle8 = High[8];
double HighCandle9 = High[9];
double LowCandle0 = Low[0];
double LowCandle1 = Low[1];
double LowCandle2 = Low[2];
double LowCandle3 = Low[3];
double LowCandle4 = Low[4];
double LowCandle5 = Low[5];
double LowCandle6 = Low[6];
double LowCandle7 = Low[7];
double LowCandle8 = Low[8];
double LowCandle9 = Low[9];
if (emaTrend0 > HighCandle0 && emaTrend1 > HighCandle1 && emaTrend2 > HighCandle2 &&emaTrend3 > HighCandle3 &&
emaTrend4 > HighCandle4 &&emaTrend5 > HighCandle5 &&emaTrend6 > HighCandle6 &&emaTrend7 > HighCandle7 &&
emaTrend8 > HighCandle8 && emaTrend9 > HighCandle9 )
signal = "downtrend";
else if (emaTrend0 < LowCandle0 && emaTrend1 < LowCandle1 && emaTrend2 < LowCandle2 &&emaTrend3 < LowCandle3 &&
emaTrend4 < LowCandle4 &&emaTrend5 < LowCandle5 &&emaTrend6 < LowCandle6 &&emaTrend7 < LowCandle7 &&
emaTrend8 < LowCandle8 && emaTrend9 < LowCandle9 )
signal = "uptrend";
return signal;
}
我想要发生的是循环变量,并且在产生诸如下降趋势或上升趋势之类的信号之前,每个条件都必须是true
。
下面的代码是我的尝试,但还是不行。当条件之一为 true
时,我的尝试产生了下降趋势 或上升趋势 的信号,但我需要的是一切都在 true
之前产生信号。
string trend() {
string signal = "";
for (int i = 0; i<=9 ; i++){
double emaTrend = iMA (NULL,i);
double highCandle = High[i];
double lowCandle = Low[i];
if (emaTrend > highCandle){
signal = "downtrend";
}
else if (emaTrend < lowCandle){
signal = "uptrend";
}
}
return signal;
}
我们将不胜感激所有回复。谢谢!
解决方法
这是一个可能的解决方案,通过计算下降和上升趋势的数量,然后比较这些计数(未测试):
std::string trend() {
const std::size_t max_count = 10;
std::size_t count_downtrend = 0;
std::size_t count_uptrend = 0;
for (std::size_t i = 0; i != max_count; ++i) {
const double emaTrend = iMA (NULL,200,MODE_EMA,PRICE_CLOSE,i);
const double highCandle = High[i];
const double lowCandle = Low[i];
if (emaTrend > highCandle) {
++count_downtrend;
} else if (emaTrend < lowCandle) {
++count_uptrend;
}
}
std::string signal;
if (count_downtrend == max_count) {
signal = "downtrend";
} else if (count_uptrend == max_count) {
signal = "uptrend";
}
return signal;
}
,
这与常规的“线性搜索循环”非常相似,如果趋势被打破,则提前突破:
enum Direction
{
Up,Down,None
};
Direction direction(double value,double low,double high)
{
return value < low ? Down : (value > high ? Up : None);
}
double emaClose(int i)
{
return iMA(NULL,i);
}
string trend()
{
const Direction d = direction(emaClose(0),Low[0],High[0]);
if (d == None)
{
return "";
}
for (int i = 1; i < 10 ; i++)
{
if (direction(emaClose(i),Low[i],High[i]) != d)
{
return "";
}
}
return d == Up ? "uptrend" : "downtrend";
}
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