有没有办法使这个C开关盒更简单?
我用逻辑回归模型得到了一个公式。因为我把每小时作为一个因素,所以这个开关箱很长,不方便修改,而且不美观。有没有办法简化它?也许我应该试着用矩阵做算术有没有办法使这个C开关盒更简单?,c,logistic-regression,C,Logistic Regression,我用逻辑回归模型得到了一个公式。因为我把每小时作为一个因素,所以这个开关箱很长,不方便修改,而且不美观。有没有办法简化它?也许我应该试着用矩阵做算术 ... #define elif else if ... // hours switch (hours) { case 0: prob[2] = prob_base[7]; break; case 1: prob[2] = prob_base[8]; break;
...
#define elif else if
...
// hours
switch (hours) {
case 0:
prob[2] = prob_base[7];
break;
case 1:
prob[2] = prob_base[8];
break;
case 2:
prob[2] = prob_base[9];
break;
case 3:
prob[2] = prob_base[10];
break;
case 4:
prob[2] = prob_base[11];
break;
case 5:
prob[2] = prob_base[12];
break;
case 6:
prob[2] = prob_base[13];
break;
case 7:
prob[2] = prob_base[14];
break;
case 8:
prob[2] = prob_base[15];
break;
case 9:
prob[2] = prob_base[16];
break;
case 10:
prob[2] = prob_base[17];
break;
case 11:
prob[2] = prob_base[18];
break;
case 12:
prob[2] = prob_base[19];
break;
case 13:
prob[2] = prob_base[20];
break;
case 14:
prob[2] = prob_base[21];
break;
case 15:
prob[2] = prob_base[22];
break;
case 16:
prob[2] = prob_base[23];
break;
case 17:
prob[2] = prob_base[24];
break;
case 18:
prob[2] = prob_base[25];
break;
case 19:
prob[2] = prob_base[26];
break;
case 20:
prob[2] = prob_base[27];
break;
case 21:
prob[2] = prob_base[28];
break;
case 22:
prob[2] = prob_base[29];
break;
case 23:
prob[2] = prob_base[30];
break;
default:
prob[2] = 0;
break;
}
// bidf
prob[3] = prob_base[31] * atof(bidf);
// isp
switch (isp) {
case 1:
prob[4] = prob_base[32];
break;
case 2:
prob[4] = prob_base[33];
break;
case 3:
prob[4] = prob_base[34];
break;
default:
prob[4] = 0;
break;
}
...
大概是这样的:
if (hours >= 0 && hours < 24)
prob[2] = prob_base[hours + 7];
else
prob[2] = 0;
prob[3] = prob_base[31] * atof(bidf);
if (isp >= 1 && isp < 4)
prob[4] = prob_base[isp + 31];
else
prob[4] = 0;
if(小时数>=0&&hours<24)
概率[2]=概率基数[h+7];
其他的
prob[2]=0;
prob[3]=prob_base[31]*atof(bidf);
如果(isp>=1&&isp<4)
prob[4]=prob_基[isp+31];
其他的
prob[4]=0;
arrx[2]=arry[i+7]也许prob[2]=prob\u base[hours+7]
?离题:这应该继续代码审查。顺便说一句,它们在C中不是关键字elif
。你看不到这里的模式吗?@Stargateur否定,这绝对不能进入代码审查,因为它不起作用。
//hours
prob[2] = prob_base[hours+7]; //no switch required