我有2个相等长度的数组.以下函数尝试使用这些数组来计算斜率.它返回每个点之间的斜率的平均值.对于以下数据集，我似乎得到的值不同于Excel和Google文档.
I have 2 arrays of equal length. The following function attempts to calculate the slope using these arrays. It returns the average of the slope between each points. For the following data set, I seem to be getting different values than Excel and Google Docs.
double[] x_values = { 1932, 1936, 1948, 1952, 1956, 1960, 1964, 1968, 1972, 1976, 1980 }; double[] y_values = { 197, 203, 198, 204, 212, 216, 218, 224, 223, 225, 236 }; public static double getSlope(double[] x_values, double[] y_values) throws Exception { if (x_values.length != y_values.length) throw new Exception(); double slope = 0; for (int i = 0; i < (x_values.length - 1); i++) { double y_2 = y_values[i + 1]; double y_1 = y_values[i]; double delta_y = y_2 - y_1; double x_2 = x_values[i + 1]; double x_1 = x_values[i]; double delta_x = x_2 - x_1; slope += delta_y / delta_x; } System.out.println(x_values.length); return slope / (x_values.length); }输出
Google:0.755
Google: 0.755
getSlope():0.962121212121212
getSlope(): 0.962121212121212
Excel:0.7501
Excel: 0.7501
推荐答案
我敢打赌其他两种方法都在计算最小二乘拟合，而您不适合.
I bet the other two methods are computing the least-squares fit, whereas you are not.
当我使用 R 验证这个猜想时，我也得到了约0.755的斜率:
When I verify this conjecture using R, I too get the slope of about 0.755:
> summary(lm(y~x)) Call: lm(formula = y ~ x) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -1.265e+03 1.793e+02 -7.053 5.97e-05 *** x 7.551e-01 9.155e-02 8.247 1.73e-05 ***相关数字为7.551e-01.还值得注意的是，这条生产线的截距约为-1265.
The relevant number is the 7.551e-01. It is also worth noting that the line has an intercept of about -1265.
这是最小二乘拟合的图片:
Here is a picture of the least-squares fit:
有关在代码中实现此操作的信息，请参见使用Java计算最小二乘法
As to implementing this in your code, see Compute least squares using java