使用 For 循环计算两个数据集的错误

Question

我有 2 个具有相同列名的数据集。

一个数据框是 predicted，每个变量有 12 个预测值。另一个数据框是 real，每个变量有 12 个实数值。

我正在尝试使这个 for 循环打印均方预测误差。

for (i in names(predicted)){
  print(paste("The Mean Squared Forecast Error for",i,"is:",mean((real$i-predicted$i)^2)))
}

这是期望的输出：

[1] "The Mean Squared Forecast Error for EMPL is: 2.22973256477949e-06"
[1] "The Mean Squared Forecast Error for CPI is: 5.15514699434535e-06"
[1] "The Mean Squared Forecast Error for FFR is: 0.0295111061402315"

我不知道为什么for循环给我的计算是NaN

提前致谢！

数据

> dput(real)
structure(list(CPI = c(0.00160433859088815, -0.00215862261991973, 
-4.59020776588659e-06, -0.000538046208090392, 0.00265772082268079, 
-0.00054730999077357, -0.00265434619923699, -0.000529520203581768, 
0.00263754860468435, 0.00208864328030867, -0.000548981800946535, 
-0.00470589103741226), EMPL = c(0.000440619330092318, 0.000505389630054864, 
-0.00064261089956652, 0.00163820296223705, 0.00123949686537728, 
0.00231633116306718, 0.00274100555496126, 0.000121807479846581, 
-0.000616354925266904, 0.00175489535365614, 0.00356476054636445, 
-0.000756181462504557), FFR = c(0.02, 0.01, -0.01, 0, 0, 0.03, 
0.23, 0.17, 0.18, 0.15, 0.17, 0.23)), class = "data.frame", row.names = c(NA, 
-12L))

> dput(predicted)
structure(list(CPI = c(-0.00185325936075232, -0.000205612911651649, 
0.000515554281526663, 0.000832059799115526, -9.59291577299353e-05, 
-0.000135397534643682, -0.000319969700328926, -0.000531443546314851, 
0.000334709759798288, 0.000149004361515839, -0.000190722371242093, 
-6.3736415706231e-05), EMPL = c(0.00180490246150916, 0.00155636832886101, 
0.00142059674176869, 0.00104389866040043, 0.0016108887176987, 
0.00152450457655764, 0.00106805136354765, 0.00151749850981702, 
0.00126307148334478, 0.00121395424088581, 0.0013782559420265, 
0.00149406940983166), FFR = c(0.0664586752084257, 0.228921245733048, 
0.135151339424908, -0.162398412959252, -0.128277890243452, -0.134292810077186, 
-0.049475881436482, 0.0442366320282881, -0.0376918344723898, 
-0.018180368465285, 0.0655870731024621, 0.0492813528000579)), class = "data.frame", row.names = c(NA, 
-12L))

Answer 1

请注意，如果i <- colname，则data$i不会产生相应的列。请改用 data[[i]]。

for (i in names(predicted)){
  print(paste("The Mean Squared Forecast Error for",i,"is:",mean((real[[i]]-predicted[[i]])^2)))
}

使用 For 循环计算两个数据集的错误

Compute error with two datasets using For loop

for-loop

r