r - 当我的数据未正确对齐时计算均方根误差-6ren

r - 当我的数据未正确对齐时计算均方根误差

转载作者：行者123 更新时间：2023-11-30 09:28:16

我有这些数据，我试图计算实际值和预测值的均方根误差:

头

# A time tibble: 6 x 4
# Index: index
  IRI_KEY index      value key   
    <dbl> <date>     <dbl> <fct> 
1  648459 2005-01-31  1.43 actual
2  648459 2005-02-07  1.16 actual
3  648459 2005-02-14  1.22 actual
4  648459 2005-02-21  1.16 actual
5  648459 2005-02-28  1.04 actual
6  648459 2005-03-07  1.45 actual

尾部

# A time tibble: 6 x 4
# Index: index
  IRI_KEY index      value key    
    <dbl> <date>     <dbl> <fct>  
1      NA 2011-12-12  1.79 predict
2      NA 2011-12-19  1.76 predict
3      NA 2011-12-26  1.76 predict
4      NA 2012-01-02  1.67 predict
5      NA 2012-01-09  1.64 predict
6      NA 2012-01-16  1.69 predict

首先，我尝试使用该列中的相同 ID 键填充 NA 值(这些 ID 键在每个数据帧上都会发生变化)。因此，“实际”结果有一个分配给它们的 ID 键，但“预测”结果由于某种原因没有分配给它们。

其次，我尝试计算“实际”和“预测”的均方根误差。在我使用 spread 函数后，由于“实际”和“预测”两列中的 NA 值，我所返回的结果为“NaN”。

如何计算均方根误差或如何配置数据以使日期匹配？

我训练了一个模型，直到 2011-01-24 日期，并从 2011-01-24 到 2012-01-16 对其进行了测试

 rmse_calculation <- 
    df %>%
      spread(key = key, value = value) %>%
    rename(truth    = actual,
           estimate = predict)
  rmse(truth, estimate)

数据:

df <- structure(list(IRI_KEY = c(648459, 648459, 648459, 648459, 648459, 
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648459, 648459, 648459, 648459, 648459, 648459, 648459, 648459, 
648459, 648459, 648459, 648459, 648459, 648459, 648459, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
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NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
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2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("actual", 
"predict"), class = "factor")), row.names = c(NA, -416L), index_quo = ~index, index_time_zone = "UTC", class = c("tbl_time", 
"tbl_df", "tbl", "data.frame"))

编辑:

该模型从 2005-01-31 开始，到 2012-01-16 结束，并且有每周周期。模型中有 364 周(364/52 = 7 年)。我在前 6 年训练了模型(从 2005-01-31 到 2011-01-17) )，并在最后一年测试了模型(从 2011-01-24 周到 2012-01-16 )。

我有去年的预测，也有这一时期的实际值。我正在尝试计算我的预测或过去 52 周的 rmse。

编辑2:

所以基本上在查看 rmse_calculation 表(第 364 行)之后，我尝试“推高”预测列，然后删除预测列中的所有 NA 值，其中我将只剩下 52 个观察值，然后我可以计算52周的rmse。

编辑3:

填写 IRI_KEY 列并不那么重要。

最佳答案

看来我们可以安全地丢弃IRI_KEY以扩展键值index。有了这个，我们可以进行左连接或扩展来获得有效的相同关联:

df %>%
  select(-IRI_KEY) %>%
  spread(key, value) %>%
  filter(complete.cases(.))
# # A tibble: 52 x 3
#    index      actual predict
#    <date>      <dbl>   <dbl>
#  1 2011-01-24   1.39    1.54
#  2 2011-01-31   1.50    1.50
#  3 2011-02-07   1.26    1.45
#  4 2011-02-14   1.40    1.50
#  5 2011-02-21   1.44    1.47
#  6 2011-02-28   1.60    1.53
#  7 2011-03-07   1.53    1.52
#  8 2011-03-14   1.55    1.51
#  9 2011-03-21   1.36    1.49
# 10 2011-03-28   1.48    1.52
# # ... with 42 more rows
df %>%
  select(-IRI_KEY) %>%
  spread(key, value) %>%
  filter(complete.cases(actual, predict)) %>%
  with(., ModelMetrics::rmse(actual, predict))
# [1] 0.3130566

我们必须使用 filter(complete.cases(actual, Predict)) 因为 rmse 不需要 NA 值，并且它不接受其他 R 函数的常用标准 na.rm=TRUE。

<小时/>

这种spread方法的缺点是它会丢弃您的IRI_KEY，因为(正如@MrFlick强调的那样)它不会在您的预测步骤中传输。另一种方法是将您的预测值左连接到相同索引行上:

df %>%
  filter(key == "predict") %>%
  select(index, value) %>%
  left_join(filter(df, key == "actual"), by="index") %>%
  rename(actual = value.y, predict = value.x)
# # A tibble: 52 x 5
#    index      predict IRI_KEY actual key   
#    <date>       <dbl>   <dbl>  <dbl> <fct> 
#  1 2011-01-24    1.54  648459   1.39 actual
#  2 2011-01-31    1.50  648459   1.50 actual
#  3 2011-02-07    1.45  648459   1.26 actual
#  4 2011-02-14    1.50  648459   1.40 actual
#  5 2011-02-21    1.47  648459   1.44 actual
#  6 2011-02-28    1.53  648459   1.60 actual
#  7 2011-03-07    1.52  648459   1.53 actual
#  8 2011-03-14    1.51  648459   1.55 actual
#  9 2011-03-21    1.49  648459   1.36 actual
# 10 2011-03-28    1.52  648459   1.48 actual
# # ... with 42 more rows

这允许我们同样使用rmse函数:

df %>%
  filter(key == "predict") %>%
  select(index, value) %>%
  left_join(filter(df, key == "actual"), by="index") %>%
  rename(actual = value.y, predict = value.x) %>%
  with(., ModelMetrics::rmse(actual, predict))
# [1] 0.3130566

注意:我没有开始使用这种方法，因为输出表明我知道预测值与我不知道的 IRI_KEY 值相关联(只有你这样做)。如果您不确定日期是否提供了足够的相关性来识别 key ，那么这种方法就是错误的，并且可能/将会导致稍后的分析管道中的错误推论。

关于r - 当我的数据未正确对齐时计算均方根误差，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/54296434/