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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1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
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1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
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。
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/