我在尝试计算实际值和预测值的均方根时拥有这些数据：
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
尾巴
# 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，我尝试在该列中使用相同的ID密钥填充NA值（这些ID密钥在每个数据帧上都会发生变化）。因此，“实际”结果已分配了一个ID密钥，但“预测”结果却由于某种原因而没有。
第二，我试图计算“实际”和“预测”的均方根值。使用spread函数后，由于两列“实际”和“预测”中的NA值，我返回的是“ NaN”。
如何计算均方根值或如何配置数据以使日期匹配？
我训练了一个模型，直到日期2011-01-24，并从2011-01-24到2012-01-16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,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
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。该模型中有364周（364/52 = 7年）。我在前6年（从2005-01-31到2011-01-17)训练了模型，并在最后一年（从2011-01-24到2012-01-16的一周）测试了模型。
我有去年的预测，我也有此期间的实际值。我正在尝试计算我有预测或过去52周的均方根值。
编辑2：
因此，基本上，在查看rmse_calculation表（第364行）之后，我试图“上推”预测列，然后删除预测列中的所有NA值，而我只剩下52观察，然后我可以计算52周的均方根值。
编辑3：
IRI_KEY列中的填充不是那么重要。
答案 0 :(得分：3)
似乎我们可以安全地丢弃IRI_KEY以便扩展键值index。这样，我们可以进行左连接或散布来有效地实现相同的关联：
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不需要任何NA值，并且它不接受其他R函数常用的标准na.rm=TRUE。
此spread方法的缺点是它会丢弃您的IRI_KEY，因为（如@MrFlick突出显示的那样）它不会在您的预测步骤中传输。另一种方法是将您的predict值左连接到相同的index行中：
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
这使我们可以与rmse函数相同的用法：
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：我不是从这种方法开始的，因为输出表明我知道预测值与IRI_KEY值相关，我不知道（只有您知道）。如果您不确定日期是否可以提供足够的相关性来识别键，则此方法是有缺陷的，并且可能/将在以后的分析流程中导致错误的推论。