使用基于最大差异的值查找对数差异

Question

这应该非常简单，但是我很难过。我使用基数lag = 15函数生成汇总列diff()，创建了一个包含15个值（max_predicted_diff）的最大差异的汇总输出。但是，我现在想使用那些用于计算max_predicted_diff的相同值，但要通过计算它们之间的对数差来进行计算。我从diffLog包中找到了dse，但随后出现了dplyr问题，原因是使用了与计算max_predicted_diff所用的相同值。

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示例

print(head(df,10))

   Sample Predicted
1   apple 0.7356986
2   apple 0.7388222
3   apple 0.7419447
4   apple 0.7450658
5   apple 0.7481857
6   apple 0.7513042
7   apple 0.7544212
8   apple 0.7575368
9   apple 0.7606509
10  apple 0.7637635

df %>% summarise(max_predicted_diff = max(diff(Predicted, lag = 15)))

  max_predicted_diff
1               0.04670478

我如何找出使用哪些值来找到0.04670478的答案？然后，如何总结所使用的这两个值的日志？我已经使用max()查找了max_predicted_diff，但是我将使用什么汇总函数来解决日志值的差异？我认为max()在这里不起作用，因为我不认为diffLog将使用与max_predicted_diff相同的值（只是log ed）？

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使用diffLog()包中的dse，我可以轻松计算对数差异，但是我不知道它使用了哪个值以及如何使用与查找max_predicted_diff相同的值。

df %>% summarise(max_predicted_diff_log = max(diffLog(Predicted, lag = 15)))
  max_predicted_diff_log

1               0.06154992

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可复制数据

df structure(list(Sample = c("apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple", "apple", "apple", "apple", 
"apple", "apple", "apple", "apple"), Predicted = c(0.735698569365871, 
0.738822222617743, 0.741944657028027, 0.74506582323819, 0.748185672193904, 
0.751304155146149, 0.754421223652273, 0.75753682957702, 0.760650925093515, 
0.76376346268421, 0.766874395141795, 0.76998367557007, 0.773091257384776, 
0.776197094314395, 0.779301140400904, 0.782403350000502, 0.785503677784295, 
0.788602078738943, 0.791698508167276, 0.794792921688872, 0.797885275240596, 
0.800975525077113, 0.804063627771357, 0.807149540214967, 0.810233219618698, 
0.813314623512785, 0.81639370974728, 0.819470436492359, 0.822544762238589, 
0.825616645797166, 0.828686046300123, 0.831752923200501, 0.83481723627249, 
0.837878945611542, 0.84093801163445, 0.843994395079395, 0.847048057005967, 
0.850098958795148, 0.853147062149278, 0.856192329091979, 0.859234721968058, 
0.862274203443374, 0.865310736504688, 0.868344284459473, 0.871374810935701, 
0.874402279881605, 0.877426655565415, 0.880447902575054, 0.883465985817829, 
0.886480870520078, 0.889492522226799, 0.892500906801256, 0.895505990424551, 
0.898507739595181, 0.901506121128565, 0.904501102156547, 0.907492650126881, 
0.910480732802683, 0.913465318261867, 0.91644637489656, 0.919423871412485, 
0.922397776828334, 0.925368060475109, 0.928334691995449)), row.names = c(NA, 
64L), class = "data.frame")

Answer 1

这是您的追求吗？

library(dplyr, warn.conflicts = FALSE)

df %>% 
  mutate(
    lag15 = lag(Predicted, n = 15),
    lag_diff = Predicted -  lag15
  ) %>%
  filter(lag_diff == max(lag_diff, na.rm = TRUE))

#>   Sample Predicted     lag15   lag_diff
#> 1  apple 0.7824034 0.7356986 0.04670478