r - 使用基于最大差异的值查找对数差异
问题描述
这应该很简单,但我很难做到。我使用生成汇总列lag = 15
的基本函数创建了一个汇总输出,其中包括跨越 15 个值 () 的值的最大差异。但是,我现在想使用那些用于计算但通过计算它们之间的对数差异的相同值。我从包中找到,但我在使用与计算.diff()
max_predicted_diff
max_predicted_diff
diffLog
dse
dplyr
max_predicted_diff
例子
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
library(dplyr)
df %>% summarise(max_predicted_diff = max(diff(Predicted, lag = 15)))
max_predicted_diff
1 0.04670478
如何找出用于找到0.04670478答案的值?然后我如何总结使用的这两个值的日志?我曾经max()
找到过,max_predicted_diff
但我会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,因为我认为diffLog
不会使用与max_predicted_diff
(just log
ed) 相同的值?
diffLog()
从包中使用,dse
我可以轻松计算对数差异,但我不知道它使用了哪些值,以及如何使用与 find 相同的值max_predicted_diff
。
library(dse)
df %>% summarise(max_predicted_diff_log = max(diffLog(Predicted, lag = 15)))
max_predicted_diff_log
1 0.06154992
可重现的数据
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")
解决方案
这就是你所追求的吗?
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
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