Rmy*_*loR 5 r filter smoothing rolling-computation rolling-average
我对用于计算R中的滚动平均值的所有软件包都比较陌生,希望您能向我展示正确的方向。
我以以下数据为例:
ms <- c(300, 300, 300, 301, 303, 305, 305, 306, 308, 310, 310, 311, 312,
314, 315, 315, 316, 316, 316, 317, 318, 320, 320, 321, 322, 324,
328, 329, 330, 330, 330, 332, 332, 334, 334, 335, 335, 336, 336,
337, 338, 338, 338, 340, 340, 341, 342, 342, 342, 342)
correct <- c(1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0,
1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1,
1, 0, 0, 1, 0, 0, 1, 1, 0, 0)
df <- data.frame(ms, correct)
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ms是时间点(以毫秒为单位),correct是特定操作是否正确执行
(1 =正确,0 =不正确)。
我现在的目标是,我要计算固定毫秒数的窗口正确率(或平均值)百分比。如您所见,某些时间点丢失了,某些时间点出现了多次。因此,我不想基于行号进行过滤。我研究了诸如“ tidyquant”之类的某些程序包,但在我看来,这类程序包需要时间/日期变量而不是数字变量来确定取平均值的窗口。有没有办法指定数值的窗口df$ms?
试用:
library(dplyr)
# count the number of values per ms
df <- df %>%
group_by(ms) %>%
mutate(Nb.values = n())
# consider a window of 1 ms and compute the percentage for each window
df2 <- setNames(aggregate(correct ~ factor(df$ms, levels = as.character(seq(min(df$ms), max(df$ms), 1))),
df, sum),
c("ms", "Count.correct"))
# complete data frame (including unused levels)
df2 <- tidyr::complete(df2, ms)
df2$ms <- as.numeric(levels(df2$ms))[df2$ms]
df2 <- df2 %>% left_join(distinct(df[, c(1, 3)]), "ms")
# compute a rolling mean of the percentage of correct, with a width of 5
df2 %>%
mutate(Window = paste(ms, ms+4, sep = "-"), # add windows
Rolling.correct = zoo::rollapply(Count.correct, 5, sum, na.rm = T,
partial = TRUE, fill = NA, align = "left") /
zoo::rollapply(Nb.values, 5, sum, na.rm = T, partial = TRUE,
fill = NA, align = "left")) # add rolling mean
# A tibble: 43 x 5
ms Count.correct Nb.values Window Rolling.correct
<dbl> <dbl> <int> <chr> <dbl>
1 300 2 3 300-304 0.40
2 301 0 1 301-305 0.00
3 302 NA NA 302-306 0.25
4 303 0 1 303-307 0.25
5 304 NA NA 304-308 0.25
6 305 0 2 305-309 0.25
7 306 1 1 306-310 0.25
8 307 NA NA 307-311 0.00
9 308 0 1 308-312 0.20
10 309 NA NA 309-313 0.25
# ... with 33 more rows
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