使用rollapply计算每周的百分比变化
使用rollapply计算每周的百分比变化
提问于 2020-11-30 19:20:29
我有一些数据，如下所示：
date col1 col2 col3
<chr> <dbl> <dbl> <dbl>
1 2020_09_01 53542. 22133. 25295.
2 2020_09_02 54157. 22505. 25327.
3 2020_09_03 54137. 23115. 24993.
4 2020_09_04 50795. 23127. 24166.
5 2020_09_05 32829. 19600. 21860.我正在尝试使用rollapply函数来计算7天的百分比变化-或者每周的百分比变化。我似乎不能让rollapply像我期望的那样工作。rollapply将每天计算与前一周相比的百分比变化。
lagPeriod = 7
matrixCalcFunction <- function(x){
(myData[[x]] - myData[[x - lagPeriod]]) / myData[[x - lagPeriod]]
}
myData %>%
pivot_longer(cols = contains("col")) %>%
tidyquant::tq_mutate(
select = value,
mutate_fun = rollapply,
width = lagPeriod ,
align = "right",
FUN = matrixCalcFunction
)预期输出：
date col1 col2 col3
<chr> <dbl> <dbl> <dbl>
1 2020_09_01 NA NA NA
2 2020_09_02 NA NA NA
3 2020_09_03 NA NA NA
4 2020_09_04 NA NA NA
5 2020_09_05 NA NA NA
6 2020_09_06 NA NA NA
7 2020_09_07 -0.065 -0.055 -0.39
8 2020_09_08 -0.058 -0.029 -0.041
9 2020_09_09 0.068 0.071 0.039
10 2020_09_10 0.023 -0.0002 0.045数据：
myData <- structure(list(date = c("2020_09_01", "2020_09_02", "2020_09_03",
"2020_09_04", "2020_09_05", "2020_09_06", "2020_09_07", "2020_09_08",
"2020_09_09", "2020_09_10", "2020_09_11", "2020_09_12", "2020_09_13",
"2020_09_14", "2020_09_15", "2020_09_16", "2020_09_17", "2020_09_18",
"2020_09_19", "2020_09_20", "2020_09_21", "2020_09_22", "2020_09_23",
"2020_09_24", "2020_09_25", "2020_09_26", "2020_09_27", "2020_09_28",
"2020_09_29", "2020_09_30"), col1 = c(53542.497, 54156.934, 54136.844,
50794.971, 32828.797, 28475.082, 50083.573, 51017.288, 57819.908,
51945.242, 27823.172, 34349.466, 28527.527, 54845.664, 56531.057,
56556.415, 55396.121, 54303.732, 37513.441, 30041.867, 52397.815,
55449.939, 56203.125, 53654.182, 53289.437, 38511.761, 28046.879,
52132.573, 56055.611, 55520.683), col2 = c(22133.29, 22504.958,
23115.242, 23126.773, 19599.718, 16752.282, 20920.38, 21844.255,
24763.05, 23121.879, 17430.447, 20110.582, 18795.882, 24027.224,
24890.61, 24408.889, 24363.402, 24582.204, 20146.731, 18376.923,
23063.298, 24221.946, 25228.194, 24658.424, 23333.315, 20066.397,
17504.372, 23561.362, 23456.284, 24101.302), col3 = c(25294.573,
25326.797, 24992.764, 24166.084, 21859.885, 17549.005, 24306.496,
24269.409, 25968.326, 25253.976, 17974.404, 22636.375, 20105.166,
27000.274, 26291.22, 27277.371, 26851.75, 26133.317, 24055.107,
19515.875, 25573.014, 31957.279, 28961.316, 26896.495, 26440.726,
22941.927, 19990.825, 26595.878, 27725.468, 25965.802)), row.names = c(NA,
-30L), class = c("tbl_df", "tbl", "data.frame"))编辑：
这段代码可以运行，但我对diff(.x))/lag(.x, 7)感到有点困惑，也不确定它是否如我所愿，因为我得到了与预期输出不同的结果。
myData %>%
column_to_rownames("date") %>%
mutate(across(everything(), ~ round(c(NA, diff(.x))/lag(.x, 7), 5),
names = "{col}_delta"))对于单个观察(col1第1行和第7行)，我可以使用类似diff(c(pull(myData[7, 2]), pull(myData[1, 2]))) = 3458.924的内容，然后我可以将其划分为：3458.924 / pull(myData[1, 2]) = 0.0646。因此，在diff函数中添加类似于diff(c(.x, lag(.x, 7)))的内容就可以得到结果。
回答 1
Stack Overflow用户
回答已采纳
发布于 2020-11-30 19:57:18
将其转换为zoo对象，然后使用diff提供一个zoo对象。可以使用fortify.zoo(x)将其转换回数据帧，其中x是diff的结果。或者，将其保留为zoo对象，这样您就可以使用zoo的其他功能。
library(zoo)
z <- read.zoo(myData, format = "%Y_%m_%d")
diff(z, 6, arith = FALSE, na.pad = TRUE) - 1要使用rollapply而不是最后一行，请使用：
rollapplyr(z, 7, function(x) x[7] / x[1] - 1, fill = NA)或者使用lag.zoo：
z / lag(z, -6, na.pad = TRUE) - 1注意，dplyr阻塞了lag，所以要么确保它没有被加载，要么如果你需要它，就使用library(dplyr, exclude = c("filter", "lag"))加载它。
关于编辑，尝试这样做：
library(dplyr, exclude = c("lag", "filter"))
myData %>% mutate(across(-1, ~ . / dplyr::lag(., 6) - 1))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/65072580
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