在 R 中，如何从更大的数据集中创建单独的时间序列（循环并运行 Mann-Kendall 测试）？

Question

我开始自动化用于水质数据时间序列数据的 Mann-Kendall 测试。以前是用Excel做的，但是复制粘贴太多了。我想创建一个 R 脚本来运行 MK 测试（包“趋势”）并计算 Sens Slope。我有以下示例表，其中包含我们的水质数据：

| Site | Program | Year | parameter1 | parameter2 |
|------|---------|------|-----|-----|
| A    | ABC     | 1990 | 5   | 100 |
| A    | ABC     | 1991 | 10  | 75  |
| A    | ABC     | 1992 | 15  | 50  |
| A    | ABC     | 1993 | 20  | 25  |
| A    | ABC     | 1994 | 25  | 5   |
| B    | ABC     | 1990 | 10  | 88  |
| B    | ABC     | 1991 | 20  | 44  |
| B    | ABC     | 1992 | 30  | 22  |
| B    | ABC     | 1993 | 40  | 11  |
| B    | ABC     | 1994 | 50  | 6   |
| C    | XYZ     | 1990 | 6   | 64  |
| C    | XYZ     | 1991 | 12  | 44  |
| C    | XYZ     | 1992 | 18  | 24  |
| C    | XYZ     | 1993 | 24  | 14  |
| C    | XYZ     | 1994 | 30  | 4   |
| D    | XYZ     | 1990 | 7   | 99  |
| D    | XYZ     | 1991 | 14  | 88  |
| D    | XYZ     | 1992 | 21  | 77  |
| D    | XYZ     | 1993 | 28  | 66  |
| D    | XYZ     | 1994 | 35  | 55  |

我需要为每个参数（ANC 和 SO4）取出数据中的每个时间序列（因此对于站点 A、B、C、D），并在 R 中运行 MannKendall 测试（代码如下）。我需要一个如下所示的输出表，但填充了 MK 统计量和 sens 斜率（不是如下所示的 1）。

| Site | Program | Parameter | MK Statistic | Sens Slope |
|------|---------|-----------|--------------|------------|
| A    | ABC     | ANC       | 1            | 1          |
| A    | ABC     | SO4       | 1            | 1          |
| B    | ABC     | ANC       | 1            | 1          |
| B    | ABC     | SO4       | 1            | 1          |
| C    | XYZ     | ANC       | 1            | 1          |
| C    | XYZ     | SO4       | 1            | 1          |
| D    | XYZ     | ANC       | 1            | 1          |
| D    | XYZ     | SO4       | 1            | 1          |

关于如何生成此输出表的任何想法？我知道它在某些时候需要一个循环，但不完全确定从哪里开始。可能是每个站点、程序，然后是 ANC 或 S04。下面的 R 代码来自一个单独的站点和参数组合，但是对于我们拥有的 100 个站点和 6 个水质参数，这将是一个痛苦的复制。

install.packages("trend")
library("trend")

#put our data in a time series (but this only creates 1 site and its time series)
time_series <- ts(Trends$parameter1, start=c(1990, 1), end=c(1994, 1), frequency=1)
print(time_series)


#Run the MK Test and Sens Slope from package trend
mk.test(time_series, alternative = c("two.sided", "greater", "less"),
continuity = TRUE)
sens.slope(time_series, conf.level = 0.95) 

输出示例 - 这些是来自我的实际数据的结果，而不是示例数据集（因为我没有在示例数据中的所有站点上成功运行 MKtest）。下面有^^^^的数字是我最终输出表需要的数字。

> mk.test(time_series , alternative = c("two.sided", "greater", "less"),
+ continuity = TRUE)

    Mann-Kendall trend test

data:  time_series
z = -5.7308, n = 26, p-value = 9.996e-09
alternative hypothesis: true S is not equal to 0
sample estimates:
           S         varS          tau 
-261.0000000 2058.3333333   -0.8030769 
^^^^^^^^^^^^

> sens.slope(time_series , conf.level = 0.95)

    Sens slope

data:  time_series
z = -5.7308, n = 26, p-value = 9.996e-09
alternative hypothesis: true z is not equal to 0
95 percent confidence interval:
 -1.3187075 -0.9495238
sample estimates:
Sens slope
  -1.136842
  ^^^^^^^^^

Answer 1

我们可以通过“站点”将数据集split 转换为data.frames 的list 并应用测试

library(trend)
lst1 <- split(Trends[c("parameter1", "parameter2")], Trends$Site)
out <- lapply(lst1, function(dat) 
           lapply(dat, function(para) {                

           time_series <- ts(para, start=c(1983, 1), end=c(2018, 1), frequency=1)
           tsout <- mk.test(time_series, alternative = c("two.sided", 
                   "greater", "less"), continuity = TRUE)
           sensout <- sens.slope(time_series, conf.level = 0.95) 
   list(tsout = tsout, sensout = sensout)
  }
  ))

-输出

out$A$parameter1
#$tsout

#   Mann-Kendall trend test

#data:  time_series
#z = 0.57147, n = 36, p-value = 0.5677
#alternative hypothesis: true S is not equal to 0
#sample estimates:
#           S         varS          tau 
#4.200000e+01 5.147333e+03 7.352146e-02 


#$sensout

#   Sen's slope

#data:  time_series
#z = 0.57147, n = 36, p-value = 0.5677
#alternative hypothesis: true z is not equal to 0
#95 percent confidence interval:
# 0.0000 0.3125
#sample estimates:
#Sen's slope 
#          0 

out$D$parameter2
#$tsout

#   Mann-Kendall trend test

#data:  time_series
#z = -0.57147, n = 36, p-value = 0.5677
#alternative hypothesis: true S is not equal to 0
#sample estimates:
#            S          varS           tau 
# -42.00000000 5147.33333333   -0.07352146 


#$sensout

#   Sen's slope

#data:  time_series
#z = -0.57147, n = 36, p-value = 0.5677
#alternative hypothesis: true z is not equal to 0
#95 percent confidence interval:
# -0.6875  0.0000
#sample estimates:
#Sen's slope 
#          0 

数据

Trends <- structure(list(Site = c("A", "A", "A", "A", "A", "B", "B", "B", 
"B", "B", "C", "C", "C", "C", "C", "D", "D", "D", "D", "D"), 
    Program = c("ABC", "ABC", "ABC", "ABC", "ABC", "ABC", "ABC", 
    "ABC", "ABC", "ABC", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ", "XYZ", 
    "XYZ", "XYZ", "XYZ", "XYZ"), Year = c(1990L, 1991L, 1992L, 
    1993L, 1994L, 1990L, 1991L, 1992L, 1993L, 1994L, 1990L, 1991L, 
    1992L, 1993L, 1994L, 1990L, 1991L, 1992L, 1993L, 1994L), 
    parameter1 = c(5L, 10L, 15L, 20L, 25L, 10L, 20L, 30L, 40L, 
    50L, 6L, 12L, 18L, 24L, 30L, 7L, 14L, 21L, 28L, 35L), parameter2 = c(100L, 
    75L, 50L, 25L, 5L, 88L, 44L, 22L, 11L, 6L, 64L, 44L, 24L, 
    14L, 4L, 99L, 88L, 77L, 66L, 55L)), class = "data.frame", row.names = c(NA, 
-20L))

Answer 2

我有一种使用 tidyverse 的不同方法。到目前为止，它可能有点超出您的经验，但我建议您检查一下，因为当您来自 excel 时，我发现它比 base R 更容易使用

library(dplyr)
library(ggplot2)
library(tidyr)
library(purrr)

install.packages("trend")
library("trend")  

out <- dat %>% gather(parameter, value, ANC, SO4) %>% 
  group_by(parameter, Site) %>% nest() %>% 
  mutate(ts_out = map(data, ~ts(.x$value, start=c(1990, 1), end=c(1994, 1), frequency=1))) %>% 
  mutate(mk_res = map(ts_out, ~mk.test(.x, alternative = c("two.sided", "greater", "less"),
                                      continuity = TRUE)),
         sens = map(ts_out, ~sens.slope(.x, conf.level = 0.95))) %>% 
  mutate(mk_stat = map_dbl(mk_res, ~.x$p.value),
         sens_stat = map_dbl(sens, ~.x$p.value)) %>% 
  select(parameter, Site, mk_stat, sens_stat)

out
# A tibble: 8 x 4
  parameter Site     mk_stat sens_stat
  <chr>     <fct>      <dbl>     <dbl>
1 ANC       " A    "  0.0275    0.0275
2 ANC       " B    "  0.0275    0.0275
3 ANC       " C    "  0.0275    0.0275
4 ANC       " D    "  0.0275    0.0275
5 SO4       " A    "  0.0275    0.0275
6 SO4       " B    "  0.0275    0.0275
7 SO4       " C    "  0.0275    0.0275
8 SO4       " D    "  0.0275    0.0275

这会在表格中给出输出。我不确定这是否是您要从测试中删除的部分，但应该更容易更改

我建议查看每个步骤的输出以了解结构。这种分析风格的好资源是R for Data Science Many Models Chapter