r - 取网格化(坐标排序)年度时间数据集的 30 年平均值

Question

我有一个网格化数据集的 .csv。每个网格(由纬度/经度坐标表示)都有一个从 1950 年到 2100 年的年度时间序列和伴随值。

格式如下:

<表类="s-表"><头><次>次 经<日>纬度 值1值2值3<正文>1950-01-01-80.7942.961231951-01-01-80.7942.962461952-01-01-80.7942.963691953-01-01-80.7942.9648121954-01-01-80.7942.96510151955-01-01-80.7942.961231956-01-01-80.7942.962461957-01-01-80.7942.963691958-01-01-80.7942.9648121959-01-01-80.7942.96510151960-01-01-80.7942.96123....................................1950-01-01-80.6342.961231951-01-01-80.7942.96246....................................1950-01-01-79.8844.291231951-01-01-79.8844.26246

您可以在 https://climatedata.ca/download/ 获得 .csv 数据集的真实示例(选择频率:每年；变量:任何人；位置:在 map 上选择2+个格子)

我正在尝试弄清楚如何为每个网格取 30 年的平均值。

例如，对于第一个网格 (-80.79, 42.96)，我想取 1950-1969、1970-2005、2006-2039、2049-2069 和 2070-2100 的平均值。然后对第二个网格做同样的事情。等等

或者，

对每个网格取 1950-1969 年的平均值，然后对每个网格取 1970-2005 年的平均值。等等

基本上，我希望能够对所有网格重复相同的计算。

Answer 1

下面是使用 dplyr 的方法。 group_by 每个时期(这里四舍五入到最接近的 30 年)和网格单元格，然后我们可以使用 summarize(across(... 来计算平均值每组中的这三个值。

library(dplyr)
df1 %>%
  group_by(yr30 = floor(lubridate::year(time)/30)*30, lon, lat) %>%
  summarize(across(value1:value3, mean), .groups = "drop")


# A tibble: 4 × 6
   yr30   lon   lat value1 value2 value3
  <dbl> <dbl> <dbl>  <dbl>  <dbl>  <dbl>
1  1950 -80.8  43.0   2.75    5.5   8.25
2  1950 -80.6  43.0   1       2     3   
3  1950 -79.9  44.3   2       4     6   
4  1950 -79.9  44.3   1       2     3 

---

使用的示例数据:

df1 <- data.frame(
        time = c("1950-01-01","1951-01-01","1952-01-01",
                 "1953-01-01","1954-01-01","1955-01-01","1956-01-01",
                 "1957-01-01","1958-01-01","1959-01-01","1960-01-01","1950-01-01",
                 "1951-01-01","1950-01-01","1951-01-01"),
         lon = c(-80.79,-80.79,-80.79,-80.79,-80.79,
                 -80.79,-80.79,-80.79,-80.79,-80.79,-80.79,-80.63,-80.79,
                 -79.88,-79.88),
         lat = c(42.96,42.96,42.96,42.96,42.96,42.96,
                 42.96,42.96,42.96,42.96,42.96,42.96,42.96,44.29,44.26),
      value1 = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 1L, 2L, 1L, 2L),
      value2 = c(2L, 4L, 6L, 8L, 10L, 2L, 4L, 6L, 8L, 10L, 2L, 2L, 4L, 2L, 4L),
      value3 = c(3L, 6L, 9L, 12L, 15L, 3L, 6L, 9L, 12L, 15L, 3L, 3L, 6L, 3L, 6L)
) 
df1$time = as.Date(df1$time)