我有经度和纬度的向量:
longDim
[1] -79.65770 -79.21761 -78.77750
latiDim
[1] -39.70588 -39.26471 -38.82353
我想并行地遍历它们的组合。为此,首先我使用expand.grid
所有可能的组合来创建一个数据框:
my.grid <- expand.grid(longDim, latiDim)
然后mclapply()
在结果数据帧的行上使用:
mclapply(1:nrow(my.grid), function(x){some_function})
其中some_function
返回一个包含两个对象的列表,每个对象的长度为139。
因此,结果,我得到了一个尺寸为9x2的嵌套列表,如下所示:
str(l1)
List of 9
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 2
..$ su.25: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ su.30: Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
现在,我需要将该列表从9x2重新调整为3x3x2尺寸。这是我正在寻找的格式:
str(l2)
List of 3
$ :List of 3
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 3
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
$ :List of 3
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
..$ :List of 2
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
.. ..$ : Named num [1:139] 0 0 0 0 0 0 0 0 0 0 ...
.. .. ..- attr(*, "names")= chr [1:139] "1961" "1962" "1963" "1964" ...
我该如何实现?
要复制的代码l1
,l2
可以在以下位置找到:https : //pastebin.com/raw/LTyZi0mp(时间太长,无法在此处发布)
我们还可以gl
用来为split
ing创建分组索引
split(lst, as.integer(gl(length(lst1), 3, length(lst1))))
lst1 <- replicate(9, list(list(x = 1:5, y = 1:5)))
本文收集自互联网,转载请注明来源。
如有侵权,请联系 [email protected] 删除。
我来说两句