我有一个名字的载体:
names <- c("a", "b", "c", "d", "e", "f", "g", "h")
我想分别将这些名称与这些数字进行一系列组合:
numbers <- c(0, 0:2, 0:3, 0:1, 0:1, 0, 0, 0)
所以我的组合是:
a0 b0 b1 b2 c0 c1 c2 c3 d0 d1 e0 e1 f0 g0 h0
我该怎么做？我宁愿使用一个载体化函数，也不愿迭代.
我try 了这些代码，但得到了错误的结果(可能是由于收件箱或其他原因):
paste0(names, numbers, recycle0 = F)
paste0(names, numbers, recycle0 = T)
paste0(names, numbers |> as.list())
purrr::map_chr(names, paste0(c(0, 0:2, 0:3, 0:1, 0:1, 0, 0, 0)))
还有一些我记不清的人.
1) cumsum(...)是名称的索引，即1 2 2 2 3 3 3 3 4 4 5 5 6 7 8，因此请索引到names中并将numbers粘贴到末尾.不使用外部包装.
paste0(names[cumsum(numbers == 0)], numbers)
## [1] "a0" "b0" "b1" "b2" "c0" "c1" "c2" "c3" "d0" "d1" "e0" "e1" "f0" "g0" "h0"
2)第二种方法，也是基于cumsum(...)，是使用rle获取行程长度，重复每个名称该次数，然后粘贴数字. 这也不使用任何包.
paste0(rep(names, rle(cumsum(numbers == 0))$lengths), numbers)
## [1] "a0" "b0" "b1" "b2" "c0" "c1" "c2" "c3" "d0" "d1" "e0" "e1" "f0" "g0" "h0"
3)这个没有使用cumsum，但它确实使用了一个包.在numbers中，将0替换为名称，将其他数字替换为NA. 然后使用na.locf0填写NA并粘贴数字.
library(zoo)
paste0(na.locf0(replace(NA * numbers, numbers == 0, names)), numbers)
## [1] "a0" "b0" "b1" "b2" "c0" "c1" "c2" "c3" "d0" "d1" "e0" "e1" "f0" "g0" "h0"
问题输入
names <- c("a", "b", "c", "d", "e", "f", "g", "h")
numbers <- c(0, 0:2, 0:3, 0:1, 0:1, 0, 0, 0)