Question

所以说我有一个列表

x <- as.list(c("john","jerry","james"))

以及值列表

y <- as.list(c(8,5,10))

我想遍历x的第一个元素，y的第一个元素的次数，并在最后添加一个后缀。因此，对于元素john，它将为john_1，john_2 ... john_8。那么jerry就是jerry_1，jerry_2 ... jerry_5。还有james_1，james_2，...，james_10

这是我尝试过的不起作用

z <- lapply(x, function(i){paste(i,"_",rep(max))})

但是那不给我我想要的东西。

也许，我不应该使用lapply或map *的某些版本，但不要对它们进行处理。

Answer 1

基于R的想法

.take(10)

Answer 2

或者：

x <- as.list(c("john","jerry","james"))
y <- as.list(c(8,5,10))

lapply(seq_along(x), function(i) paste0(x[[i]], "_", seq_len(y[[i]])))

# [[1]]
# [1] "john_1" "john_2" "john_3" "john_4" "john_5" "john_6" "john_7" "john_8"
# 
# [[2]]
# [1] "jerry_1" "jerry_2" "jerry_3" "jerry_4" "jerry_5"
# 
# [[3]]
# [1] "james_1"  "james_2"  "james_3"  "james_4"  "james_5"  "james_6"  "james_7"  "james_8"  "james_9"  "james_10"

Answer 3

如果可以为所有值使用单个向量，则可以使用paste，rep和sequence

paste(rep(x, y), sequence(y), sep = "_")

#[1] "john_1"   "john_2"   "john_3"   "john_4" "john_5"   "john_6" "john_7"  "john_8"  
#[9] "jerry_1"  "jerry_2"  "jerry_3"  "jerry_4" "jerry_5"  "james_1"  "james_2"  "james_3" 
#[17] "james_4" "james_5"  "james_6"  "james_7"  "james_8"  "james_9"  "james_10"

Answer 4

一个选项是

paste(rep(x, y), unlist(lapply(y, seq)), sep="_")

如何重复列表多次

4 个答案: