在dplyr中重复行并相应地生成一个新列

我有一个数据集，如国家/地区\u未出现\u指数

countries_not_appear_indices <- as.data.frame(c("CUW", "ARM", "VGB", "ATG", "KNA", "GRD", "VCT", "LCA", "TCA", "GNB", "GNQ", "CYM", "MNE", "MDV", "MKD", "GIB", "LIE", "COM", "NCL", "BES", "PYF")) %>%
  rename(iso = `c("CUW", "ARM", "VGB", "ATG", "KNA", "GRD", "VCT", "LCA", "TCA", "GNB", "GNQ", "CYM", "MNE", "MDV", "MKD", "GIB", "LIE", "COM", "NCL", "BES", "PYF")`) %>%
  mutate(ie = 0, ih = 0)

但是，我还必须以正确的方式生成month函数

---

有什么线索吗？

您可以使用

取消计数

将每行重复12次，并使用

行号（）创建月份列。

-

库（dplyr）
图书馆（tidyr）
国家/地区\u未\u出现\u指数%>%
未计（12）%>%
分组依据（iso）%>%
mutate（month=行号（），.after='iso'）%>%
解组
#国际标准化组织
#      
#1 CUW 1 0 0
#2 CUW 2 0 0
#3 CUW 3 0 0
#4 CUW 4 0 0
#5cuw500
#6cuw600
#7 CUW7 0 0
#8 CUW 8 0 0
#9cuw900
#10 CUW 10 0 0 0
#…还有242行

我们可以在

基本R

new <- countries_not_appear_indices[rep(seq_len(nrow(countries_not_appear_indices)), 12),]
new$month <-  with(new, ave(seq_along(iso), iso, FUN = seq_along))

新建
countries_not_appear_indices[rep(seq_len(nrow(countries_not_appear_indices)), each = 12), ]

library(dplyr)
library(tidyr)

countries_not_appear_indices %>%
  uncount(12) %>%
  group_by(iso) %>%
  mutate(month = row_number(), .after = 'iso') %>%
  ungroup 

#   iso   month    ie    ih
#   <chr> <int> <dbl> <dbl>
# 1 CUW       1     0     0
# 2 CUW       2     0     0
# 3 CUW       3     0     0
# 4 CUW       4     0     0
# 5 CUW       5     0     0
# 6 CUW       6     0     0
# 7 CUW       7     0     0
# 8 CUW       8     0     0
# 9 CUW       9     0     0
#10 CUW      10     0     0
# … with 242 more rows

new <- countries_not_appear_indices[rep(seq_len(nrow(countries_not_appear_indices)), 12),]
new$month <-  with(new, ave(seq_along(iso), iso, FUN = seq_along))