功能将名称分配给多个列表元素

Question

我有数据帧int1和int2的列表。该代码的最终目标是将名称分配给int1和int2中的元素。我工作的其余工作流程要求我多次命名列表的元素，我想知道如何创建一个函数以减少使用基本R函数下线的击键数量。有什么想法吗？

library(lubridate)
library(tidyverse)
library(purrr)

date <- rep_len(seq(dmy("01-01-2011"), dmy("31-07-2011"), by = "days"), 200)
ID <- rep(c("A","B", "C"), 200)
df <- data.frame(date = date,
                 x = runif(length(date), min = 60000, max = 80000),
                 y = runif(length(date), min = 800000, max = 900000),
                 ID)

df$Month <- month(df$date)

# Create first list
int1 <- df %>%
  # arrange(ID) %>%   # skipped for readability of result
  mutate(new = floor_date(date, '10 day')) %>%
  mutate(new = if_else(day(new) == 31, new - days(10), new)) %>% 
  group_by(ID, new) %>%
  filter(Month == "1") %>% 
  group_split()

# Create second list
int2 <- df %>%
  # arrange(ID) %>%   # skipped for readability of result
  mutate(new = floor_date(date, '10 day')) %>%
  mutate(new = if_else(day(new) == 31, new - days(10), new)) %>% 
  group_by(ID, new) %>%
  filter(Month == "2") %>% 
  group_split()

# Expected Output

# Assign names to int1
names(int1) <- sapply(int1, function(x) paste(x$ID[1],
                                              x$new[1], sep = "_"))
# Assign names to int2
names(int2) <- sapply(int2, function(x) paste(x$ID[1],
                                              x$new[1], sep = "_"))

Answer 1

使用group_split将不会命名list元素。它在？group_split中指定

它不会基于分组的列表列表列表，因为这通常会丢失信息并且令人困惑。

而是使用split从base r使用粘贴使用。从'id'返回 “新”列类似

int1 <- df %>%
  # arrange(ID) %>%   # skipped for readability of result
  mutate(new = floor_date(date, '10 day')) %>%
  mutate(new = if_else(day(new) == 31, new - days(10), new)) %>% 
  group_by(ID, new) %>%
  filter(Month == "1") %>% ungroup %>% 
  {split(., .[c('ID', 'new')])}

， int2的