在两个值/日期间隔之间按 id 填写最大值答案

【问题标题】：Fill in max by id between two values/date intervals在两个值/日期间隔之间按 id 填写最大值
【发布时间】：2020-06-27 06:49:24
【问题描述】：

我需要 R 脚本方面的帮助。

我有一个如下所示的表格：

> dput(first)
structure(list(Date = structure(c(1438387200, 1441065600, 1456790400, 
1459468800, 1462060800, 1464739200, 1467331200, 1470009600, 1472688000, 
1475280000, 1477958400, 1480550400, 1483228800, 1485907200, 1488326400, 
1491004800, 1493596800, 1464739200, 1467331200, 1470009600, 1472688000, 
1475280000, 1477958400, 1480550400, 1483228800, 1517443200, 1519862400, 
1522540800, 1525132800, 1527811200, 1530403200, 1533081600, 1535760000, 
1538352000, 1541030400), class = c("POSIXct", "POSIXt"), tzone = "UTC"), 
    ID = c("A1", "A1", "A1", "A1", "A1", "A1", "A1", "A1", "A1", 
    "A1", "A1", "A1", "A1", "A1", "A1", "A1", "A2", "A2", "A2", 
    "A2", "A2", "A2", "A2", "A2", "A2", "A3", "A3", "A3", "A3", 
    "A3", "A3", "A3", "A3", "A3", "A3"), flag = c(0, 0, 0, 1, 
    0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 
    0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0)), row.names = c(NA, -35L
), class = c("tbl_df", "tbl", "data.frame"))

我需要改变表，例如，如果从上次 flag ==1（每笔贷款）到再次为 1 的时间少于 5 个月，则应将零替换为 1，否则为保持原样。

输出看起来像这样：

> dput(second)
structure(list(Date = structure(c(1438387200, 1441065600, 1456790400, 
1459468800, 1462060800, 1464739200, 1467331200, 1470009600, 1472688000, 
1475280000, 1477958400, 1480550400, 1483228800, 1485907200, 1488326400, 
1491004800, 1493596800, 1464739200, 1467331200, 1470009600, 1472688000, 
1475280000, 1477958400, 1480550400, 1483228800, 1517443200, 1519862400, 
1522540800, 1525132800, 1527811200, 1530403200, 1533081600, 1535760000, 
1538352000, 1541030400), class = c("POSIXct", "POSIXt"), tzone = "UTC"), 
    ID = c("A1", "A1", "A1", "A1", "A1", "A1", "A1", "A1", "A1", 
    "A1", "A1", "A1", "A1", "A1", "A1", "A1", "A2", "A2", "A2", 
    "A2", "A2", "A2", "A2", "A2", "A2", "A3", "A3", "A3", "A3", 
    "A3", "A3", "A3", "A3", "A3", "A3"), flag = c(0, 0, 0, 1, 
    1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 
    1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0)), row.names = c(NA, -35L
), class = c("tbl_df", "tbl", "data.frame"))

坦率地说，我不知道从哪里开始，我对 R 有点陌生。

【问题讨论】：

我刚刚用dput编辑过，非常感谢！
或许，试试first %>% filter(as.logical(flag)) %>% group_by(ID) %>% group_by(grp = cumsum(interval(lag(Date, default = first(Date)), Date) %/% months(1) > 5), .add = TRUE) %>% complete(Date = seq(first(Date), last(Date), by = '1 month'), fill = list(flag = 1)) %>% ungroup %>% select(-grp) %>% right_join(first, by = c("ID", "Date")) %>% arrange(ID, Date) %>% transmute(flag = coalesce(flag.x, flag.y))
标志的“A2”值是否有错字？即2017-01-01 A2 in first ?
我的意思是，第一个是 0，但我发现第二个是 1
是的，实际上，我认为您的输出是正确的！非常感谢！1

标签： r date max dplyr

【解决方案1】：

我们可以filter从'first'数据中'flag'为1的行，然后按'ID'分组，根据相邻'的月差创建第二组（'grp'）日期，检查它是否大于 5，执行cumsum（每当“日期”差异大于 5 个月时，它会将组计数增加 1），通过执行一系列 ' 扩展具有 complete 的数据集来自first 和last 'Date' by '1 月' 的日期'，加入原始数据集并将'标志'列替换为coalesce

library(dplyr)
library(tidyr)
library(lubridate)
first %>% 
     filter(as.logical(flag)) %>% 
     group_by(ID) %>% 
     group_by(grp = cumsum(interval(lag(Date, 
       default = first(Date)), Date) %/% months(1) > 5), .add = TRUE) %>%
     complete(Date = seq(first(Date), last(Date), by = '1 month'), 
           fill = list(flag = 1)) %>% 
     ungroup %>% 
     select(-grp) %>% 
     right_join(first, by = c("ID", "Date")) %>% 
     arrange(ID, Date) %>% 
     transmute(Date, ID, flag = coalesce(flag.x, flag.y))

【讨论】：

非常感谢 akrun 的帮助。我会有一个后续问题。如何计算每个月的相对频率？ count(flag==1 per ID)/count(zeros per ID) 但仅适用于每个月的新标志==1？谢谢！