识别数据框中的以下日期组答案

【问题标题】：Identify groups of following dates in a dataframe识别数据框中的以下日期组
【发布时间】：2020-03-03 13:01:12
【问题描述】：

我有一个这样的df：

    mydate daynight myvar
195 2018-11-21  1   64
196 2018-11-21  2    4
197 2018-11-22  1    7
198 2018-11-22  2    2
199 2018-11-23  2    0
200 2018-11-24  2    0
201 2018-11-25  2   64
202 2018-11-26  2    6
203 2018-11-27  2    0
204 2018-11-28  2    0
205 2018-11-29  2   20
206 2018-11-30  2    4
207 2018-12-01  1   64
214 2018-12-04  2    1
215 2018-12-05  1    4
216 2018-12-05  2    0
217 2018-12-08  2   30
218 2018-12-09  2   15
219 2018-12-10  2    0
220 2018-12-11  2   14
221 2018-12-12  1   28

我正在尝试在daynight 列中识别至少有 3 个跟随 2 的组，并且我想知道这些组的开始和结束日期。有了那个df，它会是：

     beg          end         
[1,] "2018-11-22" "2018-11-30"
[2,] "2018-12-05" "2018-12-11"

【问题讨论】：

标签： r date datetime

【解决方案1】：

示例数据：

df = read.table(text = "
mydate daynight myvar
195 2018-11-21  1   64
196 2018-11-21  2    4
197 2018-11-22  1    7
198 2018-11-22  2    2
199 2018-11-23  2    0
200 2018-11-24  2    0
201 2018-11-25  2   64
202 2018-11-26  2    6
203 2018-11-27  2    0
204 2018-11-28  2    0
205 2018-11-29  2   20
206 2018-11-30  2    4
207 2018-12-01  1   64
214 2018-12-04  2    1
215 2018-12-05  1    4
216 2018-12-05  2    0
217 2018-12-08  2   30
218 2018-12-09  2   15
219 2018-12-10  2    0
220 2018-12-11  2   14
221 2018-12-12  1   28
", header=T)

一种方法是这样的：

library(dplyr)
library(data.table)

  df %>%
  group_by(group = rleid(daynight)) %>%  # group by consequtive daynight values
  summarise(val = unique(daynight),      # get the daynight value
            Count = n(),                 # count rows of that value
            Beg = first(mydate),         # get first date
            End = last(mydate)) %>%      # get last date
  filter(Count > 2 & val == 2) %>%       # keep only cases where you have 3+ daylight = 2
  select(-group, -val)                   # remove unecessary columns

# # A tibble: 2 x 3
#    Count Beg        End       
#    <int> <fct>      <fct>     
# 1     9 2018-11-22 2018-11-30
# 2     5 2018-12-05 2018-12-11

请注意，此方法假定您的日期列已排序（如您的示例中所示）。

【讨论】：

【解决方案2】：

这是一个基本的 R 想法，

i1 <- cumsum(c(TRUE, diff(d2$daynight) != 0))

do.call(rbind, 
 Filter(function(x)!is.null(x), 
          tapply(d2$mydate, i1, FUN = function(i)if(length(i) >= 3){c(i[1], i[length(i)])})))

给出，

         [,1]         [,2]        
4 "2018-11-22" "2018-11-30"
8 "2018-12-05" "2018-12-11"

数据

dput(d2)
structure(list(mydate = c("2018-11-21", "2018-11-21", "2018-11-22", 
"2018-11-22", "2018-11-23", "2018-11-24", "2018-11-25", "2018-11-26", 
"2018-11-27", "2018-11-28", "2018-11-29", "2018-11-30", "2018-12-01", 
"2018-12-04", "2018-12-05", "2018-12-05", "2018-12-08", "2018-12-09", 
"2018-12-10", "2018-12-11", "2018-12-12"), daynight = c(1L, 2L, 
1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 
2L, 2L, 1L), myvar = c(64L, 4L, 7L, 2L, 0L, 0L, 64L, 6L, 0L, 
0L, 20L, 4L, 64L, 1L, 4L, 0L, 30L, 15L, 0L, 14L, 28L)), row.names = c("195", 
"196", "197", "198", "199", "200", "201", "202", "203", "204", 
"205", "206", "207", "214", "215", "216", "217", "218", "219", 
"220", "221"), class = "data.frame")

【讨论】：