Question

考虑以下数据框（按ID和时间排序）：

df <- data.frame(id = c(rep(1,7),rep(2,5)), event = c("a","b","b","b","a","b","a","a","a","b","a","a"), time = c(1,3,6,12,24,30,32,1,2,6,17,24))
df
   id event time
1   1     a    1
2   1     b    3
3   1     b    6
4   1     b   12
5   1     a   24
6   1     b   30
7   1     a   42
8   2     a    1
9   2     a    2
10  2     b    6
11  2     a   17
12  2     a   24

我想计算每个“id”组中给定事件序列出现的次数。考虑以下具有时间限制的顺序：

seq <- c("a", "b", "a")
time_LB <- c(0, 2, 12)
time_UB <- c(Inf, 8, 18)

这意味着事件“a”可以随时开始，事件“b”必须在事件“a”之后不早于2且不晚于8开始，另一事件“a”必须不早于12开始且不事件“b”之后的18岁之后。一些创建序列的规则：

关于“时间”列，事件不需要是连续的。例如，seq可以从第1,3和5行构建。
要计算，序列必须具有不同的第一个事件。例如，如果计算seq =第8行，第10行和第11行，则不得计算seq =行8,10和12。
如果事件不违反第二条规则，则事件可以包含在许多构造的序列中。例如，我们计算两个序列：行1,3,5和行5,6,7。

预期结果：

df1
  id count
1  1     2
2  2     2

R - Identify a sequence of row elements by groups in a dataframe和Finding rows in R dataframe where a column value follows a sequence中存在一些相关问题。

这是使用“dplyr”解决问题的方法吗？

Answer 1

我相信这是你正在寻找的。它为您提供所需的输出。请注意，在Dim oExcel As Object Dim db As DAO.Database Dim rs As DAO.Recordset Dim CurrentColumn As Integer 'Make a new instance of Excel Set oExcel = CreateObject("Excel.Application") 'Have that instance open your workbook oExcel.Open("YourWorkBookName") 'Open the database Set db = CurrentDb 'Create a SQL result from a SQL string to pull data from your database Set rs = db.OpenRecordset(YourSQLString) 'Assign the value of a field in your SQL output to a cell oExcel.Workbook("YourWorkBookName").Sheets("YourSheetName").Cell(YourCellRow, YourCellColumn") = rs.("FieldName")中定义time列时，原始问题中存在拼写错误，其中您有32而不是42。我说这是一个拼写错误，因为它与df定义之下的输出不匹配。我在下面的代码中将32更改为42。

df

这是输出：

library(dplyr)

df <- data.frame(id = c(rep(1,7),rep(2,5)), event = c("a","b","b","b","a","b","a","a","a","b","a","a"), time = c(1,3,6,12,24,30,42,1,2,6,17,24))

seq <- c("a", "b", "a")
time_LB <- c(0, 2, 12)
time_UB <- c(Inf, 8, 18)

df %>% 
  full_join(df,by='id',suffix=c('1','2')) %>% 
  full_join(df,by='id') %>% 
  rename(event3 = event, time3 = time) %>%
  filter(event1 == seq[1] & event2 == seq[2] & event3 == seq[3]) %>% 
  filter(time1 %>% between(time_LB[1],time_UB[1])) %>% 
  filter((time2-time1) %>% between(time_LB[2],time_UB[2])) %>% 
  filter((time3-time2) %>% between(time_LB[3],time_UB[3])) %>%
  group_by(id,time1) %>%
  slice(1) %>%   # slice 1 row for each unique id and time1 (so no duplicate time1s)
  group_by(id) %>% 
  count()

另外，如果省略dplyr管道的最后两部分进行计数（查看它匹配的序列），则会得到以下序列：

# A tibble: 2 x 2
     id     n
  <dbl> <int>
1     1     2
2     2     2

编辑回应关于广义化的评论：是的，可以将其概括为任意长度的序列，但需要一些R伏都教。最值得注意的是，请注意Source: local data frame [4 x 7] Groups: id, time1 [4] id event1 time1 event2 time2 event3 time3 <dbl> <fctr> <dbl> <fctr> <dbl> <fctr> <dbl> 1 1 a 1 b 6 a 24 2 1 a 24 b 30 a 42 3 2 a 1 b 6 a 24 4 2 a 2 b 6 a 24的使用，它允许您在对象列表和Reduce上应用常用函数，我从foreach包借用做一些任意的循环。这是代码：

foreach

这输出以下内容：

library(dplyr)
library(foreach)

df <- data.frame(id = c(rep(1,7),rep(2,5)), event = c("a","b","b","b","a","b","a","a","a","b","a","a"), time = c(1,3,6,12,24,30,42,1,2,6,17,24))

seq <- c("a", "b", "a")
time_LB <- c(0, 2, 12)
time_UB <- c(Inf, 8, 18)

multi_full_join = function(df1,df2) {full_join(df1,df2,by='id')}
df_list = foreach(i=1:length(seq)) %do% {df} 
df2 = Reduce(multi_full_join,df_list)

names(df2)[grep('event',names(df2))] = paste0('event',seq_along(seq))
names(df2)[grep('time',names(df2))] = paste0('time',seq_along(seq))
df2 = df2 %>% mutate_if(is.factor,as.character)

df2 = df2 %>% 
  mutate(seq_string = Reduce(paste0,df2 %>% select(grep('event',names(df2))) %>% as.list)) %>% 
  filter(seq_string == paste0(seq,collapse=''))

time_diff = df2 %>% select(grep('time',names(df2))) %>%
  t %>%
  as.data.frame() %>%
  lapply(diff) %>% 
  unlist %>%  matrix(ncol=2,byrow=TRUE) %>% 
  as.data.frame

foreach(i=seq_along(time_diff),.combine=data.frame) %do%
{
  time_diff[[i]] %>% between(time_LB[i+1],time_UB[i+1])
} %>% 
  Reduce(`&`,.) %>% 
  which %>% 
  slice(df2,.) %>% 
  filter(time1 %>% between(time_LB[1],time_UB[1])) %>% # deal with time1 bounds, which we skipped over earlier
  group_by(id,time1) %>%
  slice(1) # slice 1 row for each unique id and time1 (so no duplicate time1s)

如果您只想要计数，则可以Source: local data frame [4 x 8] Groups: id, time1 [4] id event1 time1 event2 time2 event3 time3 seq_string <dbl> <chr> <dbl> <chr> <dbl> <chr> <dbl> <chr> 1 1 a 1 b 6 a 24 aba 2 1 a 24 b 30 a 42 aba 3 2 a 1 b 6 a 24 aba 4 2 a 2 b 6 a 24 aba然后group_by(id)与原始代码段一样。

Answer 2

也许将事件序列表示为字符串并使用正则表达式更容易：

df.str = lapply(split(df, df$id), function(d) {
    z = rep('-', tail(d,1)$time); z[d$time] = as.character(d$event); z })

df.str = lapply(df.str, paste, collapse='')

# > df.str
# $`1`
# [1] "a-b--b-----b-----------a-----b-----------a"
#
# $`2`
# [1] "aa---b----------a------a"


df1 = lapply(df.str, function(s) length(gregexpr('(?=a.{1,7}b.{11,17}a)', s, perl=T)[[1]]))

> data.frame(id=names(df1), count=unlist(df1))
#   id count
# 1  1     2
# 2  2     2

R - 根据数据帧中的时间约束查找行元素序列

2 个答案: