R:检测频率

Question

我必须在R中创建一个函数(或循环)来检测超频。检测超频频率的要求将在180天内到达3次，如果该要求得到满足，则该人将变得异常频繁，不仅是在未来，而且在过去的访问中，他也没有满足超频繁的要求。

pacient <- c(10,10,10,10,10,11,11,12,12,12,13, 13, 15, 14); pacient
date <- as.Date(c("01/01/2018","02/05/2018", "04/06/2018", "10/11/2018", "05/12/2018", "02/01/2018", "06/08/2018", "01/01/2018", "03/01/2018", "06/03/2018", "05/08/2018", "05/08/2019", "05/07/2019", "08/07/2017"), format = "%d/%m/%Y"); date 
DF <- data.frame(pacient, date); DF


count_visit <- function(x){
  DF <- data.table(DF)
  DTord<-DF[with(DF , order(DF $ date)), ]; DTord 
  DTord[,num_visit := order(date), by  = pacient];DTord 
  DTordID <- DTord[with(DTord, order(DTord$pacient)), ]; DTordID  
  DTordID[,max_visit := max(num_visit), by  = pacient];DTordID 
framedatos <- as.data.frame(DTordID)

  return(framedatos)}

REUP_visit <- count_visit(DF); head(REUP_visit)


   pacient    date      num_visit   max_visit
    10     01/01/2018      1           5
    10     02/05/2018      2           5
    10     04/06/2018      3           5
    10     10/11/2018      4           5
    10     05/12/2018      5           5 
    11     02/01/2018      1           2
    11     06/08/2018      2           2
    12     01/01/2018      1           3   
    12     03/01/2018      2           3
    12     06/03/2018      3           3
    13     05/08/2018      1           2
    13     05/08/2019      2           2
    14     08/07/2017      1           1
    15     05/07/2019      1           1

到目前为止，我只创建了一个功能，它告诉我每名病人的就诊次数和病人的最大探视次数(这是我需要做其他事情的)：

  pacient    date    num_visit  max_visit  days_visit   <180 future_hyperf  past_hyperf
    10     01/01/2018      1           5       0          1      no           yes
    10     02/05/2018      2           5       121        2      no           yes
    10     04/06/2018      3           5       33         3      yes          yes
    10     10/11/2018      4           5       159        4      yes          yes  
    10     05/12/2018      5           5       25         5      yes          yes
    11     02/01/2018      1           2       0          1      no           no 
    11     06/08/2018      2           2       216        1      no           no 
    12     01/01/2018      1           3       0          1      no           yes 
    12     03/01/2018      2           3       2          2      no           yes 
    12     06/03/2018      3           3       62         3      yes          yes  
    13     05/08/2018      1           2       0          1      no           no         
    13     05/08/2019      2           2       365        1      no           no 
    14     08/07/2017      1           1       0          1      no           no 
    15     05/07/2019      1           1       0          1      no           no 

我需要的输出有："day_visit“、"<180”、"future_hyperf“和"past_hyperf”。

变量"day_visit“的目标是将病人第一次到急诊室就诊的次数定为0，然后计算两次就诊之间的天数。

    DF <- DF %>%
  group_by(pacient) %>%
  arrange(date) %>%
  mutate(days_visit= date - lag(date, default = first(date)))

变量" <180“将是第一次出现的数字1、第二个变量(如果是上次访问的<180天)、第三个变量(如果是上一次访问的<180天)等等。例如，如果病人达到2，而第三次就诊不能满足<180天，则有必要再次放置1(循环将被重新启动)。

变量"future_hyperf“说是或不是。如果病人在<180的变量中达到3岁，就好像它创造了未来一样，如果探视时间晚于180天，而且不符合规定，这并不重要。一旦达到了标准，就永远如此。

变量"past_hyperf“将变量"future_hyperf”中的所有患者都转换为过去。

谢谢!

解决方案

DF3 <-  DF %>%
  arrange(pacient, date) %>%
  group_by(pacient) %>%
  mutate(days_visit = as.integer(date - lag(date, default = first(date))) ,
         less_180 = days_visit < 180) %>%
  mutate(counter = rowid(pacient, cumsum(date - shift(date, fill=first(date)) > 180)),
         future_hyperf = case_when(counter >= 3 ~ "yes",
                                   TRUE ~ "no"),
         past_hyperf = case_when(max(counter, na.rm = T) >= 3 ~ "yes",
                                 TRUE ~ "no")) 
DF3 <- DF3[with(DF3,order(pacient,date)),]

Answer 1

试试这个：

pacient <- c(10, 10, 10, 10, 10, 11, 11, 12, 12, 12, 13, 13, 15, 14)
pacient
date <-
  as.Date(
    c(
      "01/01/2018",
      "02/05/2018",
      "04/06/2018",
      "10/11/2018",
      "05/12/2018",
      "02/01/2018",
      "06/08/2018",
      "01/01/2018",
      "03/01/2018",
      "06/03/2018",
      "05/08/2018",
      "05/08/2019",
      "05/07/2019",
      "08/07/2017"
    ),
    format = "%d/%m/%Y"
  )
date
DF <- data.frame(pacient, date)
DF
#packages
library(dplyr)
library(lubridate)
#time zone
lct <- Sys.getlocale("LC_TIME")
Sys.setlocale("LC_TIME", "C")
DF <- DF %>%
  group_by(pacient) %>%
  mutate(num_visit = cumsum(pacient) / pacient) %>% # number of visits
  mutate(max_visit = max(num_visit)) %>% # max visit
  mutate(days_visit = as.Date(date, "%d/%m/%Y") - lag(as.Date(date, "%d/%m/%Y"))) %>% # days between visits
  mutate(minus_180_days = case_when(days_visit < 180 &
                                      !is.na(days_visit) ~ num_visit,
                                    TRUE ~ 1)) %>% # is days between visits < 180
  mutate(future_hyperf = case_when(minus_180_days > 3 ~ "yes",
                                   TRUE ~ "no")) %>% # future hyperf
  mutate(past_hyperf = case_when(max(minus_180_days, na.rm = T) >= 3 ~ "yes",
                                 TRUE ~ "no")) # past hyperf

希望它能帮上忙

Answer 2

我会这么做的。说明在注释中。

library(tidyverse)
DF %>% 
    group_by(pacient) %>% # group the data by "pacient"
    mutate(lag_date = lag(date, n = 2)) %>% # create the variable of lag dates by 2 visits
    mutate(date_diff = as.integer(date - lag_date)) %>% # Calculate the difference in dates 
    mutate(date_diff = case_when(is.na(date_diff) ~ 9999L, # replace NAs with 999 (cummin does not allow na.rm)
                                 TRUE ~ date_diff)) %>% #
    mutate(min_period = cummin(date_diff)) %>% # calculate the cumulative minimum of the differencce
    mutate(future_hyperf = min_period < 180) %>% # check the cumulative min is less than 180
    mutate(past_hyperf = min(min_period) < 180) %>% 
    ungroup()


## # A tibble: 14 x 7
##    pacient date       lag_date   date_diff min_period future_hyperf past_hyperf
##      <dbl> <date>     <date>         <int>      <int> <lgl>         <lgl>      
##  1      10 2018-01-01 NA              9999       9999 FALSE         TRUE       
##  2      10 2018-05-02 NA              9999       9999 FALSE         TRUE       
##  3      10 2018-06-04 2018-01-01       154        154 TRUE          TRUE       
##  4      10 2018-11-10 2018-05-02       192        154 TRUE          TRUE       
##  5      10 2018-12-05 2018-06-04       184        154 TRUE          TRUE       
##  6      11 2018-01-02 NA              9999       9999 FALSE         FALSE      
##  7      11 2018-08-06 NA              9999       9999 FALSE         FALSE      
##  8      12 2018-01-01 NA              9999       9999 FALSE         TRUE       
##  9      12 2018-01-03 NA              9999       9999 FALSE         TRUE       
## 10      12 2018-03-06 2018-01-01        64         64 TRUE          TRUE       
## 11      13 2018-08-05 NA              9999       9999 FALSE         FALSE      
## 12      13 2019-08-05 NA              9999       9999 FALSE         FALSE      
## 13      15 2019-07-05 NA              9999       9999 FALSE         FALSE      
## 14      14 2017-07-08 NA              9999       9999 FALSE         FALSE