Question

我必须在R中创建一个函数（或循环）以检测超频率。检测高频率的要求是在180天内达到3次，如果满足该要求，则该人将是高频率的，不仅是在将来，而且在过去不符合高频率要求的探视中也是

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，我仅设法创建了一个函数，该函数可以告诉我每位患者的就诊次数以及患者的最大就诊次数（这是我需要的其他功能）：

  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

我需要的输出是以下内容：“ day_visit”，“ <180”，“ future_hyperf”和“ past_hyperf”。

变量“ day_visit”的目的是将患者首次访问急诊室的次数设为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”将是第一次出现数字1，第二次是2（如果以前访问是<180天），3（如果第一次访问是<180天））等等。例如，如果患者达到2岁而第三次就诊不满足<180天，则有必要再次放置1个（循环将重新开始）。

变量“ future_hyperf”表示是或否。一旦患者的变量<180达到3，就标记为将来。如果就诊时间晚于180天且不符合要求，则无所谓。一旦满足标准，它将永远存在。

变量“ past_hyperf”会将所有在变量“ future_hyperf”中拥有的患者也转换为过去的患者。

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

这就是我要怎么做。解释在注释中。

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

Answer 2

尝试一下：

log.info("LogData Object::::::{}", StructuredArguments.keyValue("LogData", data));

希望有帮助

R：检测频率

2 个答案: