在 R 中创建一个规则来计算每个患者每天的咨询次数

create a rule in R to count number of consultations per patient per day

我使用实际数据集中的关键场景创建了以下数据集：

df <- data.frame (organisation_id  = c("1","1","2","2","2","2","2","2","3","3","3","3","3","4","4","4","4","4","4","4","4","4","4","4","4","4","4","4","4","4","4","4","4","4"),
                  patient_id = c("1230","1230","1222","1222","1244","1244","987","987","2223","2223","2247","2247","2247","1234","1234","1234","1234","1234","1234","1234","1234","1239","1239","1239","3322","3322","3322","5434","5434","4488","4488","4488","1250","1250"),
                  date = c("08-02-2018","08-02-2018","12-01-2018","12-01-2018","12-01-2018","22-02-2018","12-01-2018","22-02-2018","01-03-2019","01-03-2019","01-03-2019","01-03-2019","01-03-2019","12-07-2020","12-07-2020","12-07-2020","12-07-2020","12-07-2020","12-07-2020","12-07-2020","12-07-2020","13-07-2020","13-07-2020","13-07-2020","16-06-2021","16-06-2021","16-06-2021","14-05-2019","14-05-2019","17-03-2020","17-03-2020","17-03-2020","03-02-2019","03-02-2019"),
                  consultation_mode = c("Telephone","Face-to-Face","Telephone","Telephone","Face-to-Face","Face-to-Face","Telephone","Telephone","Home visit","Home visit","Face-to-Face","Face-to-Face","Face-to-Face","Telephone","Telephone","Telephone","Telephone","Face-to-Face","Face-to-Face","Face-to-Face","Face-to-Face","Home visit","Home visit","Home visit","Face-to-Face","Telephone","Face-to-Face","Telephone","Face-to-Face","Face-to-Face","Telephone","Telephone","Face-to-Face","Face-to-Face"),
                  professional_id = c("24","11","123","110","123","110","123","333","444","444","444","444","444","1133","12","25","26","12","34","35","38","44","44","5556","443","443","445","29","29","555","5556","12","1133","113663"),
                  professional_role = c("Doctor","Support","Doctor","Support","Doctor","Support","Doctor","Nurse","Doctor","Doctor","Doctor","Doctor","Doctor","Support","Support","Nurse","Nurse","Support","Doctor","Doctor","Nurse","Nurse","Nurse","Doctor","Doctor","Doctor","Doctor","Doctor","Doctor","Doctor","Doctor","Support","Support","Support"),
                  professional_name = c("Dr John Taylor","Mary Wright","Dr Patricia Jones","James Davies","Dr Patricia Jones","James Davies","Dr Patricia Jones","Peter Hall","Dr Mary Wilson","Dr Mary Wilson","Dr Mary Wilson","Dr Mary Wilson","Dr Mary Wilson","Mary Wright","Anthony Patel","Jennifer Walker","Jennifer Walker","Anthony Patel","Dr Carol Bell","Dr Carol Bell","Deborah Dixon","Kevin R Collins","Kevin Collins","Dr Robert Brown","Dr Mary Wilson","Dr Mary Wilson","Dr John Snow","Dr John Taylor","Dr John Taylor","Dr James Smith","Dr Robert Brown","Anthony Patel","Mary Wright","Mary TEST Wright")
)

df$organisation_id <- as.factor(df$organisation_id)
df$patient_id <- as.factor(df$patient_id)
df$date <- as.Date(df$date, "%d-%m-%Y")
df$consultation_mode <- as.factor(df$consultation_mode)
df$professional_id <- as.factor(df$professional_id)
df$professional_role <- as.factor(df$professional_role)

我想创建两个额外的列（include? 和 Nr_consultations_per_Pt_day），如下所示：

对于每个 organisation_id、patient_id、date 和 consultation_mode 检查：

1- 如果只有 1 行，include? = 1 并且 Nr_consultations_per_Pt_day = 1 professional_role.

2- 如果超过 1 行，include? = 1 对于每个不同的 professional_id 和 professional_name with consultation_role = 'Doctor' 或 'Nurse'.

注意：如果“医生”或“护士”有 2+ 个不同 professional_id 但相同 professional_name 的条目，则第一行获取include? = 1 和以下行 include? = 0。例如詹妮弗沃克的 25 / 26 个 ID。同样，如果“Doctor”或“Nurse”有 2 个以上的条目具有相同的 professional_id 但不同的 professional_name，则第一行得到 include? = 1，随后的行得到 include? = 0。例如Kevin R Collins / Kevin Collins 的 44 ID。

2.1- 如果有 0 个 'Doctor' 或 'Nurse'（全部'支持'），则第一行得到 include? = 1 和以下行 include? = 0，其中 Nr_consultations_per_Pt_day = 1 表示 professional_role.

中间数据集：

organisation_id patient_id date consultation_mode professional_id professional_role professional_name include?
1 1230 08-02-2018 Telephone 24 Doctor Dr John Taylor 1
1 1230 08-02-2018 Face-to-Face 11 Support Mary Wright 1
2 1222 12-01-2018 Telephone 123 Doctor Dr Patricia Jones 1
2 1222 12-01-2018 Telephone 110 Support James Davies 0
2 1244 12-01-2018 Face-to-Face 123 Doctor Dr Patricia Jones 1
2 1244 22-02-2018 Face-to-Face 110 Support James Davies 1
2 987 12-01-2018 Telephone 123 Doctor Dr Patricia Jones 1
2 987 22-02-2018 Telephone 333 Nurse Peter Hall 1
3 2223 01-03-2019 Home visit 444 Doctor Dr Mary Wilson 1
3 2223 01-03-2019 Home visit 444 Doctor Dr Mary Wilson 0
3 2247 01-03-2019 Face-to-Face 444 Doctor Dr Mary Wilson 1
3 2247 01-03-2019 Face-to-Face 444 Doctor Dr Mary Wilson 0
3 2247 01-03-2019 Face-to-Face 444 Doctor Dr Mary Wilson 0
4 1234 12-07-2020 Telephone 1133 Support Mary Wright 0
4 1234 12-07-2020 Telephone 12 Support Anthony Patel 0
4 1234 12-07-2020 Telephone 25 Nurse Jennifer Walker 1
4 1234 12-07-2020 Telephone 26 Nurse Jennifer Walker 0
4 1234 12-07-2020 Face-to-Face 12 Support Anthony Patel 0
4 1234 12-07-2020 Face-to-Face 34 Doctor Dr Carol Bell 1
4 1234 12-07-2020 Face-to-Face 35 Doctor Dr Carol Bell 0
4 1234 12-07-2020 Face-to-Face 38 Nurse Deborah Dixon 1
4 1239 13-07-2020 Home visit 44 Nurse Kevin R Collins 1
4 1239 13-07-2020 Home visit 44 Nurse Kevin Collins 0
4 1239 13-07-2020 Home visit 5556 Doctor Dr Robert Brown 1
4 3322 16-06-2021 Face-to-Face 443 Doctor Dr Mary Wilson 1
4 3322 16-06-2021 Telephone 443 Doctor Dr Mary Wilson 1
4 3322 16-06-2021 Face-to-Face 445 Doctor Dr John Snow 1
4 5434 14-05-2019 Telephone 29 Doctor Dr John Taylor 1
4 5434 14-05-2019 Face-to-Face 29 Doctor Dr John Taylor 1
4 4488 17-03-2020 Face-to-Face 555 Doctor Dr James Smith 1
4 4488 17-03-2020 Telephone 5556 Doctor Dr Robert Brown 1
4 4488 17-03-2020 Telephone 12 Support Anthony Patel 0
4 1250 03-02-2019 Face-to-Face 1133 Support Mary Wright 1
4 1250 03-02-2019 Face-to-Face 113663 Support Mary TEST Wright 0

最终数据集： 一个 organisation_id、patient_id、date 以及 consultation_mode 和 professional_role.

的每个类别的示例
organisation_id patient_id date consultation_mode professional_role Nr_consultations_per_Pt_day
1 1230 08-02-2018 Face-to-Face Doctor 0
1 1230 08-02-2018 Face-to-Face Nurse 0
1 1230 08-02-2018 Face-to-Face Support 1
1 1230 08-02-2018 Telephone Doctor 1
1 1230 08-02-2018 Telephone Nurse 0
1 1230 08-02-2018 Telephone Support 0
1 1230 08-02-2018 Home visit Doctor 0
1 1230 08-02-2018 Home visit Nurse 0
1 1230 08-02-2018 Home visit Support 0

等等

关于如何在 R 中以有效的方式执行此操作的任何想法？

如果我对你的描述理解正确，对于每一行我们要评估以下条件来决定是否 include? = 1:

  1. organisation_id-patient_id-date-consultation_mode 的行组大小为 1
  2. organisation_id-patient_id-date-consultation_mode 的行组大小大于 1，并且该行对应于：
    1. AND医生是第一个医生id/name
    2. 护士 AND 是 第一个 具有相同 id/name
      3. 的护士
    3. 支持并且是第一个支持并且是organisation_id-patient_id-date-consultation_mode没有医生或护士的小组的一部分

此逻辑将创建“中间”table。要创建“最终”table，我们遍历 consultation_mode 和 professional_role 的每个类别并设置 Nr_consultations_per_Pt_day = 1（如果存在 include? = 1 的对应条目）。

基于以上预期，我会这样做：

library(tidyverse)

# For each row, add the size of its 
# organisation_id-patient_id-date-consultation_mode group
df2 <- df %>% group_by(organisation_id, patient_id, date, consultation_mode) %>% 
    mutate(group_size = n()) %>% ungroup()

# For each row, indicate whether it's the first entry of 
# organisation_id-patient_id-date-consultation_mode-professional_role group 
# of people with the SAME NAME but possiblly different ID
df3 <- df2 %>% group_by(organisation_id, patient_id, date, consultation_mode, 
        professional_role, professional_name) %>% 
    mutate(first_by_name = row_number()==1) %>% 
    ungroup()

# For each row, indicate whether it's the first entry of 
# organisation_id-patient_id-date-consultation_mode-professional_role group 
# of people with the SAME ID but possiblly different name
df4 <- df3 %>% group_by(organisation_id, patient_id, date, consultation_mode, 
        professional_role, professional_id) %>% 
    mutate(first_by_id = row_number()==1) %>% 
    ungroup()

# For each row, indicate whether there's no doctor/nurse in its 
# organisation_id-patient_id-date-consultation_mode
# and indicate the first entry in such support-only group
df5 <- df4 %>% group_by(organisation_id, patient_id, date, consultation_mode) %>% 
    mutate(support_only_group = length(intersect(professional_role, c("Doctor", "Nurse"))) == 0) %>% 
    mutate(first_in_support_only = row_number()==1 & support_only_group) %>% 
    ungroup()

# Apply rules to determine the inclusion status of each row
df6 <- df5 %>% mutate(`include?` = if_else(
        group_size == 1 | 
        (professional_role %in% c("Doctor","Nurse") & (first_by_name & first_by_id)) |
        first_in_support_only, 1, 0)) 
df6

转换成最后的table:

# Convert into the final table
df7 <- df6 %>% 
    select(-c(group_size, first_by_name, first_by_id, support_only_group, first_in_support_only)) %>% 
    group_by(organisation_id, patient_id, date) %>% 
    expand(consultation_mode, professional_role) %>%
    left_join(df6) %>%
    mutate(Nr_consultations_per_Pt_day = replace_na(`include?`,0)) %>%
    select(-c(professional_id, professional_name, `include?`)) %>%
    group_by(organisation_id, patient_id, date, consultation_mode, professional_role) %>%
    summarise(Nr_consultations_per_Pt_day = sum(Nr_consultations_per_Pt_day))

df7 %>% filter(patient_id %in% c(2223, 1250, 1230))