R语言计算均值(mean)和标准差(standard deviation)时如何排除四月数据

实现方法 #

要排除4月份数据计算均值和标准差，只需在筛选计算用数据集时新增月份过滤条件即可，具体修改如下：

1. 修正原函数的两处隐性问题：
   • 函数内部多处直接调用全局数据集data，未使用传入的参数dta，统一替换为参数名保证功能正常
   • 原代码中na_if替换空值的操作未赋值，导致后续过滤空DTPE的逻辑不生效，补充赋值操作
   • 移除base绘图语法中多余的+号，避免运行警告
2. 在生成计算均值标准差用的ds_CO数据集时，新增month(date2) != 4的过滤条件，排除所有4月的记录

修改后的完整可运行代码如下：

library(dplyr)
library(tidyr)
library(lubridate)

data <- structure(
  list(Id=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1),
       date1 = c("2021-06-20","2021-06-20","2021-06-20","2021-06-20","2021-06-20",
                 "2021-06-20","2021-06-20","2021-06-20","2021-06-20","2021-06-20","2021-06-20",
                 "2021-06-20","2021-06-20","2021-06-20","2021-06-20","2021-06-20","2021-06-20",
                 "2021-06-20","2021-06-20","2021-06-20","2021-06-20"),
       date2 = c("2021-07-01","2021-07-01","2021-07-01","2021-07-01","2021-04-02",
                 "2021-04-02","2021-04-02","2021-04-02","2021-04-02","2021-04-02","2021-04-03",
                 "2021-04-03","2021-04-03","2021-04-03","2021-04-03","2021-04-08","2021-04-08",
                 "2021-07-09","2021-07-09","2021-07-10","2021-07-10"),
       Week= c("Thursday","Thursday","Thursday","Thursday","Friday","Friday","Friday","Friday",
               "Friday","Friday","Saturday","Saturday","Saturday","Saturday","Saturday","Thursday",
               "Thursday","Friday","Friday","Saturday","Saturday"),
       DTPE = c("Ho","Ho","Ho","Ho","","","","","","","","","","","","","","","","Ho","Ho"),
       D1 = c(8,1,9, 3,5,4,7,6,3,8,2,3,4,6,7,8,4,2,6,2,3), DR01 = c(4,1,4,3,3,4,3,6,3,7,2,3,4,6,7,8,4,2,6,7,3),
       DR02 = c(8,1,4,3,3,4,1,6,3,7,2,3,4,6,7,8,4,2,6,2,3), DR03 = c(7,5,4,3,3,4,1,5,3,3,2,3,4,6,7,8,4,2,6,4,3),
       DR04= c(4,5,6,7,3,2,7,4,2,1,2,3,4,6,7,8,4,2,6,4,3),DR05 = c(9,5,4,3,3,2,1,5,3,7,2,3,4,7,7,8,4,2,6,4,3)),
  class = "data.frame", row.names = c(NA, -21L))

graph <- function(dt, dta = data) {
  
  dim_data<-dim(dta)
  
  day<-c(seq.Date(from = as.Date(dta$date2[1]), by = "days",
                  length = dim_data[1]
  )) 
  
  data_grouped <- dta %>%
    mutate(across(starts_with("date"), as.Date)) %>%
    group_by(date2) %>%
    summarise(Id = first(Id),
              date1 = first(date1),
              Week = first(Week),
              DTPE = first(DTPE),
              D1 = sum(D1)) %>%
    select(Id,date1,date2,Week,DTPE,D1)

  # 补充赋值，让空值替换生效
  data_grouped <- data_grouped %>% 
    mutate(DTPE = na_if(DTPE, ""))

  df_OC<-subset(data_grouped, DTPE == "")
  # 新增排除4月的过滤条件
  ds_CO = df_OC %>% filter(weekdays(date2) %in% weekdays(as.Date(dt)), month(date2) != 4)

  mean<-mean(ds_CO$D1)
  sd<-sd(ds_CO$D1)
  
  dta %>%
    filter(date2 == ymd(dt)) %>%
    summarize(across(starts_with("DR"), sum)) %>%
    pivot_longer(everything(), names_pattern = "DR(.+)", values_to = "val") %>%
    mutate(name = as.numeric(name)) %>%
    plot(xlab = "Days", ylab = "Number", xlim = c(0, 45),cex=1.5,cex.lab=1.5, 
         cex.axis=1.5, cex.main=2, cex.sub=2, lwd=2.5, ylim = c((min(.$val) %/% 10) * 15, (max(.$val) %/% 10 + 1) * 100))
  abline(h=mean, col='blue')
  abline(h=(mean + sd), col='green',lty=2) 
  abline(h=(mean - sd), col='orange',lty=2)
  
}  
graph("2021-07-10",data)

内容的提问来源于stack exchange，提问作者Antonio