R语言如何删除datas变量中Numbers等于NA的对应行

要实现该需求，有两种常用的实现方案：

• 适配你已使用的dplyr语法，写法更统一：

datas <- datas %>% filter(!is.na(Numbers))

• 基础R语法，无需依赖额外包：

datas <- datas[!is.na(datas$Numbers), ]

完整可运行代码 #

你可以直接在原有代码末尾添加上述任意一行代码即可，完整示例如下：

library(dplyr)
library(tidyverse)
library(lubridate)
library(data.table)

df1 <- structure(
  list(date= c("2021-06-28","2021-06-28","2021-06-28","2021-06-28","2021-06-28",
       "2021-06-28","2021-06-28","2021-06-28"),
       date2 = c("2021-06-30","2021-06-30","2021-06-30","2021-07-01","2021-07-01","2021-07-01","2021-07-01","2021-07-01"),
       Code = c("ABC","CDE","FGH","ABC","CDE","FGH","ABC","CDE"),
       DR01 = c(4,1,4,3,3,4,3,6), DR02= c(4,2,6,7,3,2,7,4),DR03= c(9,5,4,3,3,2,1,5),
       DR04 = c(5,4,3,3,6,2,1,9),DR05 = c(5,4,5,3,6,2,1,9),
       DR06 = c(2,4,3,3,5,6,7,8),DR07 = c(2,5,4,4,9,4,7,8),
       DR08 = c(0,0,0,1,2,0,0,0),DR09 = c(0,0,0,0,0,0,0,0),DR010 = c(0,0,0,0,0,0,0,0),DR011 = c(4,0,0,0,0,0,0,0), 
       DR012 = c(0,0,0,3,0,0,0,5),DR013 = c(0,0,1,0,0,0,2,0),DR014 = c(0,0,0,1,0,2,0,0)),
  class = "data.frame", row.names = c(NA, -8L))

df1<-df1 %>% mutate(index=row_number()) %>%
  pivot_longer(starts_with('DR')) %>%
  mutate(rleid=rleid(value==0)) %>%
  group_by(index) %>%
  mutate(value=replace(value, value==0 & rleid==last(rleid), NA)) %>%
  select(-index, -rleid) %>%
  pivot_wider(names_from = name, values_from = value)

dmda<-"2021-07-01"


datas <- df1%>%
  filter(date2 == ymd(dmda)) %>% 
  group_by(Code) %>%
  summarize(across(starts_with("DR0"), sum)) %>%
  pivot_longer(cols = -Code, names_pattern = "DR0(.+)", 
               values_to = "val") %>% 
  mutate(name = readr::parse_number(name))
colnames(datas)[-1] <-c("Days","Numbers")

# 新增此行删除Numbers为NA的行
datas <- datas %>% filter(!is.na(Numbers))

内容的提问来源于stack exchange，提问作者user16774617