删除数字 = NA 的行答案

【问题标题】：Remove the lines that have numbers = NA删除数字 = NA 的行
【发布时间】：2021-10-04 23:03:14
【问题描述】：

我想从我的datas 变量中删除具有Numbers = NA 的行。我将在下面插入可执行代码供您查看。

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")

【问题讨论】：

原始问题的答案有帮助吗？对于这个新问题，如果您省略所有这些代码并仅分享dput(datas)，可能会更清楚

标签： r

【解决方案1】：

我们可以使用filter 和complete.cases

library(dplyr)
datas <- datas %>% 
    filter(complete.cases(Numbers))

或者也可以使用na.omit

datas <- na.omit(datas)

或者在代码中，在pivot_longer中指定values_drop_na = TRUE

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", values_drop_na = TRUE) %>% 
  mutate(name = readr::parse_number(name))
colnames(datas)[-1] <-c("Days","Numbers")

【讨论】：