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如何根据R中的日期填写缺失值?

[英]how to fill in missing values based on dates in R?

I have a data frame in the following format that represent a large data set that I have 我有以下格式的数据框,它们代表我拥有的大型数据集

F.names<-c('M','M','M','A','A')
L.names<-c('Ab','Ab','Ab','Ac','Ac')
year<-c('August 2015','September 2014','September 2016', 'August 2014','September 2013')
grade<-c(NA,'9th Grade','11th Grade',NA,'11th grade')

df.have<-data.frame(F.names,L.names,year,grade)

F.names L.names           year      grade
1       M      Ab    August 2015       <NA>
2       M      Ab September 2014  9th Grade
3       M      Ab September 2016 11th Grade
4       A      Ac    August 2014       <NA>
5       A      Ac September 2013 11th grade

The year column is in factor format in the original data set and there are several missing values for grade .Basically I want to fill in the missing grade values based on year column so that it looks like the following. year列是factor在原始数据集格式,并有几个遗漏值grade 。基本上我想填充缺失grade基于价值观year柱,使它看起来像下面这样。

F.names L.names           year      grade
1       M      Ab    August 2015 10th Grade
2       M      Ab September 2014  9th Grade
3       M      Ab September 2016 11th Grade
4       A      Ac    August 2014 12th Grade
5       A      Ac September 2013 11th grade

I was thinking that my first step would be to covert the year column which is in factor format to a date format. 我在想,我的第一个步骤是隐蔽的year柱是在factor格式的日期格式。 and then arrange the columns in order and use something like fill from tidyr to fill the missing columns. 然后安排列顺序和使用类似filltidyr ,以填补缺失的列。 How should I go about doing this, or is there a better way to approach this? 我应该如何去做,或者有更好的方法来做到这一点?

F.names<-c('M','M','M','A','A')
L.names<-c('Ab','Ab','Ab','Ac','Ac')
year<-c('August 2015','September 2014','September 2016', 'August 2014','September 2013')
grade<-c(NA,'9th Grade','11th Grade',NA,'11th grade')

df.have<-data.frame(F.names,L.names,year,grade)

library(tidyverse)

df.have %>%
  separate(year, c("m","y"), convert = T, remove = F) %>%
  separate(grade, c("num","type"), sep="th", convert = T) %>%
  arrange(F.names, y) %>%
  group_by(F.names) %>%
  mutate(num = ifelse(is.na(num), lag(num) + 1, num),
         type = "grade") %>%
  ungroup() %>%
  unite(grade, num, type, sep="th ") %>%
  select(-m, -y)

#   F.names L.names           year      grade
# 1       A      Ac September 2013 11th grade
# 2       A      Ac    August 2014 12th grade
# 3       M      Ab September 2014  9th grade
# 4       M      Ab    August 2015 10th grade
# 5       M      Ab September 2016 11th grade

This solution assumes that you won't have 2 or more consecutive NA s for a given F.names value. 此解决方案假定对于给定的F.names值,您将没有两个或多个连续的NA

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