我在下面提到了数据框:
Date Val1 Val2
2018-04-01 125 0.05
2018-04-03 458 2.99
2018-04-05 354 1.25
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我只考虑添加缺少的日期Sys.Date()(例如,此处Sys.Date()到数据框中对应的val1和val2为0,数据 2018-04-06)。
我试过了: t2<-merge(data.frame(Date= seq(min(ymd(t1$Date)), max(ymd(date)), by = "days")), t1, by = "Date", all = TRUE)
所需的数据框:
Date Val1 Val2
2018-04-01 125 0.05
2018-04-02 0 0
2018-04-03 458 2.99
2018-04-04 0 0
2018-04-05 354 1.25
2018-04-06 0 0
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这可以用 complete
library(tidyverse)
df1 %>%
complete(Date = seq(Date[1], Sys.Date(), by = "1 day"),
fill = list(Val1 = 0, Val2 = 0))
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如果我们需要为传递多个变量fill,请创建所需的列列表fill
nm1 <- setdiff(names(df1), "Date") #in this example excluding the Date
nm2 <- setNames(as.list(rep(0, length(nm1))), nm1)
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然后将其作为参数传递给 fill
df1 %>%
complete(Date = seq(Date[1], Sys.Date(), by = "1 day"), fill = nm2)
# A tibble: 35 x 3
# Date Val1 Val2
# <date> <dbl> <dbl>
# 1 2018-04-01 125 0.05
# 2 2018-04-02 0 0
# 3 2018-04-03 458 2.99
# 4 2018-04-04 0 0
# 5 2018-04-05 354 1.25
# 6 2018-04-06 0 0
# 7 2018-04-07 0 0
# 8 2018-04-08 0 0
# 9 2018-04-09 0 0
#10 2018-04-10 0 0
# ... with 25 more rows
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这是对你的方法的更正,以 R 为基础。
在您的实际应用程序中替换max(t1$Date)为Sys.Date():
t2<-merge(data.frame(Date= as.Date(min(t1$Date):max(t1$Date),"1970-1-1")),
t1, by = "Date", all = TRUE)
t2[is.na(t2)] <- 0
# Date Val1 Val2
# 1 2018-04-01 125 0.05
# 2 2018-04-02 0 0.00
# 3 2018-04-03 458 2.99
# 4 2018-04-04 0 0.00
# 5 2018-04-05 354 1.25
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数据
t1 <- read.table(text="Date Val1 Val2
'2018-04-01' 125 0.05
'2018-04-03' 458 2.99
'2018-04-05' 354 1.25",h=T,strin=F)
t1$Date <- as.Date(df$Date)
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