creating sequence of dates for each group in r

杀马特。学长 韩版系。学妹 提交于 2019-12-29 00:07:06

问题


I have a dataset that looks like this:

      ID    created_at
MUM-0001    2014-04-16
MUM-0002    2014-01-14
MUM-0003    2014-04-17
MUM-0004    2014-04-12
MUM-0005    2014-04-18
MUM-0006    2014-04-17

I am trying to introduce new column that would be all dates between start date and defined last day (say, 12th-july-2015). I used seq function in dplyr but getting an error.

data1 <- data1 %>%
         arrange(ID) %>%
         group_by(ID) %>%
         mutate(date = seq(as.Date(created_at), as.Date('2015-07-12'), by= 1))

the error which I am getting is:

Error: incompatible size (453), expecting 1 (the group size) or 1

Can you please suggest some better way to perform this task in R ?


回答1:


You could use data.table to get the sequence of Dates from 'created_at' to '2015-07-12', grouped by the 'ID' column.

 library(data.table)
 setDT(df1)[, list(date=seq(created_at, as.Date('2015-07-12'), by='1 day')) , ID]

If you need an option with dplyr, use do

 library(dplyr)
 df1 %>% 
   group_by(ID) %>% 
   do( data.frame(., Date= seq(.$created_at,
                            as.Date('2015-07-12'), by = '1 day')))

If you have duplicate IDs, then we may need to group by row_number()

df1 %>%
    group_by(rn=row_number()) %>%
     do(data.frame(ID= .$ID, Date= seq(.$created_at,
          as.Date('2015-07-12'), by = '1 day'), stringsAsFactors=FALSE))

Update

Based on @Frank's commment, the new idiom for tidyverse is

library(tidyverse)
df1 %>%
  group_by(ID) %>% 
  mutate(d = list(seq(created_at, as.Date('2015-07-12'), by='1 day')), created_at = NULL) %>%
  unnest()

In the case of data.table

setDT(df1)[, list(date=seq(created_at, 
             as.Date('2015-07-12'), by = '1 day')), by = 1:nrow(df1)] 

data

df1 <- structure(list(ID = c("MUM-0001", "MUM-0002", "MUM-0003",
 "MUM-0004", 
 "MUM-0005", "MUM-0006"), created_at = structure(c(16176, 16084, 
16177, 16172, 16178, 16177), class = "Date")), .Names = c("ID", 
"created_at"), row.names = c(NA, -6L), class = "data.frame")


来源:https://stackoverflow.com/questions/31873462/creating-sequence-of-dates-for-each-group-in-r

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