r中日期之间的增量天数

Question

我有一个很长的列表，需要日期之间的天数

ClientID <- c("00932", "00932", "00932")
Visit <- c("2018-11-10", "2018-11-20", "2018-11-25")
ClientID    Visit
00932    2018-11-10
00932    2018-11-20
00932    2018-11-25

我需要一个新的列，写

ClientID    Visit    Days
00932    2018-11-10    0
00932    2018-11-20    10
00932    2018-11-25    15

Answer 1

将Visit更改为日期类，并为每个ClientID减去Visit与Visit的最小日期。

library(dplyr)
df %>%
mutate(Visit = as.Date(Visit, '%m-%d-%Y')) %>%
group_by(ClientID) %>%
mutate(Days = as.integer(Visit - min(Visit))) %>%
ungroup
# ClientID Visit       Days
#  <chr>    <date>     <int>
#1 00932    2018-11-10     0
#2 00932    2018-11-20    10
#3 00932    2018-11-25    15

ClientID <- c("00932", "00932", "00932")
Visit <- c("11-10-2018", "11-20-2018", "11-25-2018")
df <- data.frame(ClientID, Visit)

Answer 2

考虑到您有多个ClientID，并希望在该级别上计算Days:

library(lubridate)
library(tidyverse)
ClientID <- c("00932", "00932", "00932")
Visit <- c("11-10-2018", "11-20-2018", "11-25-2018")
df <- data.frame(ClientID, Visit)
df %>%
group_by(ClientID) %>%
mutate(Visit= mdy(Visit),
Days= as.numeric(Visit-lag(Visit)))%>%
ungroup()%>%
mutate_if(is.numeric, ~replace_na(., 0))
# A tibble: 3 x 3
ClientID Visit       Days
<chr>    <date>     <dbl>
1 00932    2018-11-10     0
2 00932    2018-11-20    10
3 00932    2018-11-25     5

添加另一个ClientID和两个观察值来更好地演示它:

ClientID <- c("00932", "00932", "00932", "00935", "00935")
Visit <- c("11-10-2018", "11-20-2018", "11-25-2018",  "11-20-2019", "11-25-2019")
df <- data.frame(ClientID, Visit)
df %>%
group_by(ClientID) %>%
mutate(Visit= mdy(Visit),
Days= as.numeric(Visit-lag(Visit)))%>%
ungroup()%>%
mutate_if(is.numeric, ~replace_na(., 0))
# A tibble: 5 x 3
ClientID Visit       Days
<chr>    <date>     <dbl>
1 00932    2018-11-10     0
2 00932    2018-11-20    10
3 00932    2018-11-25     5
4 00935    2019-11-20     0
5 00935    2019-11-25     5

按注释中的要求添加函数:

days_func <- function(df){

df %>%
group_by(ClientID) %>%
mutate(Visit= mdy(Visit),
Days= as.numeric(Visit-lag(Visit)))%>%
ungroup()%>%
mutate_if(is.numeric, ~replace_na(., 0))->df

return(df)
}
df1 <- days_func(df)
df1
# A tibble: 3 x 3
ClientID Visit       Days
<chr>    <date>     <dbl>
1 00932    2018-11-10     0
2 00932    2018-11-20    10
3 00932    2018-11-25     5

Answer 3

我假设您需要日期之间的日跨度，而不是日期和最近日期之间的日跨度，我建议这样做:

dn <- as.numeric(as.Date(Visit))

带日期的文本一致地转换为日期和数字。

dn2 <- c(dn[1], dn[-length(dn)])

我们准备第二个向量进行减法，因为它将快速工作。它的成员顺序如下:

df df2

1日1日2日1日

3日2n nth-1

Days <- dn - dn2

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