dplyr::lead 或 data.table::shift 引用变量值而不是标量
dplyr::lead or data.table::shift refer to variable value rather than scalar
给定：
library(tidyverse)
df <- data.frame(id = c(1, 1, 1, 1, 1,
rep(2, 5), rep(3, 3)),
dates = as.Date(c("2015-01-01",
"2015-01-02",
"2015-01-02",
"2015-01-03",
"2015-01-04",
"2015-02-22",
"2015-02-23",
"2015-02-23",
"2015-02-23",
"2015-02-25",
"2015-03-13",
"2015-03-14",
"2015-03-15")),
indicator = c(0, 1, 0, 0, 0,
0, 1, 0, 0, 0,
0, 1, 0),
final_date = as.Date(rep(NA, 13))) %>%
group_by(id, dates) %>%
mutate(repeat_days = n())
df
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 NA 1
# 2 1 2015-01-02 1 NA 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 0 NA 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 NA 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 NA 1
# 13 3 2015-03-15 0 NA 1
基于条件 (indicator == 1)，我想通过变量 (repeat_days) 中的值 lead dates 而不是提供缩放器值，所以我想要的输出看起来像：
#df_final
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 NA 1
# 2 1 2015-01-02 1 2015-01-03 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 0 NA 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 2015-02-25 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 2015-03-15 1
# 13 3 2015-03-15 0 NA 1
如果我们想 lead 一个标量，例如1，这个有效：
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, n = 1), TRUE ~ final_date))
但是当我提供一个变量时，它不会按预期工作，因为它不是标量:
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, repeat_days), TRUE ~ final_date))
# Error: Problem with `mutate()` column `final_date`.
# i `final_date = case_when(...)`.
# x `n` must be a nonnegative integer scalar, not an integer vector of length 5.
# i The error occurred in group 1: id = 1.
这也不起作用，因为它指的是 repeat_days 在所有这些情况下都是 1 的组的第一次出现：
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, repeat_days[1]), TRUE ~ final_date))
有没有办法直接引用 repeat_days 的行级别值而不创建额外的变量？
谢谢
编辑
感谢@Maël 很好的回答：
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, repeat_days[indicator == 1]),
TRUE ~ final_date))
我应该明确表示我也可以让每个组重复 indicator == 1 所以它也需要在这个数据集上工作：
df <- data.frame(id = c(1, 1, 1, 1, 1,
rep(2, 5), rep(3, 3), 4, 4),
dates = as.Date(c("2015-01-01",
"2015-01-02",
"2015-01-02",
"2015-01-03",
"2015-01-04",
"2015-02-22",
"2015-02-23",
"2015-02-23",
"2015-02-23",
"2015-02-25",
"2015-03-13",
"2015-03-14",
"2015-03-15",
"2015-04-15",
"2015-04-16")),
indicator = c(0, 1, 0, 1, 0,
0, 1, 0, 0, 0,
0, 1, 0, 0, 1),
final_date = as.Date(c("2015-01-01", rep(NA, 14)))) %>%
group_by(id, dates) %>%
mutate(repeat_days = n()) %>%
ungroup()
df
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 2015-01-01 1
# 2 1 2015-01-02 1 NA 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 1 NA 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 NA 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 NA 1
# 13 3 2015-03-15 0 NA 1
# 14 4 2015-04-15 0 NA 1
# 15 4 2015-04-16 1 NA 1
注意 id == 4，没有提前日期，所以在这种情况下我希望它默认为他们当前的行。此外，第一行现在已经有一个 final_date 值，因此需要使用 case_when 或类似的东西。
期望的输出：
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 2015-01-01 1
# 2 1 2015-01-02 1 2015-01-03 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 1 2015-01-04 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 2015-02-25 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 2015-03-15 1
# 13 3 2015-03-15 0 NA 1
# 14 4 2015-04-15 0 NA 1
# 15 4 2015-04-16 1 2015-04-16 1
相关链接, and 但我无法在这个有条件的特殊情况下实现类似的东西。也很高兴看到 data.table (shift?) 解决方案。
我想出了一个解决方案，它使用来自基础 R 的 sapply()。
library(dplyr)
df %>%
ungroup() %>%
mutate(final_date = as.Date(sapply(1:nrow(df), function(x)
ifelse(is.na(df$final_date[x]),
ifelse(df$indicator[x] == 1,
ifelse(is.na(df$id[x] == df$id[x + df$repeat_days[x]]),
format(as.Date(df$dates[x], origin = "2020-01-01")),
ifelse(df$id[x] == df$id[x + df$repeat_days[x]],
format(as.Date(df$dates[x + df$repeat_days[x]], origin = "2020-01-01")),
NA)),
NA),
as.character(df$final_date[x])))))
# A tibble: 15 × 5
id dates indicator final_date repeat_days
<dbl> <date> <dbl> <date> <int>
1 1 2015-01-01 0 2015-01-01 1
2 1 2015-01-02 1 2015-01-03 2
3 1 2015-01-02 0 NA 2
4 1 2015-01-03 1 2015-01-04 1
5 1 2015-01-04 0 NA 1
6 2 2015-02-22 0 NA 1
7 2 2015-02-23 1 2015-02-25 3
8 2 2015-02-23 0 NA 3
9 2 2015-02-23 0 NA 3
10 2 2015-02-25 0 NA 1
11 3 2015-03-13 0 NA 1
12 3 2015-03-14 1 2015-03-15 1
13 3 2015-03-15 0 NA 1
14 4 2015-04-15 0 NA 1
15 4 2015-04-16 1 2015-04-16 1
编写一个接受 n 向量的 lead 函数可能会更简单。下面我称这个函数为lead2。其余代码保持不变。
更新：您进一步说明，如果indicator = 1但没有提前日期，则final_date应填写当前日期。这可以用 dplyr::coalesce 来实现，它会找到向量中的第一个 non-null 元素。它类似于 SQL COALESCE 运算符。
library("tidyverse")
df <- data.frame(
id = c(
1, 1, 1, 1, 1,
rep(2, 5), rep(3, 3), 4, 4
),
dates = as.Date(c(
"2015-01-01",
"2015-01-02",
"2015-01-02",
"2015-01-03",
"2015-01-04",
"2015-02-22",
"2015-02-23",
"2015-02-23",
"2015-02-23",
"2015-02-25",
"2015-03-13",
"2015-03-14",
"2015-03-15",
"2015-04-15",
"2015-04-16"
)),
indicator = c(
0, 1, 0, 1, 0,
0, 1, 0, 0, 0,
0, 1, 0, 0, 1
),
final_date = as.Date(c("2015-01-01", rep(NA, 14)))
) %>%
group_by(id, dates) %>%
mutate(repeat_days = n()) %>%
ungroup()
lead2 <- function(x, ns) {
# x: vector of values
# ns: vector of leads
# Compute the target position for each element
is <- seq_along(x) + ns
x[is]
}
xs <- c("a", "b", "c", "d", "e", "f")
ns <- c(1, 1, 2, 3, 1, 2)
lead2(xs, ns)
#> [1] "b" "c" "e" NA "f" NA
df %>%
group_by(id) %>%
mutate(
final_date = if_else(
is.na(final_date) & indicator == 1,
coalesce(lead2(dates, repeat_days), dates),
final_date
)
)
#> # A tibble: 15 × 5
#> # Groups: id [4]
#> id dates indicator final_date repeat_days
#> <dbl> <date> <dbl> <date> <int>
#> 1 1 2015-01-01 0 2015-01-01 1
#> 2 1 2015-01-02 1 2015-01-03 2
#> 3 1 2015-01-02 0 NA 2
#> 4 1 2015-01-03 1 2015-01-04 1
#> 5 1 2015-01-04 0 NA 1
#> 6 2 2015-02-22 0 NA 1
#> 7 2 2015-02-23 1 2015-02-25 3
#> 8 2 2015-02-23 0 NA 3
#> 9 2 2015-02-23 0 NA 3
#> 10 2 2015-02-25 0 NA 1
#> 11 3 2015-03-13 0 NA 1
#> 12 3 2015-03-14 1 2015-03-15 1
#> 13 3 2015-03-15 0 NA 1
#> 14 4 2015-04-15 0 NA 1
#> 15 4 2015-04-16 1 2015-04-16 1
由 reprex package (v2.0.1)
于 2022 年 3 月 14 日创建
给定：
library(tidyverse)
df <- data.frame(id = c(1, 1, 1, 1, 1,
rep(2, 5), rep(3, 3)),
dates = as.Date(c("2015-01-01",
"2015-01-02",
"2015-01-02",
"2015-01-03",
"2015-01-04",
"2015-02-22",
"2015-02-23",
"2015-02-23",
"2015-02-23",
"2015-02-25",
"2015-03-13",
"2015-03-14",
"2015-03-15")),
indicator = c(0, 1, 0, 0, 0,
0, 1, 0, 0, 0,
0, 1, 0),
final_date = as.Date(rep(NA, 13))) %>%
group_by(id, dates) %>%
mutate(repeat_days = n())
df
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 NA 1
# 2 1 2015-01-02 1 NA 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 0 NA 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 NA 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 NA 1
# 13 3 2015-03-15 0 NA 1
基于条件 (indicator == 1)，我想通过变量 (repeat_days) 中的值 lead dates 而不是提供缩放器值，所以我想要的输出看起来像：
#df_final
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 NA 1
# 2 1 2015-01-02 1 2015-01-03 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 0 NA 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 2015-02-25 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 2015-03-15 1
# 13 3 2015-03-15 0 NA 1
如果我们想 lead 一个标量，例如1，这个有效：
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, n = 1), TRUE ~ final_date))
但是当我提供一个变量时，它不会按预期工作，因为它不是标量:
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, repeat_days), TRUE ~ final_date))
# Error: Problem with `mutate()` column `final_date`.
# i `final_date = case_when(...)`.
# x `n` must be a nonnegative integer scalar, not an integer vector of length 5.
# i The error occurred in group 1: id = 1.
这也不起作用，因为它指的是 repeat_days 在所有这些情况下都是 1 的组的第一次出现：
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, repeat_days[1]), TRUE ~ final_date))
有没有办法直接引用 repeat_days 的行级别值而不创建额外的变量？
谢谢
编辑 感谢@Maël 很好的回答：
df %>%
group_by(id) %>%
mutate(final_date = case_when(is.na(final_date) & indicator == 1 ~
lead(dates, repeat_days[indicator == 1]),
TRUE ~ final_date))
我应该明确表示我也可以让每个组重复 indicator == 1 所以它也需要在这个数据集上工作：
df <- data.frame(id = c(1, 1, 1, 1, 1,
rep(2, 5), rep(3, 3), 4, 4),
dates = as.Date(c("2015-01-01",
"2015-01-02",
"2015-01-02",
"2015-01-03",
"2015-01-04",
"2015-02-22",
"2015-02-23",
"2015-02-23",
"2015-02-23",
"2015-02-25",
"2015-03-13",
"2015-03-14",
"2015-03-15",
"2015-04-15",
"2015-04-16")),
indicator = c(0, 1, 0, 1, 0,
0, 1, 0, 0, 0,
0, 1, 0, 0, 1),
final_date = as.Date(c("2015-01-01", rep(NA, 14)))) %>%
group_by(id, dates) %>%
mutate(repeat_days = n()) %>%
ungroup()
df
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 2015-01-01 1
# 2 1 2015-01-02 1 NA 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 1 NA 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 NA 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 NA 1
# 13 3 2015-03-15 0 NA 1
# 14 4 2015-04-15 0 NA 1
# 15 4 2015-04-16 1 NA 1
注意 id == 4，没有提前日期，所以在这种情况下我希望它默认为他们当前的行。此外，第一行现在已经有一个 final_date 值，因此需要使用 case_when 或类似的东西。
期望的输出：
# id dates indicator final_date repeat_days
# <dbl> <date> <dbl> <date> <int>
# 1 1 2015-01-01 0 2015-01-01 1
# 2 1 2015-01-02 1 2015-01-03 2
# 3 1 2015-01-02 0 NA 2
# 4 1 2015-01-03 1 2015-01-04 1
# 5 1 2015-01-04 0 NA 1
# 6 2 2015-02-22 0 NA 1
# 7 2 2015-02-23 1 2015-02-25 3
# 8 2 2015-02-23 0 NA 3
# 9 2 2015-02-23 0 NA 3
# 10 2 2015-02-25 0 NA 1
# 11 3 2015-03-13 0 NA 1
# 12 3 2015-03-14 1 2015-03-15 1
# 13 3 2015-03-15 0 NA 1
# 14 4 2015-04-15 0 NA 1
# 15 4 2015-04-16 1 2015-04-16 1
相关链接data.table (shift?) 解决方案。
我想出了一个解决方案，它使用来自基础 R 的 sapply()。
library(dplyr)
df %>%
ungroup() %>%
mutate(final_date = as.Date(sapply(1:nrow(df), function(x)
ifelse(is.na(df$final_date[x]),
ifelse(df$indicator[x] == 1,
ifelse(is.na(df$id[x] == df$id[x + df$repeat_days[x]]),
format(as.Date(df$dates[x], origin = "2020-01-01")),
ifelse(df$id[x] == df$id[x + df$repeat_days[x]],
format(as.Date(df$dates[x + df$repeat_days[x]], origin = "2020-01-01")),
NA)),
NA),
as.character(df$final_date[x])))))
# A tibble: 15 × 5
id dates indicator final_date repeat_days
<dbl> <date> <dbl> <date> <int>
1 1 2015-01-01 0 2015-01-01 1
2 1 2015-01-02 1 2015-01-03 2
3 1 2015-01-02 0 NA 2
4 1 2015-01-03 1 2015-01-04 1
5 1 2015-01-04 0 NA 1
6 2 2015-02-22 0 NA 1
7 2 2015-02-23 1 2015-02-25 3
8 2 2015-02-23 0 NA 3
9 2 2015-02-23 0 NA 3
10 2 2015-02-25 0 NA 1
11 3 2015-03-13 0 NA 1
12 3 2015-03-14 1 2015-03-15 1
13 3 2015-03-15 0 NA 1
14 4 2015-04-15 0 NA 1
15 4 2015-04-16 1 2015-04-16 1
编写一个接受 n 向量的 lead 函数可能会更简单。下面我称这个函数为lead2。其余代码保持不变。
更新：您进一步说明，如果indicator = 1但没有提前日期，则final_date应填写当前日期。这可以用 dplyr::coalesce 来实现，它会找到向量中的第一个 non-null 元素。它类似于 SQL COALESCE 运算符。
library("tidyverse")
df <- data.frame(
id = c(
1, 1, 1, 1, 1,
rep(2, 5), rep(3, 3), 4, 4
),
dates = as.Date(c(
"2015-01-01",
"2015-01-02",
"2015-01-02",
"2015-01-03",
"2015-01-04",
"2015-02-22",
"2015-02-23",
"2015-02-23",
"2015-02-23",
"2015-02-25",
"2015-03-13",
"2015-03-14",
"2015-03-15",
"2015-04-15",
"2015-04-16"
)),
indicator = c(
0, 1, 0, 1, 0,
0, 1, 0, 0, 0,
0, 1, 0, 0, 1
),
final_date = as.Date(c("2015-01-01", rep(NA, 14)))
) %>%
group_by(id, dates) %>%
mutate(repeat_days = n()) %>%
ungroup()
lead2 <- function(x, ns) {
# x: vector of values
# ns: vector of leads
# Compute the target position for each element
is <- seq_along(x) + ns
x[is]
}
xs <- c("a", "b", "c", "d", "e", "f")
ns <- c(1, 1, 2, 3, 1, 2)
lead2(xs, ns)
#> [1] "b" "c" "e" NA "f" NA
df %>%
group_by(id) %>%
mutate(
final_date = if_else(
is.na(final_date) & indicator == 1,
coalesce(lead2(dates, repeat_days), dates),
final_date
)
)
#> # A tibble: 15 × 5
#> # Groups: id [4]
#> id dates indicator final_date repeat_days
#> <dbl> <date> <dbl> <date> <int>
#> 1 1 2015-01-01 0 2015-01-01 1
#> 2 1 2015-01-02 1 2015-01-03 2
#> 3 1 2015-01-02 0 NA 2
#> 4 1 2015-01-03 1 2015-01-04 1
#> 5 1 2015-01-04 0 NA 1
#> 6 2 2015-02-22 0 NA 1
#> 7 2 2015-02-23 1 2015-02-25 3
#> 8 2 2015-02-23 0 NA 3
#> 9 2 2015-02-23 0 NA 3
#> 10 2 2015-02-25 0 NA 1
#> 11 3 2015-03-13 0 NA 1
#> 12 3 2015-03-14 1 2015-03-15 1
#> 13 3 2015-03-15 0 NA 1
#> 14 4 2015-04-15 0 NA 1
#> 15 4 2015-04-16 1 2015-04-16 1
由 reprex package (v2.0.1)
于 2022 年 3 月 14 日创建