英文:
How to retain class of variable in `tapply`?
问题
假设我的数据框设置如下：
X <- data.frame(
id = c('A', 'A', 'B', 'B'),
dt = as.Date(c('2020-01-01', '2020-01-02', '2021-01-01', '2021-01-02'))
)
我想要填充一个id特定的日期dt最小值的变量。
使用：X$dtmin <- with(X, tapply(dt, id, min)[id]) 返回一个数字，因为在tapply中的simplify=T将值转换为数字。为什么会这样呢？设置simplify=F会返回一个列表，其中列表中的每个元素具有所需的数据结构，但将数据框X中的变量填充回数字。然而，调用as.Date(<output>, origin='1970-01-01')似乎显得多余。如何保留dt的数据结构？
英文:
Suppose my data frame is set up like so:
X <- data.frame(
id = c('A', 'A', 'B', 'B'),
dt = as.Date(c('2020-01-01', '2020-01-02', '2021-01-01', '2021-01-02'))
)
and I want to populate a variable of the id-specific minimum value of date dt
Doing: X$dtmin <- with(X, tapply(dt, id, min)[id]) gives a numeric because the simplify=T in tapply has cast the value to numeric. Why has it done this? Setting simplify=F returns a list which each element in the list has the desired data structure, but populating the variable in my dataframe X casts these back to numeric. Yet calling as.Date(<output>, origin='1970-01-01') seems needlessly verbose. How can I retain the data structure of dt?
答案1
得分: 2
Sure, here is the translation of the provided code snippets:
我们可以使用
```R
X$dtmin <- with(X, do.call("c", tapply(dt, id, min, simplify = FALSE)[id]))
或者使用 dplyr
library(dplyr)
X %>%
mutate(dtmin = min(dt), .by = "id")
英文:
We may use
X$dtmin <- with(X, do.call("c", tapply(dt, id, min, simplify = FALSE)[id]))
Or use dplyr
library(dplyr)
X %>%
mutate(dtmin = min(dt), .by = "id")
</details>
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