将多个字符列转换为R中的as.Date和time [英] Convert multiple character columns to as.Date and time in R
问题描述
我们有一个任意数据集，称为df：
We have an arbitrary dataset, called df:
enter <- c("2017-01-01", "2018-02-02", "2018-03-03")
guest <- c("Foxtrot","Uniform","Charlie","Kilo")
disposal <- c("2017-01-05", "2018-02-05", "2018-03-09")
rating <- c("60","50","50")
date <- c("2017-04-10", "2018-04-15", "2018-04-20")
clock <- c("16:02:00", "17:02:00", "18:02:00")
rolex <- c("20:10:00", "20:49:00", "17:44:00")
df <- data.frame(enter,guest,disposal,rating,date,clock,rolex, stringsAsFactors = F)
我要完成的工作是更改列 enter ，处置 ，以及使用 dplyr 包从字符到日期的 date 。
因此，我想出了以下内容，只需将它们链接在一起即可：
What I try to accomplish is to change the columns enter, disposal, and date from character to date, using dplyr package. So, I came up with the following, simply by chaining it together:
library(dplyr)
library(chron)
df2 <- df %>% mutate(enter = as.Date(enter, format = "%Y-%m-%d"))
%>% mutate(disposal = as.Date(disposal, format = "%Y-%m-%d"))
%>% mutate(date = as.Date(date, format = "%Y-%m-%d"))
我要的是：需要从哪个 mutant 函数dplyr摆脱多重链接，即当我们有很多带有任意名称的暗示日期的列时？我想按名称指定列，然后应用as.Date函数将其从字符更改为日期。
What I am after is this: which mutate function is needed from dplyr to get rid of the multiple chaining, i.e. when we have lots of columns with arbitrary namings that imply dates? I want to specify the columns by name, and then apply the as.Date function to change them from character to date.
某些不适用于这种情况的操作的解决方案：
Some solutions to different operations that are not applicable to this case:
3: change multiple character columns to date
例如，我已经尝试过，但是没有运气：
For example, I've tried, but with no luck:
df2 <- df %>% mutate_at(data = df, each_of(c(enter, disposal, date)) = as.Date(format = "%Y-%m-%d"))
如以下所示： dplyr更改许多数据类型
注意 clock 和 rolex 列。使用chron软件包只需将它们转换为正确的格式，即时间而不是字符
Notice the clock and rolex columns. Using the chron package simply converts them to the right format, i.e. time instead of character
df2 <- df %>% mutate(clock = chron(times = clock)) %>% mutate(rolex = chron(times = rolex))
如此处建议：
将字符转换为r中的时间
现在，是否可以使用没有所有链接的相同解决方案，尤其是当我们有任意数量的具有不同命名的列等时？
Now, is the same solution available without all the chaining, especially when we have an arbitrary amount of columns with different namings etc.?
推荐答案
您只需要调整 mutate_at 的参数。将 as.Date 的所有其他参数指定为 mutate_at 的参数。
You just need to tweak the arguments of mutate_at. Any additional arguments to as.Date are specified as arguments to mutate_at.
df2 <- df %>% mutate_at(vars(enter,disposal,date), as.Date, format="%Y-%m-%d")
问题的第二部分也有类似的解决方案。
The second part of your question has a similar solution.
df2 <- df2 %>% mutate_at(vars(clock, rolex), function(x) chron(times. = x))
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