Question

我的问题采用了一般性方面，比较了How to remove time-field string from a date-as-character variable?提出的问题。

事实上，假设我有这个日期类型变量：

> head(DataDia$Date)
[1] "2016-09-13 15:56:30.827" "2016-12-12 13:39:17.537" "2016-09-16 21:57:24.977" "2016-09-23 11:19:22.010"
[5] "2017-01-11 20:06:58.490" "2016-10-21 23:40:43.927"

如何删除所有时间字段字符串并保留日期格式。所以我明白了：

> head(DataDia$Date)
[1] "2016-09-13" "2016-12-12" "2016-09-16" "2016-09-23"
[5] "2017-01-11" "2016-10-21"

请注意，我正在处理数据表。所以我需要一种使用data.table的方法操作

Answer 1

只需使用as.Date(DataDia$Date)。

Answer 2

您可以使用：

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Answer 3

合并as.Date和as.character

x = c("2016-09-13 15:56:30.827", "2016-12-12 13:39:17.537", "2016-09-16 21:57:24.977", "2016-09-23 11:19:22.010",
        "2017-01-11 20:06:58.490", "2016-10-21 23:40:43.927")

y = as.character(as.Date(x, format = "%Y-%m-%d"))

y
[1] "2016-09-13" "2016-12-12" "2016-09-16" "2016-09-23" "2017-01-11" "2016-10-21"

如何在没有时间字符串的情况下保留日期变量

3 个答案: