如何从R中的日期中提取月份 [英] How to extract Month from date in R

问题描述

我正在使用lubridate包并应用month函数从日期中提取月份.我在日期字段上运行了str命令，我得到了

I am using the lubridate package and applying the month function to extract month from date. I ran the str command on date field and I got

Factor w/ 9498 levels "01/01/1979","01/01/1980",..: 5305 1 1 1 1 1 1 1 1 1 ...

> v1$Date<-month(v1$Date)
Error in as.POSIXlt.character(as.character(x), ...) : 
character string is not in a standard unambiguous format

这是我的数据框示例

https://drive.google.com/file/d/0B6cqWmwsEk20Q2dHblhXZi14Wk0/edit?usp =共享

我不知道我在做什么错.

I don't know what I am doing wrong.

推荐答案

?month状态:

日期时间必须是POSIXct，POSIXlt，日期，时间段，时间，年份， yearqtr，zoo，zooreg，timeDate，xts，其，ti，jul，timeSeries和fts 对象.

Date-time must be a POSIXct, POSIXlt, Date, Period, chron, yearmon, yearqtr, zoo, zooreg, timeDate, xts, its, ti, jul, timeSeries, and fts objects.

您的对象是一个因素，甚至不是字符向量(可能是由于stringsAsFactors = TRUE所致).您必须将向量转换为某个日期时间类，例如转换为POSIXlt:

Your object is a factor, not even a character vector (presumably because of stringsAsFactors = TRUE). You have to convert your vector to some datetime class, for instance to POSIXlt:

library(lubridate)
some_date <- c("01/02/1979", "03/04/1980")
month(as.POSIXlt(some_date, format="%d/%m/%Y"))
[1] 2 4

还有一个便捷功能dmy，可以执行相同的操作(@Henrik提出的提示):

There's also a convenience function dmy, that can do the same (tip proposed by @Henrik):

month(dmy(some_date))
[1] 2 4

更进一步，@ IShouldBuyABoat还提供了另一个提示，即接受dd/mm/yyyy字符格式而无需任何显式强制转换:

Going even further, @IShouldBuyABoat gives another hint that dd/mm/yyyy character formats are accepted without any explicit casting:

month(some_date)
[1] 2 4

有关格式的列表，请参见?strptime.您会发现标准明确格式"代表

For a list of formats, see ?strptime. You'll find that "standard unambiguous format" stands for

默认格式遵循ISO 8601国际标准 该标准将日期表示为"2001-02-28"，将时间表示为 "14:01:02"在此处使用前导零.

The default formats follow the rules of the ISO 8601 international standard which expresses a day as "2001-02-28" and a time as "14:01:02" using leading zeroes as here.

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