让 lubridate 减法只返回一个数值 [英] Have lubridate subtraction return only a numeric value

问题描述

我有一个名为 Started 的变量，它是人类受试者参加研究的日期，另一个名为 dos1 的变量是受试者最后一次参加研究的日期手术.我想计算从他们最后一次手术到入学之日的几个月.我试过了:

I have one variable called Started which is the date on which human subjects enrolled in a study and another variable called dos1 which is the date upon which the subject last had surgery. I want to work out how many months since their last surgery to the day of enrollment. I tried:

as.period(syrrupan$Started-syrrupan$dos1,units=c("month"))

我希望这会给我类似的东西:

I expected this to give me something like:

14, 18, 1, 26 

每个数字都是月数.

相反，我得到:

1 year, -4 months, -5 days and -1 hours   1 year, -5 months, -23 days and -1 hours   1 year, -7 months, 2 days and -1 hours   1 year, -8 months, -28 days and 1 hour   1 year, -7 months, -23 days and 1 hour.   

我怎样才能得到月份的数值?

How can I get just the numeric value of months?

推荐答案

你可以尝试使用 difftime 代替，即:

You could try using difftime instead, ie:

difftime(syrrupan$Started,syrrupan$dos1,units="days")

请注意，这将为您提供一个 difftime 类的对象，如果您想要一个数字向量，请在其周围包裹一个 as.numeric.另请注意，您不能选择月份作为单位选项，但您应该坚持使用具有固定长度的时间单位.

Note that this will give you an object of class difftime, if you want a numeric vector, wrap an as.numeric around it. Note also that you can't choose months as an option for units, but you should really stick with a time unit that has a fixed length.

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