将日期时间字符串解析为Date类的最快方法 [英] Fastest way to parse a date-time string to class Date

问题描述

我有一列以日期为字符的列，格式为 10/17/2017 12:00:00 AM .我想解析字符串并仅将日期部分保留为类 Date ，即 2017-10-17 .我正在使用-

I have a column with dates as character in the format 10/17/2017 12:00:00 AM. I want parse the string and keep only the date part as class Date, i.e. 2017-10-17. I am using -

df$ReportDate = as.Date(df$ReportDate, format = "%m/%d/%Y %I:%M:%S %p") 
df$ReportDate = as.Date(format(df$ReportDate, "%Y-%m-%d"))

这行得通，但是数据帧有超过500万行，因此大约需要2分钟.

this works, but the dataframe has over 5 million rows so this takes close to two mins.

  user  system elapsed 
104.73    0.55  105.46 

是否有更快，更有效的方法来做到这一点?

Is there a faster and more efficient way to do this?

推荐答案

请注意，日期后 as.Date 会忽略垃圾，因此在我使用的不是特别快的笔记本电脑上，此时间不到10秒:/p>

Note that as.Date will ignore junk after the date so this takes less than 10 seconds on my not particularly fast laptop:

xx <- rep("10/17/2017 12:00:00 AM", 5000000) # test input
system.time(as.Date(xx, "%m/%d/%Y"))
## user  system elapsed 
## 9.57    0.20    9.82 

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