删除小于指定值的所有时间 [英] Delete all times less than a specified value

问题描述

这可能是一个简单的问题，但我是R的新手，无法找到答案（或谷歌搜索错误的东西）。
我目前正在处理一个项目，该项目涉及删除少于5分钟的所有时间值。使用lubridate包创建时间的数据示例如下。

This is probably a simple question however I am new to R and couldn't find an answer (or was googling the wrong thing). I am currently working on a project which involves deleting all Time values that are less than 5 minutes. An example of the data is as follows with the times created using the "lubridate" package.

Time
19S
1M 24S
7M 53S
11M 6S
.
.
.

现在我希望删除所有小于5分钟的值。因此，我希望得到的最终数据集是：

Now I wish to delete all values which are less than 5 minutes. Therefore the final dataset I wish to get is:

Time
7M 53S
11M 6S
.
.
.

任何帮助都会令人惊叹！
谢谢！

Any help would be amazing! Thanks!

推荐答案

你可以这样做：

df <- df[df$time > ms('5:00'), ]

结果：

> df
    time value
3 7M 53S     3
4 11M 6S     4

足够苛刻，将其转换为dplyr代码;它不起作用：

Strangly enough, converting this to dplyr code; it doesn't work:

filter(df, time > ms('5:00'))

结果：

   time
1   53S
2 1M 6S
Warning message:
In format.data.frame(x, digits = digits, na.encode = FALSE) :
  corrupt data frame: columns will be truncated or padded with NAs

我问了一个关于它的问题并找到了答案此处。你得到了很好的解决方案：

I asked a question about that and found a answer here. you get the good solution with:

df %>% 
  mutate(time = as.numeric(time)) %>% 
  filter(time > as.numeric(ms('5:00'))) %>% 
  mutate(time = ms(paste0(floor(time/60),':',round((time/60 - floor(time/60))*60))))

数据：

df <- data.frame(time = ms(c('0:19','1:24','7:53','11:6')), value = 1:4)

这篇关于删除小于指定值的所有时间的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！