从R的工期列中拆分小时和分钟

Question

在我的数据集中有一个名为duration的列。从中我想将小时和分钟分为2个单独的列。如果没有小时或分钟，则要相应地添加0h或0m。

在下面的图像中提供了相同的现有列详细信息以及预期的新列：

train <- read.csv("sampledata.csv", stringsAsFactors = F)
train$Duration

编辑：

sampledata <- data.frame(
   emp_id = c (1:5), 
   Duration = c("10h 50m","5h 34m","9h","4h 15m","23m"),
   stringsAsFactors = FALSE
)

sampledata$Duration

Answer 1

使用sub（）和gsub的解决方案如下

# first identify strings with "h"
h_in_str <- grepl("h", sampledata$Duration)
# if string has "h", then return all before "h" or else return 0
sampledata$Hours <- ifelse(h_in_str, sub("h.*", "", sampledata$Duration), 0)

# identify strings with "m"
m_in_str <- grepl("m", sampledata$Duration)
# if string has "m", return all numbers without those preceding "h" or else return 0
sampledata$Minutes <- ifelse(m_in_str, 
gsub("([0-9]+).*$", "\\1", sub(".*h", "", sampledata$Duration)), 0)

这将为您提供所需的数据

sampledata
emp_id Duration Hours Minutes
1      1  10h 50m   10      50
2      2   5h 34m    5      34
3      3       9h    9       0
4      4   4h 15m    4      15
5      5      23m    0      23

Answer 2

我不会说最好的答案，但是一种方法是

#Get numbers next to hours and minutes
hour_minute <- sub("(\\d+)h (\\d+)m", "\\1-\\2", sampledata$Duration)

sampledata[c("hour", "minutes")] <- t(sapply(strsplit(hour_minute, "-"), 
function(x) {
  if (length(x) == 2) x 
  else if (endsWith(x, "h")) c(sub("h", "", x), 0)
  else c(0, sub("m", "", x))
}))

sampledata
  emp_id Duration hour minutes
1      1  10h 50m   10      50
2      2   5h 34m    5      34
3      3       9h    9       0
4      4   4h 15m    4      15
5      5      23m    0      23