Question

我有一个像这样的数据框

TXT

我正在尝试将2小时的小时列与小时及其单位转换为2列。

我想要的输出

ID <- c("A","A","A","A","A","A","B","B","B","B","B","B")
Day <- c("Mon","Mon","Mon","Fri","Fri","Fri","Tue","Tue","Tue","Wed","Wed","Wed")
minute <- c(49,32,15,38,18,16,06,16,26,31,33,38)
second <- c(12,22,08,16,21,42,41,48,32,21,26,18)
hour0 <- c(0,0,0,60,0,0,0,0,0,0,0,0)
hour1 <- c(0,100,0,0,0,0,68,0,0,0,0,0)
hour2 <- c(0,0,0,0,0,0,0,92,0,0,0,72)
hour3 <- c(0,0,92,0,62,0,0,0,81,0,0,0)
hour4 <- c(110,0,0,0,0,0,0,0,0,93,0,0)
hour5 <- c(0,0,0,0,0,112,0,0,0,0,0,0)
hour6 <- c(0,0,0,0,0,0,0,0,0,0,105,0)

df <- data.frame(ID,Day,minute,second,hour0,hour1,hour2,hour3,hour4,hour5,hour6,
                 stringsAsFactors=FALSE)

我正在尝试以这种方式进行操作，但这并不是我想要的那样

   ID Day minute second hour unit
1   A Mon     49     12    4  110
2   A Mon     32     22    1  100
3   A Mon     15      8    3   92
4   A Fri     38     16    0   60
5   A Fri     18     21    3   62
6   A Fri     16     42    5  112
7   B Tue      6     41    1   68
8   B Tue     16     48    2   92
9   B Tue     26     32    3   81
10  B Wed     31     21    4   93
11  B Wed     33     26    6  105
12  B Wed     38     18    2   72

有人可以指出我正确的方向吗？

Answer 1

它可以与starts_with或matches匹配，而不是指定'hour0'，'hour1'等，然后在'unit'为0的行中filter进行匹配。从“小时”（parse_number）中提取数字部分

library(tidyverse)
df %>% 
  gather(hour, unit, starts_with("hour")) %>% 
  filter(unit != 0) %>% 
  mutate(hour = readr::parse_number(hour)) %>%
  arrange(ID, 
       factor(Day, levels = c("Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun")))

将几个小时的专栏重塑为关键和价值

1 个答案: