R拆分列取决于值

Question

我有以下数据,我想计算总分钟数,我想知道是否可以将列拆分为两列,其中一列中的分钟和另一列中的秒数？

> q
       time
1   0m 22s 
2    1m 7s 
3   3m 35s 
4  11m 43s 
5    1m 8s 
6   2m 21s 
7   9m 33s 
8   0m 56s 
9    0m 2s 
10   0m 2s 
11  0m 50s 
12  0m 25s 
13  0m 33s 
14  2m 26s 
15  0m 20s 
16  1m 47s 
17  0m 36s 
18   0m 3s 
19   0m 2s 
20   0m 5s 

==>给:

> q
    min    seconds
1   0     22
2   1     7

等等

Answer 1

我对日期不熟悉,但你可以查看函数as.Date或strptime.使用您的data.frame:

df <- data.frame(time = c("0m 22s", "1m 7s", "3m 35s", "11m 43s", "1m 8s", "2m 21s", "9m 33s", "0m 56s", "0m 2s", "0m 2s", "0m 50s", "0m 25s", "0m 33s", "2m 26s", "0m 20s", "1m 47s", "0m 36s", "0m 3s", "0m 2s", "0m 5s"))

df$time.2 <- strptime(df$time, "%Mm %Ss")

现在您可以选择具体的值,只需看看即可

attributes(df[, "time.2"])

并指派

df$min <- df[, "time.2"][["min"]]
df$sec <- df[, "time.2"][["sec"]]

这给了:

R> df
      time              time.2 min sec
1   0m 22s 2010-12-02 00:00:22   0  22
2    1m 7s 2010-12-02 00:01:07   1   7
3   3m 35s 2010-12-02 00:03:35   3  35
4  11m 43s 2010-12-02 00:11:43  11  43
5    1m 8s 2010-12-02 00:01:08   1   8
6   2m 21s 2010-12-02 00:02:21   2  21
7   9m 33s 2010-12-02 00:09:33   9  33
8   0m 56s 2010-12-02 00:00:56   0  56
9    0m 2s 2010-12-02 00:00:02   0   2
10   0m 2s 2010-12-02 00:00:02   0   2
11  0m 50s 2010-12-02 00:00:50   0  50
12  0m 25s 2010-12-02 00:00:25   0  25
13  0m 33s 2010-12-02 00:00:33   0  33
14  2m 26s 2010-12-02 00:02:26   2  26
15  0m 20s 2010-12-02 00:00:20   0  20
16  1m 47s 2010-12-02 00:01:47   1  47
17  0m 36s 2010-12-02 00:00:36   0  36
18   0m 3s 2010-12-02 00:00:03   0   3
19   0m 2s 2010-12-02 00:00:02   0   2
20   0m 5s 2010-12-02 00:00:05   0   5

编辑:因为您只想拆分data.frame以便能够计算总分钟数,所以您甚至不需要创建新列min,sec并且可以简单地使用该列time.2.这两个步骤已经足够了

df$time.2 <- strptime(df$time, "%Mm %Ss")
sum(df[, "time.2"][["min"]])

R> [1] 30