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r通过因子削减数据帧

[英]r cut dataframe by factors

Let's say I have this 假设我有这个

+-------+-----+------+
| Month | Day | Hour |
+-------+-----+------+
|     1 |   1 |    1 |
|     1 |   1 |    2 |
|     1 |   1 |    3 |
|     1 |   1 |    4 |
|     1 |   2 |    1 |
|     1 |   2 |    2 |
|     1 |   2 |    3 |
|     1 |   2 |    4 |
|     2 |   1 |    1 |
|     2 |   1 |    2 |
|     2 |   1 |    3 |
|     2 |   1 |    4 |
+-------+-----+------+

I would like to cut by month and day factors to have this 我想cut月份和日期的因素

+-------+-----+------+-------+
| Month | Day | Hour | Block |
+-------+-----+------+-------+
|     1 |   1 |    1 | [1,2] |
|     1 |   1 |    2 | [1,2] |
|     1 |   1 |    3 | [3,4] |
|     1 |   1 |    4 | [3,4] |
|     1 |   2 |    1 | [1,2] |
|     1 |   2 |    2 | [1,2] |
|     1 |   2 |    3 | [3,4] |
|     1 |   2 |    4 | [3,4] |
|     2 |   1 |    1 | [1,2] |
|     2 |   1 |    2 | [1,2] |
|     2 |   1 |    3 | [3,4] |
|     2 |   1 |    4 | [3,4] |
+-------+-----+------+-------+

I thought that maybe using by or tapply could be a way but I cannot figure how. 我认为也许使用bytapply是一种方法,但我不知道如何使用。

We can create a sequence for each hour of the day with cut and replace parantheticals with brackets: 我们可以使用cut来为一天中的每个小时创建一个序列,并用方括号替换括号中的内容:

df1$Block <- cut(df1$Hour, c(1,seq(2,24, by=2)), include.lowest=TRUE)
df1$Block <- sub("(", "[", df1$Block, fixed=T)
df1
#    Month Day Hour Block
# 1      1   1    1 [1,2]
# 2      1   1    2 [1,2]
# 3      1   1    3 [2,4]
# 4      1   1    4 [2,4]
# 5      1   2    1 [1,2]
# 6      1   2    2 [1,2]
# 7      1   2    3 [2,4]
# 8      1   2    4 [2,4]
# 9      2   1    1 [1,2]
# 10     2   1    2 [1,2]
# 11     2   1    3 [2,4]
# 12     2   1    4 [2,4]

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