[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
by
或tapply
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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