max=aggregate(cbind(a$VALUE,Date=a$DATE) ~ format(a$DATE, "%m") + cut(a$CLASS, breaks=c(0,2,4,6,8,10,12,14)) , data = a, max)[-1]
max$DATE=as.Date(max$DATE, origin = "1970-01-01")
示例数据：
DATE GRADE VALUE
2008-09-01 1 20
2008-09-02 2 30
2008-09-03 3 50
.
.
2008-09-30 2 75
.
.
2008-10-01 1 95
.
.
2008-11-01 4 90
.
.
2008-12-01 1 70
2008-12-02 2 40
2008-12-28 4 30
2008-12-29 1 40
2008-12-31 3 50
根据上表仅我的第一个月的预期输出是：
DATE GRADE VALUE
2008-09-30 (0,2] 75
2008-09-02 (2,4] 50
我的实际数据输出：
format(DATE, "%m")
1 09
2 10
3 11
4 12
5 09
6 10
7 11
cut(a$GRADE, breaks = c(0, 2, 4, 6, 8, 10, 12, 14)) value
1 (0,2] 0.30844444
2 (0,2] 1.00000000
3 (0,2] 1.00000000
4 (0,2] 0.73333333
5 (2,4] 0.16983488
6 (2,4] 0.09368000
7 (2,4] 0.10589335
Date
1 2008-09-30
2 2008-10-31
3 2008-11-28
4 2008-12-31
5 2008-09-30
6 2008-10-31
7 2008-11-28
输出不是根据样本数据，因为数据太大。一个简单的逻辑是有1到10的等级，所以我想在相应的等级组中找到一个月的最高值。例如：我需要每组最高值（0,2），（0,4）等
我使用了函数max的聚合条件，并将两个列分为Date和Grade。现在当我运行代码并显示max的值时，我依次得到 3个表作为输出。现在我想绘制这个输出，但由于这个原因我无法做到这一点。那我怎么能合并所有这些输出呢？
答案 0 :(得分：1)
尝试：
library(dplyr)
a %>%
group_by(MONTH=format(DATE, "%m"), GRADE=cut(GRADE, breaks=seq(0,14,by=2))) %>%
summarise_each(funs(max))
# MONTH GRADE DATE VALUE
#1 09 (0,2] 2008-09-30 75
#2 09 (2,4] 2008-09-03 50
#3 10 (0,2] 2008-10-01 95
#4 11 (2,4] 2008-11-01 90
#5 12 (0,2] 2008-12-29 70
#6 12 (2,4] 2008-12-31 50
或使用data.table
library(data.table)
setDT(a)[, list(DATE=max(DATE), VALUE=max(VALUE)),
by= list(MONTH=format(DATE, "%m"),
GRADE=cut(GRADE, breaks=seq(0,14, by=2)))]
# MONTH GRADE DATE VALUE
#1: 09 (0,2] 2008-09-30 75
#2: 09 (2,4] 2008-09-03 50
#3: 10 (0,2] 2008-10-01 95
#4: 11 (2,4] 2008-11-01 90
#5: 12 (0,2] 2008-12-29 70
#6: 12 (2,4] 2008-12-31 50
或使用aggregate
res <- transform(with(a,
aggregate(cbind(VALUE, DATE),
list(MONTH=format(DATE, "%m") ,GRADE=cut(GRADE, breaks=seq(0,14, by=2))), max)),
DATE=as.Date(DATE, origin="1970-01-01"))
res[order(res$MONTH),]
# MONTH GRADE VALUE DATE
#1 09 (0,2] 75 2008-09-30
#4 09 (2,4] 50 2008-09-03
#2 10 (0,2] 95 2008-10-01
#5 11 (2,4] 90 2008-11-01
#3 12 (0,2] 70 2008-12-29
#6 12 (2,4] 50 2008-12-31
a <- structure(list(DATE = structure(c(14123, 14124, 14125, 14152,
14153, 14184, 14214, 14215, 14241, 14242, 14244), class = "Date"),
GRADE = c(1L, 2L, 3L, 2L, 1L, 4L, 1L, 2L, 4L, 1L, 3L), VALUE = c(20L,
30L, 50L, 75L, 95L, 90L, 70L, 40L, 30L, 40L, 50L)), .Names = c("DATE",
"GRADE", "VALUE"), row.names = c(NA, -11L), class = "data.frame")
如果您想在分组中加入YEAR
library(dplyr)
a %>%
group_by(MONTH=format(DATE, "%m"), YEAR=format(DATE, "%Y"), GRADE=cut(GRADE, breaks=seq(0,14, by=2)))%>%
summarise_each(funs(max))
# MONTH YEAR GRADE DATE VALUE
#1 09 2008 (0,2] 2008-09-30 75
#2 09 2008 (2,4] 2008-09-03 50
#3 09 2009 (0,2] 2009-09-30 75
#4 09 2009 (2,4] 2009-09-03 50
#5 10 2008 (0,2] 2008-10-01 95
#6 10 2009 (0,2] 2009-10-01 95
#7 11 2008 (2,4] 2008-11-01 90
#8 11 2009 (2,4] 2009-11-01 90
#9 12 2008 (0,2] 2008-12-29 70
#10 12 2008 (2,4] 2008-12-31 50
#11 12 2009 (0,2] 2009-12-29 70
#12 12 2009 (2,4] 2009-12-31 50
a <- structure(list(DATE = structure(c(14123, 14124, 14125, 14152,
14153, 14184, 14214, 14215, 14241, 14242, 14244, 14488, 14489,
14490, 14517, 14518, 14549, 14579, 14580, 14606, 14607, 14609
), class = "Date"), GRADE = c(1L, 2L, 3L, 2L, 1L, 4L, 1L, 2L,
4L, 1L, 3L, 1L, 2L, 3L, 2L, 1L, 4L, 1L, 2L, 4L, 1L, 3L), VALUE = c(20L,
30L, 50L, 75L, 95L, 90L, 70L, 40L, 30L, 40L, 50L, 20L, 30L, 50L,
75L, 95L, 90L, 70L, 40L, 30L, 40L, 50L)), .Names = c("DATE",
"GRADE", "VALUE"), row.names = c("1", "2", "3", "4", "5", "6",
"7", "8", "9", "10", "11", "12", "21", "31", "41", "51", "61",
"71", "81", "91", "101", "111"), class = "data.frame")
答案 1 :(得分：0)
使用基数R的代码可能会有所帮助（使用来自akrun答案的'a'数据框）：
xx = strsplit(as.character(a$DATE), '-')
a$month = sapply(strsplit(as.character(a$DATE), '-'),'[',2)
gradeCats = cut(a$GRADE, breaks = c(0, 2, 4, 6, 8, 10, 12, 14))
aggregate(VALUE~month+gradeCats, data= a, max)
month gradeCats VALUE
1 09 (0,2] 75
2 10 (0,2] 95
3 12 (0,2] 70
4 09 (2,4] 50
5 11 (2,4] 90
6 12 (2,4] 50