在R中合并输出
样本数据:在R中合并输出,r,aggregate,R,Aggregate,样本数据: 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") 根据上表,仅第一个月我的预期产量为: DATE GRADE VALUE 2008-09-01 1
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")
根据上表,仅第一个月我的预期产量为:
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
在我的真实数据中输出:
DATE GRADE VALUE
2008-09-30 (0,2] 75
2008-09-02 (2,4] 50
输出不符合样本数据,因为数据太大。一个简单的逻辑是有从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。我需要每组的最大值(0,2),(0,4)等等
我使用了函数max的聚合条件,并用两列Date和Grade对其进行分组。现在,当我运行代码并显示max的值时,我得到了一个接一个的3个表作为输出。现在我想绘制这个输出,但由于这个原因,我无法这样做。那么,我如何合并所有这些输出?尝试:
format(DATE, "%m")
1 09
2 10
3 11
4 12
5 09
6 10
7 11
cut(a$GRADE, breaks = c(0, 2, 4, 6, 8, 10, 12, 14)) value
1 (0,2] 0.30844444
2 (0,2] 1.00000000
3 (0,2] 1.00000000
4 (0,2] 0.73333333
5 (2,4] 0.16983488
6 (2,4] 0.09368000
7 (2,4] 0.10589335
Date
1 2008-09-30
2 2008-10-31
3 2008-11-28
4 2008-12-31
5 2008-09-30
6 2008-10-31
7 2008-11-28
或使用data.table
library(dplyr)
a %>%
group_by(MONTH=format(DATE, "%m"), GRADE=cut(GRADE, breaks=seq(0,14,by=2))) %>%
summarise_each(funs(max))
# MONTH GRADE DATE VALUE
#1 09 (0,2] 2008-09-30 75
#2 09 (2,4] 2008-09-03 50
#3 10 (0,2] 2008-10-01 95
#4 11 (2,4] 2008-11-01 90
#5 12 (0,2] 2008-12-29 70
#6 12 (2,4] 2008-12-31 50
或者使用聚合
library(data.table)
setDT(a)[, list(DATE=max(DATE), VALUE=max(VALUE)),
by= list(MONTH=format(DATE, "%m"),
GRADE=cut(GRADE, breaks=seq(0,14, by=2)))]
# MONTH GRADE DATE VALUE
#1: 09 (0,2] 2008-09-30 75
#2: 09 (2,4] 2008-09-03 50
#3: 10 (0,2] 2008-10-01 95
#4: 11 (2,4] 2008-11-01 90
#5: 12 (0,2] 2008-12-29 70
#6: 12 (2,4] 2008-12-31 50
数据
a以下使用base R的代码可能有用(使用akrun答案中的“a”数据帧):
您的意思是合并输出吗?@user3923765。您能显示您的预期输出吗?因为第二列名称太长,没有空间打印相邻的所有列。实际上只有一个“表”(即数据帧)。例如,如果执行dim(max)
,您会得到什么?(作为旁注,max
是一个基本的R函数,因此最好为输出选择另一个名称)我得到了14,2,而我应该得到14,3?而且我不能为不同的变量绘制值dates@user3923765.对于创建的示例数据,我每年和每月都会得到预期的输出。我还有5年的数据要处理,但是当我运行此代码时,它只会给我去年月份的值。例如,如果我输入2008年的数据-2013年。我在2013年的12个月内只得到12个值,而我希望得到60个值,每年12个。@user3923765。在这种情况下,您仅按月和年级进行分组,您可能还必须将年也包括在分组中。可能按格式(日期,“%Y”)进行分组
。未测试。无论我使用%Y%还是%M%@user3923765,结果都是一样的。您是否检查了我的新更新。顺便说一句,我没有更改2008年和2009年的值
列
res <- transform(with(a,
aggregate(cbind(VALUE, DATE),
list(MONTH=format(DATE, "%m") ,GRADE=cut(GRADE, breaks=seq(0,14, by=2))), max)),
DATE=as.Date(DATE, origin="1970-01-01"))
res[order(res$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")
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")
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