I have some monthly data and I want to add a column to my data frame that associates the smallest value in the the first column to the largest value in the first column. The second smallest value in the first column to the second largest value in the first column, ect…
我有一些月度数据，我想在我的数据框中添加一列，将第一列中的最小值与第一列中的最大值相关联。第一列中的第二个最小值到第一列中的第二个最大值，例如......
Here is some sample data
这是一些示例数据
x1<-c(100,151,109,59,161,104,170,101)
dat<-data.frame(x1)
rownames(dat)<-c('Apr','May', 'Jun','Jul', 'Aug', 'Sep', 'Oct', 'Nov')
x1
Apr 100
May 151
Jun 109
Jul 59
Aug 161
Sep 104
Oct 170
Nov 101
I m trying to get my data to look like this
我试图让我的数据看起来像这样
x1 x2
Apr 100 161
May 151 101
Jun 109 104
Jul 59 170
Aug 161 100
Sep 104 109
Oct 170 59
Nov 101 151
I'm going in circles with rank, sort, and order. Any help would be appreciated.
我正在排名，排序和秩序。任何帮助，将不胜感激。
4 个解决方案
#1
6
It's reasonably straightforward if you create a temporary object that has the ascending and descending values paired up:
如果您创建一个具有升序和降序值配对的临时对象，则相当简单：
> temp <- data.frame(asc =x1[order(x1)],desc=x1[rev(order(x1))])
> dat$x2 <- temp$desc[ match(dat$x1, temp$asc) ]
> dat
x1 x2
Apr 100 161
May 151 101
Jun 109 104
Jul 59 170
Aug 161 100
Sep 104 109
Oct 170 59
Nov 101 151
The match function is designed to construct integer indexing values that are used as arguments to "[". It's the fundamental function inside merge.
匹配函数用于构造整数索引值，用作“[”的参数。这是合并中的基本功能。
#2
4
Similar idea as @BondedDust:
与@BondedDust类似的想法：
library(dplyr)
dat %>% mutate(x2 = x1[match(row_number(desc(x1)), row_number(x1))])
Which gives:
这使：
# x1 x2
#1 100 161
#2 151 101
#3 109 104
#4 59 170
#5 161 100
#6 104 109
#7 170 59
#8 101 151
#3
4
Making use of base R's month.abb
利用基数R的month.abb
df = dat[order(dat$x1),, drop = FALSE]
df$x2 = sort(x1,decreasing = T)
df[match(month.abb, rownames(df), nomatch = 0),]
# x1 x2
#Apr 100 161
#May 151 101
#Jun 109 104
#Jul 59 170
#Aug 161 100
#Sep 104 109
#Oct 170 59
#Nov 101 151
Using data.table simply
简单地使用data.table
library(data.table)
df = setDF(setDT(dat, keep.rownames=T)[order(x1), x2 := sort(x1, decreasing = T)])
rownames(df) = df$rn; df[,1] = NULL
# x1 x2
#Apr 100 161
#May 151 101
#Jun 109 104
#Jul 59 170
#Aug 161 100
#Sep 104 109
#Oct 170 59
#Nov 101 151
#4
3
Approach in data.table:
在data.table中的方法：
library(data.table)
setDT(dat,keep.rownames=T)[order(x1),x2:=rev(x1)]
rn x1 x2
1: Jul 59 170
2: Apr 100 161
3: Nov 101 151
4: Sep 104 109
5: Jun 109 104
6: May 151 101
7: Aug 161 100
8: Oct 170 59
If you want to end up with the rows in order, I think the easiest way is to use month.abb as the levels of rn as a factor:
如果你想按顺序结束行，我认为最简单的方法是使用month.abb作为因子的水平：
setDT(dat,keep.rownames=T)[order(x1),x2:=rev(x1)
][order(factor(rn,levels=month.abb))]
rn x1 x2
1: Apr 100 161
2: May 151 101
3: Jun 109 104
4: Jul 59 170
5: Aug 161 100
6: Sep 104 109
7: Oct 170 59
8: Nov 101 151
Could also use order(match(rn,month.abb)) instead, if that suits you; if you're going to be re-sorting a lot by month, it may make sense to define rn as a factor so you don't have to do the match or factor dog and pony show repeatedly: dat[,rn:=factor(rn,levels=month.abb)][order(rn)]
也可以使用order（match（rn，month.abb））代替，如果这适合你;如果你要按月重新排序，那么将rn定义为一个因子可能是有意义的，所以你不必重复进行匹配或因子狗和小马显示：dat [，rn：= factor （RN，水平= month.abb）] [顺序（RN）]
#1
6
It's reasonably straightforward if you create a temporary object that has the ascending and descending values paired up:
如果您创建一个具有升序和降序值配对的临时对象，则相当简单：
> temp <- data.frame(asc =x1[order(x1)],desc=x1[rev(order(x1))])
> dat$x2 <- temp$desc[ match(dat$x1, temp$asc) ]
> dat
x1 x2
Apr 100 161
May 151 101
Jun 109 104
Jul 59 170
Aug 161 100
Sep 104 109
Oct 170 59
Nov 101 151
The match function is designed to construct integer indexing values that are used as arguments to "[". It's the fundamental function inside merge.
匹配函数用于构造整数索引值，用作“[”的参数。这是合并中的基本功能。
#2
4
Similar idea as @BondedDust:
与@BondedDust类似的想法：
library(dplyr)
dat %>% mutate(x2 = x1[match(row_number(desc(x1)), row_number(x1))])
Which gives:
这使：
# x1 x2
#1 100 161
#2 151 101
#3 109 104
#4 59 170
#5 161 100
#6 104 109
#7 170 59
#8 101 151
#3
4
Making use of base R's month.abb
利用基数R的month.abb
df = dat[order(dat$x1),, drop = FALSE]
df$x2 = sort(x1,decreasing = T)
df[match(month.abb, rownames(df), nomatch = 0),]
# x1 x2
#Apr 100 161
#May 151 101
#Jun 109 104
#Jul 59 170
#Aug 161 100
#Sep 104 109
#Oct 170 59
#Nov 101 151
Using data.table simply
简单地使用data.table
library(data.table)
df = setDF(setDT(dat, keep.rownames=T)[order(x1), x2 := sort(x1, decreasing = T)])
rownames(df) = df$rn; df[,1] = NULL
# x1 x2
#Apr 100 161
#May 151 101
#Jun 109 104
#Jul 59 170
#Aug 161 100
#Sep 104 109
#Oct 170 59
#Nov 101 151
#4
3
Approach in data.table:
在data.table中的方法：
library(data.table)
setDT(dat,keep.rownames=T)[order(x1),x2:=rev(x1)]
rn x1 x2
1: Jul 59 170
2: Apr 100 161
3: Nov 101 151
4: Sep 104 109
5: Jun 109 104
6: May 151 101
7: Aug 161 100
8: Oct 170 59
If you want to end up with the rows in order, I think the easiest way is to use month.abb as the levels of rn as a factor:
如果你想按顺序结束行，我认为最简单的方法是使用month.abb作为因子的水平：
setDT(dat,keep.rownames=T)[order(x1),x2:=rev(x1)
][order(factor(rn,levels=month.abb))]
rn x1 x2
1: Apr 100 161
2: May 151 101
3: Jun 109 104
4: Jul 59 170
5: Aug 161 100
6: Sep 104 109
7: Oct 170 59
8: Nov 101 151
Could also use order(match(rn,month.abb)) instead, if that suits you; if you're going to be re-sorting a lot by month, it may make sense to define rn as a factor so you don't have to do the match or factor dog and pony show repeatedly: dat[,rn:=factor(rn,levels=month.abb)][order(rn)]
也可以使用order（match（rn，month.abb））代替，如果这适合你;如果你要按月重新排序，那么将rn定义为一个因子可能是有意义的，所以你不必重复进行匹配或因子狗和小马显示：dat [，rn：= factor （RN，水平= month.abb）] [顺序（RN）]