我已经浏览了很多关于订购因素的帖子,但还没找到与我的问题匹配的帖子.不幸的是,我对R的了解还很不成熟.
我有一个我正在使用的考古工件目录的子集.我正在尝试交叉制表诊断历史工件类型和站点测试位置.使用ddply或tapply足够容易.
我的问题是我想通过它们的平均诊断日期(数字/年)对工件类型(一个因子)进行排序,并且我会按字母顺序获取它们.我知道我需要让它成为一个有序因子,但无法弄清楚如何通过另一列中的年份值来订购它.
IDENTIFY MIDDATE
engine-turned fine red stoneware 1769
white salt-glazed stoneware, scratch blue 1760
wrought nail, 'L' head 1760
yellow lead-glazed buff earthenware 1732
...
需要订购的是:
IDENTIFY MIDDATE
yellow lead-glazed buff earthenware 1732
white salt-glazed stoneware, scratch blue 1760
wrought nail, 'L' head 1760
engine-turned fine red stoneware 1769
...
因子(IDENTIFY)需要按日期(MIDDATE)排序.我以为我有它
Catalog$IDENTIFY<-factor(Catalog$IDENTIFY,levels=Catalog$MIDDATE,ordered=TRUE)
但得到警告:
In `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels)
else paste0(labels,: duplicated levels will not be allowed
in factors anymore
IDENTIFY具有~130个因子级别,并且许多具有相同的MIDDATE值,因此我需要通过MIDDATE和另一个列TYPENAME来订购IDENTIFY.
更多细节:
我有一个数据帧Catalog,它分解为(即str(Catalog)):
> str(Catalog)
'data.frame': 2211 obs. of 15 variables:
$ TRENCH : Factor w/ 7 levels "DRT 1","DRT 2",..: 1 1 1 1 1 1 1 1 1 1 ...
$ U_TYPE : Factor w/ 3 levels "EU","INC","STP": 1 1 1 1 1 1 1 1 1 1 ...
$ U_LBL : Factor w/ 165 levels "001","005","007",..: 72 72 72 72 72 72 ...
$ L_STRAT : Factor w/ 217 levels "#2-7/25","[3]",..: 4 4 4 4 4 4 89 89 89 89 ...
$ START : num 0 0 0 0 0 0 39.4 39.4 39.4 39.4 ...
$ END : num 39.4 39.4 39.4 39.4 39.4 39.4 43.2 43.2 43.2 43.2 ...
$ Qty : int 1 1 3 5 1 1 6 8 1 1 ...
$ MATNAME : Factor w/ 6 levels "Ceramics","Chipped Stone",..: 1 1 1 5 5 6 ...
$ TYPENAME: Factor w/ 9 levels "Architectural Hardware",..: 9 9 9 1 1 3 9 ...
$ CATNAME : Factor w/ 32 levels "Biface","Bottle Glass",..: 24 29 29 6 24 ...
$ IDENTIFY: Factor w/ 112 levels "amethyst bottle glass",..: 17 91 96 71 103 ...
$ BEGDATE : int 1820 1820 1830 1835 1700 NA 1670 1762 1800 1720 ...
$ ENDDATE : int 1900 1970 1860 1875 1820 NA 1795 1820 1820 1780 ...
$ OCC_LBL : Ord.factor w/ 5 levels "Late 19th Century"<..: 2 1 2 2 4 5 4 3 ...
$ MIDDATE : num 1860 1895 1845 1855 1760 ...
我需要制作IDENTIFY一个有序因子,然后通过MIDDATE- > TYPENAME- > alpha by 重新排序IDENTIFY.
我真正得到的是如何通过多列的组合订单重新排序.
我会在数据库中执行此操作,但我正在运行的很多内容都是在各种交叉表格中的加权平均值(例如,按地点划分的工件类别的地表下加权平均深度)...
......可以在Access中使用,但是很麻烦,而且不可预测.在R中管理起来更容易和更清晰,但我宁愿不必手动重新排序结果表.
我想要产生的是以下几点:
>xtab.Catalog<-tapply(Catalog$Qty,list(Catalog$IDENTIFY,Catalog$TRENCH),sum)
IDENTIFY DRT1 DRT2 DRT3 DRT4 DRT5 DRT6
Staffordshire stoneware 4 NA NA NA NA NA
undecorated delftware 6 4 NA NA NA NA
unidentified wrought nail 15 9 3 1 3 NA
white salt-glazed stoneware 6 1 1 NA 2 1
white salt-glazed scratch blue 1 NA NA NA NA NA
white stoneware, slip-dipped NA NA NA NA NA NA
wrought nail, 'L' head 2 NA NA NA NA NA
wrought nail, 'rose' head 62 21 4 NA 1 1
wrought nail, 'T' head 2 NA 1 NA NA 1
yellow lead-glazed 12 NA NA NA 1 3
...
...但我需要它们按逻辑(即按时间顺序/类型)顺序排序,而不是按字母排序.
Mat*_*erg 41
这是一个可重复的样本,有解决方案:
set.seed(0)
a = sample(1:20,replace=F)
b = sample(1:20,replace=F)
f = as.factor(letters[1:20])
> a
[1] 18 6 7 10 15 4 13 14 8 20 1 2 9 5 3 16 12 19 11 17
> b
[1] 16 18 4 12 3 5 6 1 15 10 19 17 9 11 2 8 20 7 13 14
> f
[1] a b c d e f g h i j k l m n o p q r s t
Levels: a b c d e f g h i j k l m n o p q r s t
现在为新因素:
fn = factor(f, levels=unique(f[order(a,b,f)]), ordered=TRUE)
> fn
[1] a b c d e f g h i j k l m n o p q r s t
20 Levels: k < l < o < f < n < b < c < i < m < d < s < q < g < h < e < ... < j
在'a',下一个'b'和最后'f'本身上排序(尽管在这个例子中,'a'没有重复的值).
zac*_*ach 23
我建议使用以下基于dplyr的方法(h/t daattali),它可以扩展到任意数量的列:
library(dplyr)
Catalog <- Catalog %>%
arrange(MIDDATE, TYPENAME) %>% # sort your dataframe
mutate(IDENTIFY = factor(IDENTIFY, unique(IDENTIFY))) # reset your factor-column based on that order
该函数fct_reorder2就是这样做的。
fct_reorder请注意按升序排序和按fct_reordering2降序排序的微妙之处。
文档中的代码：
df0 <- tibble::tribble(
~color, ~a, ~b,
"blue", 1, 2,
"green", 6, 2,
"purple", 3, 3,
"red", 2, 3,
"yellow", 5, 1
）
df0$color <- factor(df0$color)
fct_reorder(df0$color, df0$a, min)
#> [1] blue green purple red yellow
#> Levels: blue red purple yellow green
fct_reorder2(df0$color, df0$a, df0$b)