3

我正在创建一个包含另一个变量之和的 2 个因子的交叉表

b<-xtabs(utib~qpl+i , data=data )

这给了我一张像

                   <=2.5 2.5-5 5-10 10-50 50-150   150-250   >250     NA
  aaaa               3231   211   64    10      1       0    0       32
  bbbbbbbbb          442    78    7      3      0       0    0        5
  cccccc             6462   466   76     6      0       0    0      157

我需要在最后一行添加我需要它保持“表类”的每一列的总和，因为我必须用代码导出它：

   library(PerformanceAnalytics))
    win.metafile(file)    
    PerformanceAnalytics:::textplot(c)
    dev.off()
4

2 回答 2

3

您可以使用该addmargin()功能

#sample data
data<-data.frame(
   utib=rpois(50,10), 
   qpl=sample(letters[1:3], 50, replace=T),
   i=cut(runif(50,0,250), breaks=c(0,2.5,5,10,50,150))
)

(b<-xtabs(utib~qpl+i , data=data ) )

#    i
# qpl (0,2.5] (2.5,5] (5,10] (10,50] (50,150]
#   a       0      12     24      46       53
#   b       0       0      0       0      100
#   c       0       0      0      46       42

(bb<-addmargins(b, margin=1))

#    i
# qpl   (0,2.5] (2.5,5] (5,10] (10,50] (50,150]
#   a         0      12     24      46       53
#   b         0       0      0       0      100
#   c         0       0      0      46       42
 #  Sum       0      12     24      92      195
于 2014-07-15T16:32:24.893 回答
0

使用弗利克先生的数据框，这是一种带有dcast边距参数的方法：

(b <- dcast(data, utib~qpl+i, fun.aggregate = length, margins = TRUE))

Using i as value column: use value.var to override.
    utib a_(50,150] a_NA a_(all) b_(10,50] b_(50,150] b_NA b_(all) c_(10,50] c_(50,150] c_NA c_(all) (all)_(all)
1      4          1    0       1         0          0    1       1         0          1    0       1           3
2      5          0    0       0         0          1    0       1         0          1    0       1           2
3      6          1    0       1         1          0    1       2         0          1    0       1           4
4      7          0    2       2         0          1    0       1         0          0    1       1           4
5      8          2    1       3         0          1    0       1         0          1    1       2           6
6      9          2    1       3         1          0    2       3         0          0    2       2           8
7     10          3    0       3         0          2    0       2         0          0    0       0           5
8     11          1    1       2         0          1    0       1         0          0    1       1           4
9     12          1    1       2         0          0    0       0         1          0    0       1           3
10    13          1    1       2         1          0    0       1         0          0    0       0           3
11    14          0    0       0         1          0    0       1         0          0    0       0           1
12    15          0    1       1         0          0    0       0         0          0    1       1           2
13    16          1    0       1         0          0    0       0         0          0    0       0           1
14    17          0    0       0         0          1    0       1         0          0    1       1           2
15    19          0    1       1         0          0    0       0         0          0    0       0           1
16    20          0    0       0         0          0    1       1         0          0    0       0           1
17 (all)         13    9      22         4          7    5      16         1          4    7      12          50
于 2014-07-15T21:04:09.750 回答