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我正在尝试根据数据集中的 7 个变量(col9-15)中的一个或多个是否具有特定值(35),在我的数据集中有效地创建一个二进制虚拟变量(1/0),但是我不想测试所有列。

虽然 as.numeric 通常是理想的,但我一次只能使用一列:

data$indicator <- as.numeric(data$col1 == 35)

知道如何修改上述代码,以便如果data$col9-data$col15是“35”,那么我的指标变量取 1?

谢谢!!!

4

3 回答 3

5

您可以rowSums像这样使用(矢量化解决方案):

set.seed(123)
dat <- matrix(sample(c(35,1:100),size=15*20,rep=T),ncol=15,byrow=T)
cbind(dat,rowSums(dat[,9:15] == 35) > 0)
   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16]
 [1,]   29   79   41   89   94    4   53   90   55    46    96    45    68    57    10     0
 [2,]   90   24    4   33   96   89   69   64  100    66    71    54    60    29    14     0
 [3,]   97   91   69   80    2   48   76   21   32    23    14    41    41    37    15     0
 [4,]   14   23   47   26   86    4   44   80   12    56    20    12    76    90    37     0
 [5,]   67    9   38   27   82   45   81   82   80    44    76    63    71    35    48     1
 [6,]   22   38   61   35   11   24   67   42   79    10    43    99    90    89    17     0
 [7,]   13   65   34   66   32   18   79    9   47    51    60    33    49    96    48     0
 [8,]   89   92   61   41   14   94   30    6   95    72    14    55    96    59    40     0
 [9,]   65   32   31   22   37   99   15    9   14    69    62    90    67    74    52     0
[10,]   66   83   79   98   44   31   41    1   18    85    23    24     7    24    73     0
[11,]   85   50   39   24   11   39   57   21   44    22    50    35    65    37    35     1
[12,]   53   74   22   41   26   63   18   87   75    67    62    37    53    88    58     0
[13,]   84   31   71   26   60   48   26   57   92    91    27    32    99    62    94     0
[14,]   47   41   66   15   57   24   97   60   52    40    88    36    29    17    17     0
[15,]   48   25   21   68    4   70   35   41   82    92    28    97    73    69     5     0
[16,]   39   48   56   70   92   62   43   54    5    26    40    19    84    15    81     0
[17,]   55   66   17   63   31   73   40   97   97    73    25    22    59    27    53     0
[18,]   79   16   40   47   87   93   89   68   95    52    58    33    35     2    50     1
[19,]   87   35    7   16   77   74   98   47    7    65    76    13    40    22     5     0
[20,]   39    6   22    5   67   30   10    7   88    76    82    99    10    10    80     0

编辑

我替换cbindby transform。由于该列将是布尔值,因此我将其强制为 0/1。

 transform(dat,x=as.numeric((rowSums(dat[,9:15] == 35) > 0)))

结果是一个data.frame。(通过变换从矩阵强制转换)

EDIT2(由@flodel 建议)

data$indicator <- as.integer(rowSums(data[paste0("col", 9:15)] == 35) > 0)

dataOP的data.frame在哪里。

于 2013-02-22T02:28:15.590 回答
0

申请救援:)

# this sample data frame is pre-loaded
mtcars

# test whether any of the values in the
# 2nd - 5th columns of mtcars equal four..

# save the result into a new vector..
indicator.col <- 
    apply( 
        mtcars[ , 2:5 ] , 
        1 ,
        FUN = function( x ) max( x == 4 ) 
    )

# ..that quickly binds onto mtcars
# and bind it with the original mtcars
mtcars2 <- cbind( mtcars , indicator.col )

# look at your result
mtcars2
于 2013-02-22T02:30:16.310 回答
0

你也可以试试这个(从agstudy的答案中借用样本数据)

> set.seed(123)
> dat <- matrix(sample(c(35,1:100),size=15*20,rep=T),ncol=15,byrow=T)


#Create indicator initialized with 0.
> indicator <- rep(0, nrow(dat))
#Replace the elements at indices which are equal to rows in dat where you find 35
> indicator[which(dat[,9:15]==35)%%nrow(dat)] <- 1
#bind the indicator to original data
> cbind(dat, indicator)
                                                    indicator
 [1,] 29 79 41 89 94  4 53 90  55 46 96 45 68 57 10         0
 [2,] 90 24  4 33 96 89 69 64 100 66 71 54 60 29 14         0
 [3,] 97 91 69 80  2 48 76 21  32 23 14 41 41 37 15         0
 [4,] 14 23 47 26 86  4 44 80  12 56 20 12 76 90 37         0
 [5,] 67  9 38 27 82 45 81 82  80 44 76 63 71 35 48         1
 [6,] 22 38 61 35 11 24 67 42  79 10 43 99 90 89 17         0
 [7,] 13 65 34 66 32 18 79  9  47 51 60 33 49 96 48         0
 [8,] 89 92 61 41 14 94 30  6  95 72 14 55 96 59 40         0
 [9,] 65 32 31 22 37 99 15  9  14 69 62 90 67 74 52         0
[10,] 66 83 79 98 44 31 41  1  18 85 23 24  7 24 73         0
[11,] 85 50 39 24 11 39 57 21  44 22 50 35 65 37 35         1
[12,] 53 74 22 41 26 63 18 87  75 67 62 37 53 88 58         0
[13,] 84 31 71 26 60 48 26 57  92 91 27 32 99 62 94         0
[14,] 47 41 66 15 57 24 97 60  52 40 88 36 29 17 17         0
[15,] 48 25 21 68  4 70 35 41  82 92 28 97 73 69  5         0
[16,] 39 48 56 70 92 62 43 54   5 26 40 19 84 15 81         0
[17,] 55 66 17 63 31 73 40 97  97 73 25 22 59 27 53         0
[18,] 79 16 40 47 87 93 89 68  95 52 58 33 35  2 50         1
[19,] 87 35  7 16 77 74 98 47   7 65 76 13 40 22  5         0
[20,] 39  6 22  5 67 30 10  7  88 76 82 99 10 10 80         0
于 2013-02-22T04:01:29.793 回答