How to Create a Single Dummy Variable with conditions in multiple columns?

前端 未结 3 999
执念已碎
执念已碎 2020-12-11 12:56

I am trying to efficiently create a binary dummy variables (1/0) in my data set based on whether or not one or more of 7 variables (col9-15) in the data set take on a specif

相关标签:
3条回答
  • 2020-12-11 13:06

    You can use rowSums (vectorized solution) like this :

    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
    

    EDIT

    I replace the cbind by transform. Since the column will be boolean I coerce it to get 0/1.

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

    The result is a data.frame.( coerced from matrix by transform)

    EDIT2 ( as suggested by @flodel)

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

    where data is the OP's data.frame.

    0 讨论(0)
  • 2020-12-11 13:06

    apply to the rescue :)

    # 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
    
    0 讨论(0)
  • 2020-12-11 13:13

    you can also try this (borrowing sample data from agstudy's answer)

    > 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
    
    0 讨论(0)
提交回复
热议问题