CODEWITHSUNDEEP
rdplyr
user3022875
user3022875

Reputation: 9018

adding column with dplyr using mutate then averaging

I have the d2 data frame below and I'd like to add a a column that is the average of x by group and flag. But I am adding the flag column in the mutate code and I am not sure how to then add an average by group AND flag.

    d = data.frame(x=c(seq(1,5,1),seq(11,15,1),100,1000),group= c(rep("A",5),rep("B",5),"A","B")) 
    d
    d2 = d%>%
      group_by(group)  %>%  
      mutate(    
                   U=quantile(x, 0.75) + 1.5*IQR(x),
                   L=quantile(x, 0.25) - 1.5*IQR(x),
                   flag = ifelse(x>U | x<L,1,0),
                    mu = mean(x)
                   ) 
    as.data.frame(d2)

I've added a "result" vector for what the result should be

      x group    U    L flag        mu    result
1     1     A  8.5 -1.5    0  19.16667     3
2     2     A  8.5 -1.5    0  19.16667      3
3     3     A  8.5 -1.5    0  19.16667        3
4     4     A  8.5 -1.5    0  19.16667      3
5     5     A  8.5 -1.5    0  19.16667      3
6    11     B 18.5  8.5    0 177.50000     13
7    12     B 18.5  8.5    0 177.50000     13
8    13     B 18.5  8.5    0 177.50000      13
9    14     B 18.5  8.5    0 177.50000     13
10   15     B 18.5  8.5    0 177.50000     13
11  100     A  8.5 -1.5    1  19.16667   100
12 1000     B 18.5  8.5    1 177.50000   1000


note having     group_by(group,flag) 

returns

Error: unknown variable to group by : flag

Upvotes: 0

Views: 1099

Answers (1)

Steven Beaupr&#233;
Steven Beaupr&#233;

Reputation: 21621

Simply add group_by(group, flag) to the chain after your initial operations and then mutate():

d %>%
  group_by(group)  %>%  
  mutate( 
    U = quantile(x, 0.75) + 1.5 * IQR(x),
    L = quantile(x, 0.25) - 1.5 * IQR(x),
    flag = ifelse(x > U | x < L, 1, 0),
    mu = mean(x)) %>%
  group_by(group, flag) %>%
  mutate(result = mean(x))

Which gives:

#Source: local data frame [12 x 7]
#Groups: group, flag [4]
#
#       x  group     U     L  flag        mu result
#   <dbl> <fctr> <dbl> <dbl> <dbl>     <dbl>  <dbl>
#1      1      A   8.5  -1.5     0  19.16667      3
#2      2      A   8.5  -1.5     0  19.16667      3
#3      3      A   8.5  -1.5     0  19.16667      3
#4      4      A   8.5  -1.5     0  19.16667      3
#5      5      A   8.5  -1.5     0  19.16667      3
#6     11      B  18.5   8.5     0 177.50000     13
#7     12      B  18.5   8.5     0 177.50000     13
#8     13      B  18.5   8.5     0 177.50000     13
#9     14      B  18.5   8.5     0 177.50000     13
#10    15      B  18.5   8.5     0 177.50000     13
#11   100      A   8.5  -1.5     1  19.16667    100
#12  1000      B  18.5   8.5     1 177.50000   1000

Upvotes: 2

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