CODEWITHSUNDEEP
r
SA2018
SA2018

Reputation: 367

Apply a function for each row according to its column's number

3I have this function:

nombre_points<-function(typeAction)
{
  if(typeAction==1)
  {return(4)}
  else if (typeAction==2)
  {return(3)}
  else if (typeAction==3)
  {return(2)}
}

And I want to applicate it to each columns of a data frame

df>
ID 1 2 3
XX 0 1 2
YY 1 2 3
EE 2 2 1 
ZZ 0 3 4

I want to applicate the above function to this data frame where the input of the nombre_points function is the column's number.

So as result I need to get for each row a data frame like this:

df>

    ID Result
    XX 7
    YY 14
    EE 16
    ZZ 17

In fact for the observation ZZ for example 17=0*4+3*3+4*2 also for the XX observation we have a score :7=0*4+3*1+2*2

I tried to used apply function :

dd<-apply(df,1,nombre_points)

But it doesn't give the needed result

Upvotes: 0

Views: 48

Answers (1)

Ronak Shah
Ronak Shah

Reputation: 389335

Update

Since, the values might not be related to column number we can create a named vector or a lookup dataframe to get the corresponding values.

With lookup table : 

named_df <- data.frame(col = c(1, 2, 3), value = c(4, 3, 2))
rowSums(df[-1] * named_df$value[col(df[-1])])
#[1]  7 16 16 17

With named vector : 

named_vec <- c("1" = 4, "2" = 3, "3" = 2)
rowSums(named_vec[col(df[-1])] * df[-1])
#[1]  7 16 16 17

The logic for both the approach is the same , we match the column number with the name of named_vec or col of named_df and get the corresponding value and then multiply it with the actual value in the dataframe and then take sum of each row using rowSums.

Original Answer

You could multiply each value in the dataframe with it's column number - 1 to get your desired result. Here, df[-1] is to ignore the first column (ID).

df$result <- rowSums(df[-1] * (col(df) - 1))
df

#  ID X1 X2 X3 result
#1 XX  0  1  2      5
#2 YY  1  2  3      8
#3 EE  2  2  1      4
#4 ZZ  0  3  4     11

For understanding purposes col(df) gives

#     [,1] [,2] [,3] [,4]
#[1,]    1    2    3    4
#[2,]    1    2    3    4
#[3,]    1    2    3    4
#[4,]    1    2    3    4

Upvotes: 2

Related Questions