CODEWITHSUNDEEP
rdataframeloopsvector
BZH
BZH

Reputation: 11

Avoid the for loops-R

I have two data frames, x and y. For each value of x[,2], I look if the value is equal to the value of the elements of y[,1]. If so, I add a third column in the first data frame that contains the values of y[,2].

I managed to do that with loops, but how can I do this using vectors?

x=data.frame(1:15,15:1)
y=data.frame(3:5,c(7.2,8.5,0.3))

for ( i in 1:nrow(x)) { 

  for (j in 1:nrow(y)) {

    if (x[i,2]==y[j,1]){

      x[i,3]=y[j,2]

    }else{

    }
  }
}

Upvotes: 1

Views: 77

Answers (2)

TarJae
TarJae

Reputation: 79184

Update: Many thanks to @TrainingPizza (who has drawn my attention to the false output of my first answer and also provided how it could work:

library(dplyr)
x %>% 
  rowwise() %>% 
  mutate(col3 = ifelse(col2 %in% y$col1, y$col2[y$col1==col2], NA))

col1  col2  col3
   <int> <int> <dbl>
 1     1    15  NA  
 2     2    14  NA  
 3     3    13  NA  
 4     4    12  NA  
 5     5    11  NA  
 6     6    10  NA  
 7     7     9  NA  
 8     8     8  NA  
 9     9     7  NA  
10    10     6  NA  
11    11     5   0.3
12    12     4   8.5
13    13     3   7.2
14    14     2  NA  
15    15     1  NA

First answer (not correct) Here is dplyr way how to avoid the for - loop:

library(dplyr)
x %>% 
  mutate(V3 = ifelse(V2 %in% y$V1, y$V2, NA))

   V1 V2  V3
1   1 15  NA
2   2 14  NA
3   3 13  NA
4   4 12  NA
5   5 11  NA
6   6 10  NA
7   7  9  NA
8   8  8  NA
9   9  7  NA
10 10  6  NA
11 11  5 8.5
12 12  4 0.3
13 13  3 7.2
14 14  2  NA
15 15  1  NA

Upvotes: 2

akrun
akrun

Reputation: 887641

Use a join instead of loops - based on the loop comparision, the second column of 'x' is compared with the first column of 'y', thus those columns are used in the on, assign (:=) the second column (col2) from the second dataset to create the new column 'col3' in first data

library(data.table)
setDT(x)[y, col3 := i.col2, on = .(col2 = col1)]

-output

> x
    col1 col2 col3
 1:    1   15   NA
 2:    2   14   NA
 3:    3   13   NA
 4:    4   12   NA
 5:    5   11   NA
 6:    6   10   NA
 7:    7    9   NA
 8:    8    8   NA
 9:    9    7   NA
10:   10    6   NA
11:   11    5  0.3
12:   12    4  8.5
13:   13    3  7.2
14:   14    2   NA
15:   15    1   NA

data

x <- data.frame(col1 = 1:15, col2 = 15:1)
y <- data.frame(col1 = 3:5, col2 = c(7.2,8.5,0.3))

Upvotes: 3

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