How to find the proportion of row values in R dataframe?

The proportion of row value in a data frame is equivalent to the cell value divided by the summation of the cell values belonging to that entire row. The sum of all the row proportion values in a data frame is equivalent to 1. In this article, we are going to see how to find the proportion of row values in a data frame in R Programming Language.

Example 1: An iteration over the matrix is carried out, using two for loops. We compute the row sum for each row while performing row iteration, and then divide the cell value by the row sum. This value is reassigned to the data frame original cell value. The time complexity required to perform this is equivalent to O(n * m), where n is the no. of rows and m is the number of columns in the data frame. 

The following code snippet illustrates the application of this approach : 

R

# declaring a data frame in R
data_frame = data.frame(C1= c(0,1,2,3),
                        C2 = c(1:4),
                        C3 = c(9:12))

print("Original data frame")
print(data_frame)

# looping over the rows of data frame
for (i in 1:nrow(data_frame)){

    # looping over the columns of data frame
    for (j in 1:ncol(data_frame)){

        # computing sum of row i
        row_sum <- sum(data_frame[i,])

        # calculating row proportion of the cell
        # value
        data_frame[i,j] <- data_frame[i,j]/row_sum
    }
}

# printing modified data frame
print ("Modified data frame")
print (data_frame)

Output:

[1] "Original data frame"
 C1 C2 C3
1  0  1  9
2  1  2 10
3  2  3 11
4  3  4 12
[1] "Modified data frame"
         C1        C2        C3
1 0.00000000 0.1000000 0.9890110
2 0.07692308 0.1656051 0.9763215
3 0.12500000 0.2123894 0.9702410
4 0.15789474 0.2475570 0.9673166

Example 2 : Using rowSums() method

This method loops over the data frame and iteratively computes the sum of each row in the data frame. For the application of this method, the input data frame must be numeric in nature. However, this method is also applicable for complex numbers. The following syntax in R can be used to compute the row proportion of cell values, wherein the output has to be explicitly stored into a new data frame :

Syntax: mdf<-df/rowSums(df)

Arguments : df – The data frame to compute the proportion of row values

Code:

R

# declaring a data frame in R
data_frame = data.frame(C1= c(0,1,2,3),
                        C2 = c(2,3,2,3),
                        C3 = c(9:12))

print("Original data frame")
print(data_frame)

# divides each cell value with corresponding
# row sum value
data_frame<-data_frame/rowSums(data_frame)

# printing modified data frame
print ("Modified data frame")
print (data_frame)

Output:

[1] "Original data frame"
 C1 C2 C3
1  0  2  9
2  1  3 10
3  2  2 11
4  3  3 12
[1] "Modified data frame"
         C1        C2        C3
1 0.00000000 0.1818182 0.8181818
2 0.07142857 0.2142857 0.7142857
3 0.13333333 0.1333333 0.7333333
4 0.16666667 0.1666667 0.6666667

The following code snippet illustrates the calculation of row proportion on the complex numbers’ data frame : 

R

# declaring a data frame in R
data_frame = data.frame(C1= c(1+2i,3i,6+5i,1+2i),
                        C2 = c(2,3,2,3),
                        C3 = c(9:12))

print("Original data frame")
print(data_frame)

# divides each cell value with corresponding row sum value
data_frame<-data_frame/rowSums(data_frame)

# printing modified data frame
print ("Modified data frame")
print (data_frame)

Output

[1] "Original data frame"
   C1 C2 C3
1 1+2i  2  9
2 0+3i  3 10
3 6+5i  2 11
4 1+2i  3 12
[1] "Modified data frame"
                   C1                    C2                   C3
1 0.1081081+0.1486486i 0.1621622-0.02702703i 0.7297297-0.1216216i
2 0.0505618+0.2191011i 0.2191011-0.05056180i 0.7303371-0.1685393i
3 0.3601036+0.1683938i 0.0984456-0.02590674i 0.5414508-0.1424870i
4 0.0769231+0.1153846i 0.1846154-0.02307692i 0.7384615-0.0923077i

All the values are evaluated in the form of a whole number + 0i and the corresponding row proportion value is returned.