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Standard Deviation of Rows in R can be calculated using the rowSds() function within the ‘matrixStats’ package. This function takes a matrix as its argument and returns the standard deviation of each row as a vector. The standard deviation is calculated using the formula: SD = sqrt((sum((x-mean)^2))/n-1), where x is each value in the row, mean is the mean of the row, and n is the number of values in the row.
You can use the following basic syntax to calculate the standard deviation of rows in R:
row_stdev <- apply(df, 1, sd, na.rm=TRUE)
The following example shows how to use this syntax in R.
Example: Calculate Standard Deviation of Rows in R
Suppose we have the following data frame in R:
#create data frame df <- data.frame(game1=c(12, 15, 15, 18, 29, 30, 31), game2=c(15, 17, 17, 16, 29, 8, 14), game3=c(8, 22, 27, 35, 29, 22, 17)) #view data frame df game1 game2 game3 1 12 15 8 2 15 17 22 3 15 17 27 4 18 16 35 5 29 29 29 6 30 8 22 7 31 14 17
We can use the following syntax to calculate the standard deviation of the values in each row:
#calculate standard deviation of each row
row_stdev <- apply(df, 1, sd, na.rm=TRUE)
#view standard deviation of each row
row_stdev
[1] 3.511885 3.605551 6.429101 10.440307 0.000000 11.135529 9.073772
From the output we can see:
- The standard deviation of values in the first row is 3.511885.
- The standard deviation of values in the second row is 3.605551.
- The standard deviation of values in the third row is 6.429101.
And so on.
If we’d like, we can also use the function to add a new column to the data frame that shows the standard deviation of values in each row:
#add column that displays standard deviation of each row df <- transform(df, row_stdev=apply(df, 1, sd, na.rm=TRUE)) #view updated data frame df game1 game2 game3 row_stdev 1 12 15 8 3.511885 2 15 17 22 3.605551 3 15 17 27 6.429101 4 18 16 35 10.440307 5 29 29 29 0.000000 6 30 8 22 11.135529 7 31 14 17 9.073772
The new column called row_stdev displays the standard deviation of values in each row.
Note: The standard deviation of values in row 5 is equal to zero because each of the values is the same, thus there is no “deviation” at all in the values.
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