Sorting with order() in R Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the vector nums in ascending order using order().

R Programming

nums <- c(5, 2, 9, 1, 7)
sorted_nums <- nums[[1](nums)]
sorted_nums

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Asort

Brank

Cwhich

Dorder

Attempts:

3 left

💡 Hint

Common Mistakes

Using sort() inside the brackets instead of order().

Using rank() which gives ranks, not indices.

Using which() which finds indices of TRUE values.

The order() function returns the indices that would sort the vector. Using these indices to subset nums gives the sorted vector.

2fill in blank

medium

Complete the code to sort the data frame df by the column age in ascending order using order().

R Programming

df <- data.frame(name = c("Anna", "Bob", "Cara"), age = c(25, 22, 30))
sorted_df <- df[[1](df$age), ]
sorted_df

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Arank

Bsort

Corder

Dwhich

Attempts:

3 left

💡 Hint

Common Mistakes

Using sort() which returns sorted values, not indices.

Using rank() which gives ranks, not indices.

Forgetting the comma after the index to subset rows.

Using order(df$age) returns the row indices that sort the age column. Subsetting df by these indices sorts the data frame by age.

3fill in blank

hard

Fix the error in the code to sort the vector vals in descending order using order().

R Programming

vals <- c(3, 8, 1, 6)
sorted_vals <- vals[[1](-vals)]
sorted_vals

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Aorder

Bsort

Crank

Dwhich

Attempts:

3 left

💡 Hint

Common Mistakes

Using sort(-vals) inside brackets which returns values, not indices.

Using rank() which does not give sorting indices.

Using which() which is unrelated here.

Using order(-vals) returns indices that sort vals in descending order. Subsetting vals by these indices gives the descending sorted vector.

4fill in blank

hard

Fill both blanks to sort the data frame df first by score descending, then by age ascending using order().

R Programming

df <- data.frame(name = c("Tom", "Sue", "Jim"), score = c(90, 90, 85), age = c(20, 22, 21))
sorted_df <- df[order([1], [2]), ]
sorted_df

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A-df$score

Bdf$score

Cdf$age

D-df$age

Attempts:

3 left

💡 Hint

Common Mistakes

Using df$score instead of -df$score for descending order.

Using -df$age which sorts age descending instead of ascending.

Forgetting the comma after order() arguments.

To sort score descending, use -df$score. To sort age ascending, use df$age. The order() function sorts by the first argument, then the second.

5fill in blank

hard

Fill all three blanks to create a named vector sorted_vec sorted by values ascending, keeping names, using order().

R Programming

vec <- c(b = 3, a = 1, c = 2)
sorted_vec <- vec[[1](vec)]
names_sorted <- names([2])
result <- setNames([3], names_sorted)
result

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Aorder

Bsorted_vec

Cvec

Dnames_sorted

Attempts:

3 left

💡 Hint

Common Mistakes

Using sort() instead of order() for indices.

Mixing up which vector to get names from.

Not using setNames() to keep names with values.

Use order(vec) to get indices that sort vec. Subset vec by these indices to get sorted_vec. Then get names from sorted_vec and assign them back to the sorted values sorted_vec using setNames().