Suppose you have a data frame df about everyone enrolled in a class with 30 students. As it just so happens, no two students in the class share the same last name. In other words, there is no more than one student with the same last name in the class. There are three columns in the data frame, which are all character data type: UBC_ID, lastname, and firstname. Assume that UBC_ID can uniquely identify each person in this dataset. That is, each student has a unique UBC_ID. You open the dataset, and then run the following code: dfa % arrange(lastname, firstname) Right after that, you run the following line of code: dfb % arrange(lastname, UBC_ID) [ Select ] True False Not enough information The order of the observations in dfa is the same as the order of the observations in dfb. If not enough information is provided, choose “not enough information”.

Question

Suppose you have a data frame df about everyone enrolled in a class with 30 students. As it just so happens, no two students in the class share the same last name. In other words, there is no more than one student with the same last name in the class. There are three columns in the data frame, which are all character data type: UBC_ID, lastname, and firstname. Assume that UBC_ID can uniquely identify each person in this dataset. That is, each student has a unique UBC_ID. You open the dataset, and then run the following code: dfa <- df %>% arrange(lastname, firstname) Right after that, you run the following line of code: dfb <- df %>% arrange(lastname, UBC_ID) [ Select ] True False Not enough information The order of the observations in dfa is the same as the order of the observations in dfb. If not enough information is provided, choose “not enough information”.

BlackTom AI · Accepted Answer

We start by restating the scenario: there are 30 students with unique last names, and we have three character columns UBC_ID, lastname, and firstname. UBC_ID uniquely identifies each person. The code sorts as follows: dfa <- df %>% arrange(lastname, firstname) dfb <- df %>% arrange(lastname, UBC_ID) Option analysis: Option 1: True. Since every lastname is unique, the primary sort key (lastname) already determines a unique order for all rows. The secondary keys (firstname in dfa, and UBC_ID in dfb) only come into play if there are ties on lastname. Because there are no ties on lastname, both sorts produce the same overall ordering: ascending by lastname. Therefore, the order of observations in dfa is the same as in dfb. Option 2: False. This would suggest that the two sorts could yield different orders, which is incorrect given the uniqueness of last names; there are no tied last names to cause any secondary-key differences. Option 3: Not enough information. The problem provides sufficient information: last names are unique, so the two sorts will coincide in order, making the comparison determinate. Option 4: The order of the observations in dfa is the same as the order of the observations in dfb. This statement is consistent with the reasoning above and is therefore correct; it restates the conclusion drawn from the unique last names. In summary, the critical fact is the uniqueness of last names ensuring that sorting by (lastname, firstname) and by (lastname, UBC_ID) yield the same observation order, since there are no last-name ties to resolve differently with the secondary keys.

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