← QCE Digital Solutions Unit 3

Merging two data sources in pseudocode

Loops through one data source, finds the record with the matching key in the other, then combines them or appends unmatched records, avoiding duplicate keys.

Part of: QCE Digital Solutions Unit 3 — Real-World Problems and Solution Requirements — Module 88

Understanding Merging two data sources in pseudocode in QCE Digital Solutions Unit 3

What it does: Handling records with no match means adding them to the merged result so no data is lost. Checking the key before appending stops the same key appearing twice in the merged result. A merge works by matching on a key field. Loop through one source and search the other for the same key. The key must appear in both sources and identify the same entity. studentId does this; mark does not. Comparing uses a comparison operator such as =. The arrow <- is for assignment, which would overwrite a value instead of testing it. Comparing the key fields (here studentId) is how the merge decides that two records describe the same thing.

Common mistakes: The score must be copied from the matching record in the other source (b.score). Assigning a.score to itself changes nothing, so nothing is merged. It should be a.score <- b.score. The inner loop must search the other data source, listB, for the matching key. Looping through listA again only compares listA with itself. It should be for each b in listB.

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Learning

Duo

Which describes the basic merge algorithm for two data sources?

Join the two sources end to end without checking any keysLoop through one source and look up the record with the matching key in the other

Matches

for each s in students append s to merged append r to merged

↓

Both records are appended without matching on the key, so the same student appears twice with duplicate keys instead of one combined record

Applying

Meanings

Term

merge algorithm

Meaning

Loops through one data source, finds the record with the matching key in the other, then combines them or appends unmatched records, avoiding duplicate keys.

Practice

for each n in newRecords

Testing

Gaps

if a.id b.id then

Check Answer

Spots

3 if a.id = b.id then

4 a.score <- a.score

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Quick Questions

1. Which describes the basic merge algorithm for two data sources?

A. Loop through one source and look up the record with the matching key in the other ✓

B. Join the two sources end to end without checking any keys

A merge works by matching on a key field. Loop through one source and search the other for the same key.

2. A student has two lists, students and results, both with a studentId. Which should be the key field for the merge?

A. mark, because it is a number

B. studentId, because it identifies the same person in both sources ✓

The key must appear in both sources and identify the same entity. studentId does this; mark does not.

3. A record in source A has key 7, but source B has no record with key 7. What should the merge do?

A. Append the unmatched record to the merged result (or handle it by a clear rule) ✓

B. Ignore the problem and let the program crash

Records with no match must be handled deliberately. Typically you append them, or log or skip them, but never leave this case undefined.

4. Which pseudocode line correctly tests whether two records match on the key?

A. if a.id <- b.id then

B. if a.id = b.id then ✓

Comparing uses a comparison operator such as =. The arrow <- is for assignment, which would overwrite a value instead of testing it.

5. What is the risk if you append every record from source B without checking whether its key is already in the merged result?

A. Duplicate keys appear in the merged result ✓

B. The result will be sorted in the wrong order

Appending without checking makes matched records appear twice. Only append records whose key has no match.

Match the Pairs

for each s in students → Loops through every record in the first data source so each one can be checked against the second source

if s.studentId = r.studentId then → Matches records from the two sources on the key field

A record with no match in the other source is appended to the merged result instead of being dropped → if found = false then append s to merged

Before adding a record, check that its key is not already in the merged result, so there are no duplicate keys → if key not in merged then append record to merged

for each s in students append s to merged append r to merged → Both records are appended without matching on the key, so the same student appears twice with duplicate keys instead of one combined record

for each s in students if s.studentId = r.studentId then append combined to merged endif next s → There is no handling for records with no match, so unmatched students are lost from the merged result

Key Term

merge algorithm

Loops through one data source, finds the record with the matching key in the other, then combines them or appends unmatched records, avoiding duplicate keys.

Fill in the Blanks

1. for each a in sourceA for each b in sourceB if a.id = b.id then

Hint: Type the comparison operator that checks whether the two key fields are equal.

2. found ← false if found = false then append b to merged

Hint: Type the Boolean value that means no matching record was found.

3. if a.id = b.id then newRec ← combine(a, b) append newRec to merged

Hint: Type the function name that joins the two matching records into one.

4. for each b in sourceB if not exists(b.id, merged) then append b to merged

Hint: Type the Boolean operator that stops a key already in the merged result being added again.

5. merged ← [] for each a in sourceA append a to merged

Hint: Type the keyword phrase that loops through every record in the first data source.

Spot the Error

Question 1

A for each a in listA

B for each b in listB

C if a.id = b.id then

D a.score <- a.score

E end if

Line D — The score must be copied from the matching record in the other source (b.score). Assigning a.score to itself changes nothing, so nothing is merged. It should be a.score <- b.score.

Question 2

A for each a in listA

B found <- false

C for each b in listA

D if a.id = b.id then

E found <- true

F end if

Line C — The inner loop must search the other data source, listB, for the matching key. Looping through listA again only compares listA with itself. It should be for each b in listB.

Question 3

A for each n in newList

B if n.id not in master then

C append n to newList

D end if

Line C — An unmatched record has to be added to the merged data source (master). Appending it to newList, the list being looped through, does not merge anything. It should be append n to master.

Question 4

A for each n in newList

B if n.id in master then

C append n to master

D else

E append n to master

F end if

Line C — When the key already exists in master, appending the record creates a duplicate key. A matched record should update or combine with the existing record, for example update master record.

Question 5

A merged <- empty array

B for each a in merged

C for each b in listB

D if a.id = b.id then

E append a + b to merged

F end if

Line B — merged starts empty, so looping through it never runs and nothing is ever merged. The outer loop must go through the source data, for each a in listA.

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