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not an item in an array or list. An array or list item is accessed using its index."},{"text":"Its value.","correct":false,"explanation":"A value is the data being stored or retrieved, rather than the means of accessing it. An array or list item is accessed using its index."},{"text":"Its memory address.","correct":false,"explanation":"Memory addresses are managed at a lower level and are not normally used for direct list access. An array or list item is accessed using its index."}]},{"id":"e66b92e8-8a00-4543-a5d0-05703b8d6d3b","cardType":"cloze","points":null,"front":"A dictionary stores [ ] and retrieves each value using its [ ].","back":"A dictionary stores key-value pairs and retrieves each value using its key.","mcqOptions":[{"text":"A dictionary stores key-value pairs and retrieves each value using its key.","correct":true,"explanation":"A dictionary associates each key with a value, forming key-value pairs. Supplying a key allows the corresponding value to be retrieved without using a positional index."},{"text":"A dictionary stores index-value pairs and retrieves each value using its index.","correct":false,"explanation":"This confuses dictionaries with indexed sequences such as arrays. A dictionary stores key-value pairs and retrieves each associated value using its key, not a positional index."},{"text":"A dictionary stores key-value pairs and retrieves each value using its value.","correct":false,"explanation":"A value is the result of a dictionary lookup, not normally the identifier used to perform one. Dictionaries store key-value pairs and retrieve values using keys."},{"text":"A dictionary stores keys only and retrieves each value using its key.","correct":false,"explanation":"A dictionary must retain both keys and their associated values, rather than keys alone. It therefore stores key-value pairs and retrieves each value using its key."}]},{"id":"f5663ac9-54d8-4ce3-af5a-e3578ab75263","cardType":"basic","points":1,"front":"What operation uses a key to retrieve its value?","back":"A lookup.","mcqOptions":[{"text":"A lookup.","correct":true,"explanation":"A lookup supplies a key to a dictionary and returns the value associated with that key, provided that the key exists in the dictionary."},{"text":"An insertion.","correct":false,"explanation":"An insertion adds a new key-value pair or assigns a value, rather than retrieving an existing value. Retrieving a value by its key is a lookup."},{"text":"A deletion.","correct":false,"explanation":"A deletion removes a key and its associated value from the dictionary. Using a key to retrieve the associated value is instead called a lookup."},{"text":"An iteration.","correct":false,"explanation":"An iteration processes multiple dictionary entries in turn rather than directly retrieving one value from a supplied key. That direct retrieval operation is a lookup."}]},{"id":"a56e93cf-5153-4bd4-bbe1-c19595e6c546","cardType":"basic","points":1,"front":"In 'green' : 2, what is the key?","back":"'green'.","mcqOptions":[{"text":"'green'.","correct":true,"explanation":"In a dictionary entry, the expression before the colon is the key and the expression after it is the value. Therefore, 'green' is the key."},{"text":"2.","correct":false,"explanation":"The item after the colon is the value associated with the key. Here, 2 is the value, while 'green' is the key."},{"text":"'green' : 2.","correct":false,"explanation":"The complete expression represents a key-value pair rather than the key alone. The part before the colon, 'green', is the key."},{"text":"':'.","correct":false,"explanation":"The colon separates a dictionary entry’s key from its value; it is not itself the key. The key in this pair is 'green'."}]},{"id":"e70d28c3-0d31-4cf6-99b7-1dac9e9c0e2d","cardType":"basic","points":1,"front":"In wordCount['green'], what does 'green' act as?","back":"The key.","mcqOptions":[{"text":"The key.","correct":true,"explanation":"Inside square brackets after a dictionary name, 'green' identifies the entry to access. It therefore acts as the key whose associated value is retrieved."},{"text":"The value.","correct":false,"explanation":"The value is what the dictionary expression retrieves, not what appears inside the brackets. Here, 'green' is the key used to retrieve the value."},{"text":"The index.","correct":false,"explanation":"An index denotes a numerical position in an array or list. In this dictionary access, 'green' identifies an entry by acting as the key."},{"text":"The item.","correct":false,"explanation":"The dictionary itself contains items, but the bracketed expression identifies one of them. In this expression, 'green' specifically acts as the key."}]},{"id":"c86f3ce9-98bc-4409-a3cd-2e167538b30e","cardType":"basic","points":1,"front":"For a word-frequency dictionary, what should the key be?","back":"The word.","mcqOptions":[{"text":"The word.","correct":true,"explanation":"Each distinct word must identify its own dictionary entry, so the word is used as the key. Its associated value records how often it occurs."},{"text":"The count.","correct":false,"explanation":"The count is the information associated with each word, so it belongs in the value. The word itself must be the dictionary key."},{"text":"The index.","correct":false,"explanation":"A word-frequency dictionary groups occurrences by word rather than by position in the source text. Each word is therefore a key, while its count is the value."},{"text":"The dictionary.","correct":false,"explanation":"The dictionary is the entire data structure, not an identifier for one entry. Each distinct word should be a key associated with its frequency count."}]},{"id":"05b2391b-6d18-4140-957e-936959f80242","cardType":"basic","points":1,"front":"For a word-frequency dictionary, what should the value be?","back":"The word’s count.","mcqOptions":[{"text":"The word’s count.","correct":true,"explanation":"The value records how many times the key word appears in the text. Each occurrence updates this count, while the word itself remains the key."},{"text":"The word’s position.","correct":false,"explanation":"The word’s position does not represent its total frequency and may differ for every occurrence. The value should be the word’s count across the text."},{"text":"The word’s length.","correct":false,"explanation":"Word length measures the number of characters, not the number of occurrences. In a frequency dictionary, the value must store the word’s count."},{"text":"The word’s spelling.","correct":false,"explanation":"The spelling is already represented by the word used as the dictionary key. The associated value should instead record that word’s count."}]},{"id":"d735d209-5ecf-4b42-9c67-db218aa9e90b","cardType":"basic","points":1,"front":"Which data structure is best when order by position matters?","back":"An array/list.","mcqOptions":[{"text":"An array/list.","correct":true,"explanation":"Arrays and lists maintain an ordered sequence whose items can be identified by positional indices. They are therefore appropriate when position and order matter."},{"text":"A dictionary.","correct":false,"explanation":"A dictionary accesses entries by keys rather than by their numerical positions. When positional order is the main requirement, an array or list is more suitable."},{"text":"A set.","correct":false,"explanation":"A set is designed for unique membership and does not provide ordinary position-based access. An array or list should be used when order by position matters."},{"text":"A record.","correct":false,"explanation":"A record groups named fields, so its components are accessed by field names rather than sequential positions. An array or list is designed for positional order."}]},{"id":"78f99280-b5be-4a64-944e-bb8ae0b59b0e","cardType":"basic","points":1,"front":"Which data structure is best for username to user record lookup?","back":"A dictionary.","mcqOptions":[{"text":"A dictionary.","correct":true,"explanation":"A dictionary maps each unique username key directly to its associated user record, providing efficient lookup without searching sequentially through every stored record."},{"text":"A list.","correct":false,"explanation":"A list normally requires a sequential search to locate a username. A dictionary is preferable because it maps each username key directly to its user record."},{"text":"A stack.","correct":false,"explanation":"A stack restricts access to the most recently added item, so it does not support efficient username-based retrieval. A dictionary provides direct key-based lookup."},{"text":"A queue.","correct":false,"explanation":"A queue retrieves records in first-in, first-out order rather than by username. A dictionary is the appropriate structure because usernames can act as lookup keys."}]},{"id":"acebe24b-0915-4953-9cc8-83968e304e7a","cardType":"cloze","points":null,"front":"To insert into a dictionary means adding a [ ] with an associated [ ].","back":"To insert into a dictionary means adding a new key with an associated value.","mcqOptions":[{"text":"To insert into a dictionary means adding a new key with an associated value.","correct":true,"explanation":"Dictionary insertion creates a new key and associates that key with a value, allowing the value to be retrieved later by looking up the key."},{"text":"To insert into a dictionary means adding a existing key with an associated value.","correct":false,"explanation":"An existing key is already present, so assigning to it normally updates its value rather than performing a new insertion. Insertion adds a new key with an associated value."},{"text":"To insert into a dictionary means adding a new value with an associated key.","correct":false,"explanation":"This reverses the roles of dictionary components: keys identify entries, while values hold associated data. Insertion adds a new key with an associated value."},{"text":"To insert into a dictionary means adding a new index with an associated element.","correct":false,"explanation":"An index is characteristic of an array or list, not the key-value model of a dictionary. Dictionary insertion adds a new key with an associated value."}]},{"id":"ed4ed69d-37dd-4e7d-a932-56338b6fdac8","cardType":"basic","points":1,"front":"When counting words, what should happen if the word is already a key?","back":"Increment its count by 1.","mcqOptions":[{"text":"Increment its count by 1.","correct":true,"explanation":"Because the word has appeared before, its dictionary entry already stores the previous frequency. Adding 1 updates that value to include the latest occurrence."},{"text":"Reset its count to 1.","correct":false,"explanation":"Resetting the count to 1 discards all earlier occurrences, a common mistake when insertion and updating are confused. An existing word’s count must be incremented by 1."},{"text":"Decrement its count by 1.","correct":false,"explanation":"Decreasing the count would treat a new occurrence as though a previous occurrence had disappeared. An existing word’s stored frequency must instead be incremented by 1."},{"text":"Delete its count from the dictionary.","correct":false,"explanation":"Deleting the entry loses both the word and its accumulated frequency. Since the word has occurred again, its existing count should be incremented by 1."}]},{"id":"88ec3f53-d58d-4828-bbd6-a362735dd72a","cardType":"basic","points":1,"front":"When counting words, what should happen if the word is not already a key?","back":"Insert it with value 1.","mcqOptions":[{"text":"Insert it with value 1.","correct":true,"explanation":"A word’s first occurrence requires a new dictionary entry. Inserting the word as a key with value 1 records that it has appeared once."},{"text":"Increment it by value 1.","correct":false,"explanation":"A missing key has no existing count to increment, and direct access may cause an error. Its first occurrence should be inserted with value 1."},{"text":"Insert it with value 0.","correct":false,"explanation":"Starting at 0 fails to count the occurrence currently being processed, causing every new word’s frequency to be one too small. It should be inserted with value 1."},{"text":"Delete it with value 1.","correct":false,"explanation":"Deleting a word that is not already a key cannot record its first occurrence. A new entry should instead be inserted with value 1."}]},{"id":"44c3c9e9-2ba9-4ea3-9e65-c1264bd91980","cardType":"cloze","points":null,"front":"Before increasing a count, use a membership test because looking up a [ ] may cause an [ ].","back":"Before increasing a count, use a membership test because looking up a missing key may cause an error.","mcqOptions":[{"text":"Before increasing a count, use a membership test because looking up a missing key may cause an error.","correct":true,"explanation":"Directly requesting the value for a key that is absent can raise a key-related error. A membership test confirms the key exists before its count is accessed."},{"text":"Before increasing a count, use a membership test because looking up a existing key may cause an error.","correct":false,"explanation":"Looking up an existing key should retrieve its associated value rather than cause an error. The membership test is needed because looking up a missing key may cause an error."},{"text":"Before increasing a count, use a membership test because looking up a missing value may cause an error.","correct":false,"explanation":"Dictionaries are searched using keys, not by treating an absent value as the lookup identifier. The risk arises when looking up a missing key, which may cause an error."},{"text":"Before increasing a count, use a membership test because looking up a missing key may cause an automatic insertion.","correct":false,"explanation":"A failed lookup does not normally create a value automatically; it may instead raise an error. The membership test checks whether the requested key is missing."}]},{"id":"b85e8350-c938-4c79-9ba2-7e16caee82d7","cardType":"basic","points":1,"front":"What does IF word IN frequencies test?","back":"Whether word is already a key.","mcqOptions":[{"text":"Whether word is already a key.","correct":true,"explanation":"The membership operator checks the dictionary’s keys, so this condition is true precisely when word already exists as a key in frequencies."},{"text":"Whether word is already a value.","correct":false,"explanation":"Dictionary membership tests examine keys rather than searching the associated values. The expression checks whether word is already a key in frequencies."},{"text":"Whether frequencies is already non-empty.","correct":false,"explanation":"The condition does not test whether the dictionary contains any entries; it searches for one specific key. It checks whether word is already a key."},{"text":"Whether word has a count greater than zero.","correct":false,"explanation":"Membership does not compare a word with its stored frequency count. It tests whether the value of word occurs among the dictionary’s keys."}]},{"id":"f32e3338-8340-4886-a033-7b86d6896f49","cardType":"cloze","points":null,"front":"Normalisation may convert text to [ ] and remove [ ].","back":"Normalisation may convert text to lower case and remove punctuation.","mcqOptions":[{"text":"Normalisation may convert text to lower case and remove punctuation.","correct":true,"explanation":"Converting all text to lower case merges capitalisation variants, while removing punctuation prevents forms such as “word” and “word,” from being counted separately."},{"text":"Normalisation may convert text to lower case and remove letters.","correct":false,"explanation":"Removing letters would destroy the words being analysed rather than standardise them. Normalisation commonly converts text to lower case and removes punctuation."},{"text":"Normalisation may convert text to numbers and remove words.","correct":false,"explanation":"Converting ordinary text into numbers does not standardise capitalisation for word counting, and removing words destroys the data. Normalisation may use lower case and remove punctuation."},{"text":"Normalisation may convert text to upper case and remove vowels.","correct":false,"explanation":"Removing vowels changes the spelling and meaning of words rather than merely standardising their representation. Normalisation commonly converts text to lower case and removes punctuation."}]},{"id":"616759e5-bd0c-446e-895d-52496cdc8690","cardType":"basic","points":1,"front":"After processing the, green, green, what is the count for green?","back":"2.","mcqOptions":[{"text":"2.","correct":true,"explanation":"The sequence contains green twice. Its first occurrence creates a count of 1, and its second occurrence increments that count to 2."},{"text":"1.","correct":false,"explanation":"A count of 1 records only the first occurrence and misses the required increment for the repeated word. Since green appears twice, its count is 2."},{"text":"3.","correct":false,"explanation":"Counting all three input words confuses total tokens with the frequency of one particular word. Only two of the tokens are green, so its count is 2."},{"text":"0.","correct":false,"explanation":"A count of 0 would mean green never appeared, ignoring both occurrences in the sequence. Processing them produces a final count of 2."}]},{"id":"759c37ba-09b2-45b9-8781-de24d97d7bc3","cardType":"basic","points":1,"front":"What is the count for grass in The green, green grass grows after normalisation?","back":"1.","mcqOptions":[{"text":"1.","correct":true,"explanation":"Normalisation makes the text lowercase and removes punctuation, leaving one occurrence of the word grass in the sentence."},{"text":"2.","correct":false,"explanation":"The repeated word is green, not grass. After normalisation, grass appears once, so its count is 1."},{"text":"0.","correct":false,"explanation":"Punctuation removal does not remove grass; it only standardises the text. The word appears once, so the correct count is 1."},{"text":"3.","correct":false,"explanation":"This counts several words rather than occurrences of grass. Only one normalised token is grass, so the correct count is 1."}]},{"id":"3374baf0-b037-46af-8e23-edc9c6743e36","cardType":"cloze","points":null,"front":"Dictionary display order is not the important part; the [ ] matter.","back":"Dictionary display order is not the important part; the key-value associations matter.","mcqOptions":[{"text":"key-value associations","correct":true,"explanation":"A dictionary is defined by which keys are associated with which values. The order in which entries are displayed does not change those key-value associations."},{"text":"key positions","correct":false,"explanation":"Dictionaries are accessed using keys rather than fixed positions, so display order is not the essential feature. The important concept is the key-value associations."},{"text":"value positions","correct":false,"explanation":"Values do not identify dictionary entries by their position. Each key is associated with a value, so the important part is the key-value associations."},{"text":"entry display labels","correct":false,"explanation":"Display labels describe how entries may appear, not how dictionary data is organised. The meaningful relationship is between each key and its associated value."}]},{"id":"5c75fdc2-c73c-4da1-a15d-f33378f8bdae","cardType":"basic","points":1,"front":"After wordCount['green'] ← 1 then wordCount['green'] ← 2, what is stored for green?","back":"Only 'green' : 2.","mcqOptions":[{"text":"Only 'green' : 2.","correct":true,"explanation":"Assigning 2 to the existing key replaces its earlier value of 1. Dictionaries store one current value for each key, leaving only 'green' : 2."},{"text":"Only 'green' : 1.","correct":false,"explanation":"The second assignment does not get ignored; it updates the existing key. The stored entry therefore becomes 'green' : 2, not the earlier value 1."},{"text":"'green' : 1 and 'green' : 2.","correct":false,"explanation":"A dictionary cannot normally contain two separate entries with the same key. The later assignment replaces the first, leaving only 'green' : 2."},{"text":"The key 'green' is deleted.","correct":false,"explanation":"Assigning a new value does not delete the key; it updates its value. The resulting dictionary contains only 'green' : 2."}]},{"id":"87f1f17a-c7cb-447e-b07b-69ef6c633379","cardType":"basic","points":1,"front":"Can different dictionary keys have the same value?","back":"Yes - values do not have to be unique.","mcqOptions":[{"text":"Yes - values do not have to be unique.","correct":true,"explanation":"Dictionary keys must be unique so each key identifies one entry, but multiple different keys may refer to equal values. Therefore, values do not have to be unique."},{"text":"No - values must be unique.","correct":false,"explanation":"This reverses the uniqueness rule. Keys must be unique, but several different keys can store the same value, so the correct answer is yes."},{"text":"Yes - keys do not have to be unique.","correct":false,"explanation":"The conclusion is right but the reason is wrong: dictionary keys normally must be unique. It is values that may be repeated for different keys."},{"text":"No - only keys can be duplicated.","correct":false,"explanation":"Dictionary keys identify entries and therefore cannot normally be duplicated. Values can be shared by different keys, making yes the correct answer."}]},{"id":"939de258-b1aa-4ac7-97f7-1e311dc6fa36","cardType":"cloze","points":null,"front":"For query green grass, Document A scores [ ] and Document B scores [ ].","back":"For query green grass, Document A scores 3 and Document B scores 1.","mcqOptions":[{"text":"For query green grass, Document A scores 3 and Document B scores 1.","correct":true,"explanation":"Document A matches both query terms with the higher combined score of 3, while Document B matches less strongly and scores 1."},{"text":"For query green grass, Document A scores 1 and Document B scores 3.","correct":false,"explanation":"The document scores have been assigned to the wrong documents. Document A is the stronger match and scores 3, whereas Document B scores 1."},{"text":"For query green grass, Document A scores 2 and Document B scores 1.","correct":false,"explanation":"This undercounts Document A by treating the two matching terms as contributing only one point each or one combined point. The specified scores are 3 for A and 1 for B."},{"text":"For query green grass, Document A scores 3 and Document B scores 2.","correct":false,"explanation":"Document B has been given one extra point, likely by counting a partial or repeated match incorrectly. Its score is 1, while Document A correctly scores 3."}]}]}],"examCountdown":{"examDate":"2027-06-07","paperLabel":"Paper 1"},"heading":"Dictionaries","title":"Dictionaries Flashcards","topbar":{"backHref":"/a-level/computer-science/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"3d3da702-be5b-52ef-8fa0-b478713280d5","walled":"$L38"}],"$L39","$L3a"]