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SECTION 09 / 31
Build a word-length frequency dictionary
REAL COURSE DATA · KEYS, COUNTS AND CONNECTIONS
Build the word-length frequency dictionary
Follow the dictionary as its keys and values change. Predict each lookup or update before revealing the next step.
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Predict the next key, whether it already exists, and the value that will be stored.
lengthCount = {}
foreach X in words {
k = X.LetterCount
if (isKey(lengthCount,k)) {
lengthCount[k] = lengthCount[k] + 1
}
else {
lengthCount[k] = 1
}
}| LetterCount | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|---|---|
| Word-card frequency | 11 | 16 | 10 | 2 | 7 | 4 | 4 | 5 | 4 | 2 |
Use the provided LetterCount field. The Word field retains punctuation, so counting the visible characters of “morning.” is a different calculation. Repeated word occurrences each count once. There are 10 observed length keys and 65 word records. The most frequent length is 3, with 16 occurrences.
A bar chart can put observed lengths in numerical order and frequencies on the vertical axis. That ordering is a presentation step; it does not require dictionary storage to be sorted. A missing length key means no observed occurrence under this counting rule.
Try it yourself
What does lengthCount[3]=16 mean?
Show explained answer
Sixteen word cards have supplied LetterCount 3. It is not the third word’s length.
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What should the frequencies sum to?
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65, the number of word cards.
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If “the” appears many times, do we count it once?
Show explained answer
No. This is a frequency of word-card occurrences, not a count of distinct spellings.