| Age | Commit message (Collapse) | Author | |
|---|---|---|---|
| 2020-08-12 | main.py: update comments in difficulty.py. | Hui Lan | |
| 2020-08-12 | main.py: update comments in difficulty.py. | Hui Lan | |
| 2020-08-12 | main.py: update comments in difficulty.py. | Hui Lan | |
| 2020-08-12 | main.py: refactor code. | Hui Lan | |
| 2020-08-12 | README.md: force reclaim unused Docker objects by appending -f. | Hui Lan | |
| 2020-01-27 | signup.html: add a info line for the form | Hui Lan | |
| 2020-01-27 | englishpal: updated gitignore file | Hui Lan | |
| 2019-11-02 | app/difficulty.py: adjust difficulty level to reflect test level | Hui Lan | |
| If a word appears in CET4 test, it has a difficulty level of 4. If a word appears in CET6 test, it has a difficulty level of 6. If a word does not appear in the above two tests, then its difficulty level is determined by its frequency. If this word is not in the frequency table, then its level is 1. If this word is in the frequency tables, then its level is determined as follows. log2( ((F+1) / (f+1)) ), where F is the largest frequency, and f is the word's frequency. If f=F, then log2(1) = 0, i.e., its difficulty level is 0. If f<F, then log2(1) is a positive value. | |||
| 2019-11-02 | app/difficulty.py: make the computed difficulty level more reasonable | Hui Lan | |
| If a word belongs to CET4, then it has level 1. If a word belongs to CET6, then it has level 2. If a word does not belong to either CET4 or CET6, then we need to decide its level using word frequency. | |||
| 2019-11-01 | python-files: add copyright notice | Hui Lan | |
| 2019-11-01 | englishpal: first commit | Hui Lan | |
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index : EnglishPal | |
| Unnamed repository; edit this file 'description' to name the repository. | Git |
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