forked from mrlan/EnglishPal
fix BUG543
parent
cb576b40ed
commit
db66c8ed86
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@ -7,7 +7,7 @@
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import pickle
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import pickle
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import math
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import math
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from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order
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from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order, map_percentages_to_levels
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import snowballstemmer
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import snowballstemmer
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@ -94,10 +94,16 @@ def revert_dict(d):
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return d2
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return d2
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def user_difficulty_level(d_user, d):
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def user_difficulty_level(d_user, d, calc_func=0):
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'''
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two ways to calculate difficulty_level
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set calc_func!=0 to use sqrt, otherwise use weighted average
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'''
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if calc_func != 0:
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# calculation function 1: sqrt
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d_user2 = revert_dict(d_user) # key is date, and value is a list of words added in that date
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d_user2 = revert_dict(d_user) # key is date, and value is a list of words added in that date
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geometric = 0
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count = 0
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count = 0
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geometric = 1
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for date in sorted(d_user2.keys(),
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for date in sorted(d_user2.keys(),
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reverse=True): # most recently added words are more important while determining user's level
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reverse=True): # most recently added words are more important while determining user's level
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lst = d_user2[date] # a list of words
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lst = d_user2[date] # a list of words
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@ -112,12 +118,34 @@ def user_difficulty_level(d_user, d):
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word = t[0]
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word = t[0]
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hard = t[1]
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hard = t[1]
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# print('WORD %s HARD %4.2f' % (word, hard))
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# print('WORD %s HARD %4.2f' % (word, hard))
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geometric = geometric * (hard)
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geometric = geometric + math.log(hard)
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count += 1
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count += 1
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if count >= 10:
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return math.exp(geometric / max(count, 1))
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return geometric ** (1 / count)
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# calculation function 2: weighted average
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d_user2 = revert_dict(d_user) # key is date, and value is a list of words added in that date
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count = {} # number of all kinds of words
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percentages = {} # percentages of all kinds of difficulties
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total = 0 # total words
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for date in d_user2.keys():
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lst = d_user2[date] # a list of words
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for word in lst:
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if word in d:
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if d[word] not in count:
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count[d[word]] = 0
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count[d[word]] += 1
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total += 1
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if total == 0:
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return 1
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for k in count.keys():
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percentages[k] = count[k] / total
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weight = map_percentages_to_levels(percentages)
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sum = 0
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for k in weight.keys():
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sum += weight[k] * k
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return sum
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return geometric ** (1 / max(count, 1))
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def text_difficulty_level(s, d):
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def text_difficulty_level(s, d):
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@ -10,6 +10,32 @@ import operator
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import os, sys # 引入模块sys,因为我要用里面的sys.argv列表中的信息来读取命令行参数。
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import os, sys # 引入模块sys,因为我要用里面的sys.argv列表中的信息来读取命令行参数。
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import pickle_idea
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import pickle_idea
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def map_percentages_to_levels(percentages):
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'''
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功能:按照加权平均难度,给生词本计算难度分,计算权重的规则是(10 - 该词汇难度) * 该难度词汇占总词汇的比例,再进行归一化处理
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输入:难度占比字典,键代表难度3~8,值代表每种难度的单词的占比
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输出:权重字典,键代表难度3~8,值代表每种难度的单词的权重
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'''
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# 已排序的键
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sorted_keys = sorted(percentages.keys())
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# 计算权重和权重总和
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sum = 0 # 总和
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levels_proportions = {}
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for k in sorted_keys:
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levels_proportions[k] = 10 - k
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for k in sorted_keys:
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levels_proportions[k] *= percentages[k]
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sum += levels_proportions[k]
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# 归一化权重到权重总和为1
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for k in sorted_keys:
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levels_proportions[k] /= sum
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return levels_proportions
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def freq(fruit):
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def freq(fruit):
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'''
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'''
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功能: 把字符串转成列表。 目的是得到每个单词的频率。
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功能: 把字符串转成列表。 目的是得到每个单词的频率。
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