Python 2.7 如何创建一个嵌套在python中的字典，它将表示与特定作品相关的单词出现的次数？

我有这样一个清单：-

new_list=[['happy', 'people', 'need', 'rest'], ['rest', 'apple', 'good', 'happy'],['kepler', 'people', 'happy', 'apple'],['need', 'good', 'happy', 'good']]

还有另一个类似的列表

a=['good', 'apple', 'people', 'rest', 'kepler', 'need', 'happy']

它包含给定的新_列表的不同单词数。 现在，我想遍历列表“a”中的每个元素，并在新列表中找到包含该元素的子列表，然后在这些子列表中，我想计算其中单词的出现率。例如，对于列表“a”中的单词“good”，新列表中包含“good”的子列表如下

现在计算这个列表中每个单词的出现率

{"good":{"rest":1,"apple":1,"happy":2,"need":1}}

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类似地，我想要“a”列表中所有元素的总体输出。如何执行tht？

您可以对s和ifs使用嵌套的

new_list=[['happy', 'people', 'need', 'rest'], ['rest', 'apple', 'good', 'happy'],['kepler', 'people', 'happy', 'apple'],['need', 'good', 'happy', 'good']]
a=['good', 'apple', 'people', 'rest', 'kepler', 'need', 'happy']

dct = {}
for item in a:
    mid_dct = {}
    for lst in new_list:
        if item in lst: #checks if distinct item is in list
            for word in lst: #iterates over that list to store words and their number of occurences
                if word != item:
                    if word in mid_dct: #if that word in dct, adds one, if not adds that to list
                        mid_dct[word] += 1
                    else:
                        mid_dct[word] = 1
    dct[item] = mid_dct        
print dct

new_list=[['happy', 'people', 'need', 'rest'], ['rest', 'apple', 'good', 'happy'],['kepler', 'people', 'happy', 'apple'],['need', 'good', 'happy', 'good']]
a=['good', 'apple', 'people', 'rest', 'kepler', 'need', 'happy']

dct = {}
for item in a:
    mid_dct = {}
    for lst in new_list:
        if item in lst: #checks if distinct item is in list
            for word in lst: #iterates over that list to store words and their number of occurences
                if word != item:
                    if word in mid_dct: #if that word in dct, adds one, if not adds that to list
                        mid_dct[word] += 1
                    else:
                        mid_dct[word] = 1
    dct[item] = mid_dct        
print dct