Question

def main():
    file=[['mississippi', 'worth', 'reading','river'], ['commonplace', 'river', 'contrary', 'ways', 'remarkable']]
    print(set_and_count(file))

def set_and_count(common_deleted):
    sets=list(set(common_deleted[0]))
    for i in range(len(common_deleted)):
        ## make a non-repeated word list
        sets=list(set(sets+common_deleted[i]))
    ## initialize dict
    dict_wordloc={}
    for j in range(len(sets)):
        sublist=[]
        count_time=0
        for k in range(len(common_deleted)):
            if sets[j]  in common_deleted[k]:
                count_time+=1
                sublist.append(k)
                dict_wordloc[sets[k]]=count_time,sublist
    return(dict_wordloc)

main()

问题1：代码只是返回答案的一部分

在示例输入文件中，字典中的键必须为 'mississippi'，'worth'，'reading'，'river'，'complaceplace，'contrary'，'ways'，'perfectable'

但是当我运行代码时，它返回：

{'remarkable': (2, [0, 1]), 'ways': (1, [0, 1])}

每次我运行它时，返回的东西都不一样

例如，当我运行两次时，结果是

{'contrary': (2, [0, 1]), 'ways': (1, [0, 1])}

问题2：答案是错误的，在问题1的输出中，键remarkable的值必须为(1,[1])

元组中的第一项是该单词有多少个句子（嵌套列表就是一个句子）

元组中的第二项是包含该单词的句子NO

Answer 1

您可以改为使用嵌套的for循环遍历单词，并使用单词作为键将enumerate生成的索引附加到相应的dict条目：

d = {}
for i, l in enumerate(file):
    for w in l:
        d.setdefault(w, [0, []])[0] += 1
        d[w][1].append(i)

d将变为：

{'mississippi': [3, [0, 7, 8]], 'worth': [1, [0]], 'reading': [1, [0]], 'commonplace': [1, [1]], 'river': [4, [1, 2, 3, 6]], 'contrary': [1, [1]], 'ways': [1, [1]], 'remarkable': [1, [1]], 'considering': [1, [2]], 'missouri': [1, [2]], 'main': [1, [2]], 'branch': [1, [2]], 'longest': [1, [2]], 'seems': [1, [3]], 'safe': [1, [3]], 'crookedest': [1, [3]], 'part': [1, [3]], 'journey': [1, [3]], 'uses': [1, [3]], 'cover': [1, [3]], 'ground': [1, [3]], 'crow': [1, [3]], 'fly': [1, [3]], 'six': [1, [3]], 'seventy-five': [1, [3]], 'discharges': [1, [4]], 'water': [3, [4, 6, 7]], 'st': [1, [4]], 'lawrence': [1, [5]], 'twenty-five': [1, [5]], 'rhine': [1, [5]], 'thirty-eight': [1, [5]], 'thames': [1, [5]], 'vast': [1, [6]], 'drainage-basin': [2, [6, 8]], 'draws': [1, [6]], 'supply': [1, [6]], 'twenty-eight': [1, [6]], 'states': [1, [6]], 'territories': [1, [6]], 'delaware': [1, [6]], 'atlantic': [1, [6]], 'seaboard': [1, [6]], 'country': [1, [6]], 'idaho': [1, [6]], 'pacific': [1, [6]], 'slope': [1, [6]], 'spread': [1, [6]], 'forty-five': [1, [6]], 'degrees': [1, [6]], 'longitude': [1, [6]], 'receives': [1, [7]], 'carries': [1, [7]], 'gulf': [1, [7]], 'fifty-four': [1, [7]], 'subordinate': [1, [7]], 'rivers': [1, [7]], 'navigable': [2, [7, 7]], 'steamboats': [1, [7]], 'hundreds': [1, [7]], 'flats': [1, [7]], 'keels': [1, [7]], 'area': [1, [8]], 'combined': [1, [8]], 'areas': [1, [8]], 'england': [1, [8]], 'wales': [1, [8]], 'scotland': [1, [8]], 'ireland': [1, [8]], 'france': [1, [8]], 'spain': [1, [8]], 'portugal': [1, [8]], 'germany': [1, [8]], 'austria': [1, [8]], 'italy': [1, [8]], 'turkey': [1, [8]], 'almost': [1, [8]], 'wide': [1, [8]], 'region': [1, [8]], 'fertile': [1, [8]], 'valley': [1, [8]], 'proper': [1, [8]], 'exceptionally': [1, [8]]}

如果您希望将商品值设为元组，则可以使用以下方式转换d：

d = {k: tuple(v) for k, v in d.items()}

字典只返回答案的一部分（也许是关于缩进的）

1 个答案: