Question

在我的代码中，我在字典中有键（概念）。运行时，与（概念）中的键匹配的输入将打印字典（文件）中的匹配值。现在，我尝试在当前输出的底部打印，字典（文件）中的值与输入中输入的确切值匹配。在这种情况下，如果您输入“机器学习游戏”，您将获得-

输入概念想法：机器学习游戏 file0004.txt file0009.txt file0009.txt file0009.txt file0004.txt file0008.txt file0004.txt file0009.txt file0004.txt file0009.txt file0004.txt file0008.txt http://file0001.txt file0002.txt file0004.txt file0003.txt file0003.txt file0008.txt

但是，files0008.txt是唯一与在dictionary（concept）值中存在的确切特定键-（（number“ 8”）））匹配的唯一值。我想要一个值或任何精确值（可能有多个匹配的精确值）并将其列出在其他单个输出下面。像-

唯一值输出：file0008.txt

========================代码===================== =====

def concept(phrase):

    # split var(phrase) at spaces and assign to var(words)
    words = phrase.split()

    # use this to list python file titles and links to open them
    files    = {1:"http://file0001.txt",
                2:"file0002.txt",
                3:"file0003.txt",
                4:"file0004.txt",
                5:"file0005.txt",
                6:"file0006.txt",
                7:"file0007.txt",    
                8:"file0008.txt",
                9:"file0009.txt"}

    # change keys to searchable simple keyword phrases. 
    concepts = {'GAMES':[1,2,4,3,3,8],
                'BLACKJACK':[5,3,5,3,5],
                'MACHINE':[4,9,9,9,4,8],
                'DATABASE':[5,3,3,3,5],
                'LEARNING':[4,9,4,9,4,8]}

    # iterate through all var(words) found in var(word)
    for word in words:
    # convert to uppercase, search var(word) in dict 'concepts', if not found return not found"
        if word.upper() not in concepts:
            print("\n'{}':Not Found in Database \n" .format(word)) not in concepts
        else:
    # for matching keys in dict 'concept' list values in dict 'files'
            for pattern in concepts[word.upper()]:
                print(files[pattern])

# return input box at end of query        
while True:
    concept(input("Enter Concept Idea: "))

Answer 1

如果我理解您的想法，那么如果短语中的所有单词都与该文件相关联，那么您想要的是concept（）函数仅打印文件名，并且您希望它打印所有文件名

如果是这样，则使用集合代替列表很有意义。通过采用两个集合的交集，您最终得到的集合中只有两个集合中的元素。

除了一些间距问题之外，您的思维中似乎还存在两个错误：不仅文件8匹配，文件4也匹配；并且您写了“机器学习游戏”，但您可能指的是“机器学习游戏”。

遵循脚本的模式，但是使用集合而不是列表：

def concept(phrase):
    words = phrase.upper().split()

    files = {
        1:"http://file0001.txt",
        2:"file0002.txt",
        3:"file0003.txt",
        4:"file0004.txt",
        5:"file0005.txt",
        6:"file0006.txt",
        7:"file0007.txt",
        8:"file0008.txt",
        9:"file0009.txt"
    }

    concepts = {
        'GAMES': {1, 2, 4, 3, 3, 8},
        'BLACKJACK': {5, 3, 5, 3, 5},
        'MACHINE': {4, 9, 9, 9, 4, 8},
        'DATABASE': {5, 3, 3, 3, 5},
        'LEARNING': {4, 9, 4, 9, 4, 8}
    }

    # start with all files
    matching_files = set(files.keys())
    for word in words:
        if word in concepts:
            matching_files = matching_files & concepts[word]

    # print the resulting files
    for match in matching_files:
        print(files[match])


while True:
    concept(input("Enter Concept Idea: "))

当所有键与dict＃1完全匹配时如何从dict＃2打印特定值

1 个答案: