python3 setdefault的

发布时间:2019-10-16 17:29:36编辑:auto阅读(526)

    当字典 d[k]找不到正确的键时,Python会抛出异常,有没有一种优雅的方法来避免这种情况呢?答案是肯定的.

    index0.py 从索引中获取单词出现的频率信息,并写入列表 --没有使用dict.setdefault

    #!/usr/bin/env python
    # coding=utf-8
    import sys, re
    
    WORD_RE = re.compile(r'\w+')
    
    index = {}
    with open(sys.argv[1], encoding='utf-8') as fp:
        for line_no, line in enumerate(fp, 1):
            for match in WORD_RE.finditer(line):
                word = match.group()
                column_no = match.start()+1
                location = (line_no, column_no)
                occurrences = index.get(word, [])
                occurrences.append(location)
                index[word] = occurrences
    
    for word in sorted(index, key=str.upper):
        print(word, index[word])

    zen.txt

    The Zen of Python, by Tim Peters
    
    Beautiful is better than ugly.
    Explicit is better than implicit.
    Simple is better than complex.
    Complex is better than complicated.
    Flat is better than nested.
    Sparse is better than dense.
    Readability counts.
    Special cases aren't special enough to break the rules.
    Although practicality beats purity.
    Errors should never pass silently.
    Unless explicitly silenced.
    In the face of ambiguity, refuse the temptation to guess.
    There should be one-- and preferably only one --obvious way to do it.
    Although that way may not be obvious at first unless you're Dutch.
    Now is better than never.
    Although never is often better than *right* now.
    If the implementation is hard to explain, it's a bad idea.
    If the implementation is easy to explain, it may be a good idea.
    Namespaces are one honking great idea -- let's do more of those!

    执行 python3 index0.py zen.txt


    a [(19, 48), (20, 53)]
    Although [(11, 1), (16, 1), (18, 1)]
    ambiguity [(14, 16)]
    and [(15, 23)]
    are [(21, 12)]
    aren [(10, 15)]
    at [(16, 38)]
    bad [(19, 50)]
    be [(15, 14), (16, 27), (20, 50)]
    beats [(11, 23)]
    Beautiful [(3, 1)]
    better [(3, 14), (4, 13), (5, 11), (6, 12), (7, 9), (8, 11), (17, 8), (18, 25)]
    break [(10, 40)]
    by [(1, 20)]
    cases [(10, 9)]
    ...

    index.py 使用了dict.setdefault 只用了一行就解决了获取和更新单词的出现情况列表

    #!/usr/bin/env python
    # coding=utf-8
    import sys, re
    
    WORD_RE = re.compile(r'\w+')
    
    index = {}
    with open(sys.argv[1], encoding='utf-8') as fp:
        for line_no, line in enumerate(fp, 1):
            for match in WORD_RE.finditer(line):
                word = match.group()
                column_no = match.start()+1
                location = (line_no, column_no)
                index.setdefault(word, []).append(location)
    
    for word in sorted(index, key=str.upper):
        print(word, index[word])

    也就是说:

    my_dict.setdefault(key, []).append(new_value)

    等价于

    if key not in my_dict:
        my_dict[key] = []
    my_dict[key].append(new_value)

    二者效果相同,只是setdefault只需一次就完成整个操作,而后者需要进行两次查询

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