值的 Python 字典索引答案

【问题标题】：Python Dictionary index of a value值的 Python 字典索引
【发布时间】：2013-04-23 02:14:43
【问题描述】：

我目前正在尝试从字典中获取值的索引，但不确定要尝试什么。

我的字典是：

    midi2notes = {                                  
        'c':("0","12","24","36","48","60","72","84","96","108","120"),
        'des':("1","13","25","37","49","61","73","85","97","109","121"),
        'd':("2","14","26","38","50","62","74","86","98","110","122"),
        'ees':("3","15","27","39","51","63","75","87","99","111","123"),
        'e':("4","16","28","40","52","64","76","88","100","112","124"),
        'f':("5","17","29","41","53","65","77","89","101","113","125"),
        'ges':("6","18","30","42","54","66","78","90","102","114","126"),
        'g':("7","19","31","43","55","67","79","91","103","115","127"),
        'aes':("8","20","32","44","56","68","80","92","104","116"),
        'a':("9","21","33","45","57","69","81","93","105","117"),
        'bes':("10","22","34","46","58","70","82","94","106","118"),
        'b':("11","23","35","47","59","71","83","95","107","119")
    }

示例：我想从 key-'c' 获取 value-"60" 的索引。应该是 5。最有效/最快的方法是什么。

【问题讨论】：

标签： python dictionary indexing key key-value

【解决方案1】：

我愿意：

midi2notes['c'].index('60')

如果未找到索引，这将引发 ValueError，但如果数字通常在元组中，则很难击败优化的 C 后端。

>>> midi2notes = {                                  
...         'c':("0","12","24","36","48","60","72","84","96","108","120"),
...         'des':("1","13","25","37","49","61","73","85","97","109","121"),
...         'd':("2","14","26","38","50","62","74","86","98","110","122"),
...         'ees':("3","15","27","39","51","63","75","87","99","111","123"),
...         'e':("4","16","28","40","52","64","76","88","100","112","124"),
...         'f':("5","17","29","41","53","65","77","89","101","113","125"),
...         'ges':("6","18","30","42","54","66","78","90","102","114","126"),
...         'g':("7","19","31","43","55","67","79","91","103","115","127"),
...         'aes':("8","20","32","44","56","68","80","92","104","116"),
...         'a':("9","21","33","45","57","69","81","93","105","117"),
...         'bes':("10","22","34","46","58","70","82","94","106","118"),
...         'b':("11","23","35","47","59","71","83","95","107","119")
...     }
>>> midi2notes['c'].index('60')
5
>>> midi2notes['c'].index('180')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: tuple.index(x): x not in tuple
>>>

【讨论】：

另外，请注意，如果元组中有另一个 "60" （在这种情况下没有），这将不起作用。但是，+1 :)。
@Haidro：在这种情况下，没有任何代码可以工作，因为当60s 可以有多个时，“60 的索引”不是一个明确定义的东西……
@Haidro -- 当然，如果它不是唯一的，那么整个前提都是错误的。也就是说，从算法上讲，如果您 bisect 元组（因为它们似乎都已排序），您可以做得更好。不过，对于这种大小的元组，我怀疑它会表现得更好。编辑二等分不起作用，因为它们没有排序，因为它们是字符串而不是整数。
如果这个操作被执行了无数次（技术术语），那么创建一个像mi = {(k, v): i for k,vv in midi2notes.items() for i,v in enumerate(vv)} 这样的索引字典可能是值得的。
@mgilson：我猜bisect 实际上会更慢。即使它只是 O(log N) 而不是 O(N)，但对于 OP 的输入数据，您正在谈论 11 个超级平凡的 C 循环与 4 个不那么平凡的循环……

【解决方案2】：

如果您需要经常这样做，而不是每次都查找索引，只需创建一个 dict 将值映射回索引：

midi2notes_map = {key: {value: index for index, value in enumerate(tup)}
                  for key, tup in midi2notes.items()}

那么你可以这样做：

midi2notes_map['c']['60']

或者，或者：

midi2notes_map = {(key, value): index
                  for key, tup in midi2notes.items()
                  for index, value in enumerate(tup)}

midi2notes_map['c', 60]

每次进行查找时，这应该会提高 5 倍的效率（因为您只需要对键进行哈希处理，而不是线性搜索 10 个项目的元组）……但当然，构建地图需要一些时间，因此，如果您只进行几次查找，则可能不值得。

当然，在担心它之前，您应该确保此查找的性能确实相关。（我猜你能想到的最慢的实现，使用一个显式的嵌套循环，仍然足够快来播放一个 MIDI 序列......）如果是这样，使用timeit 和你的实际数据设置一个测试和使用模式来找出什么是真正最快的。

【讨论】：

@mgilson：真的，这取决于他的应用程序的其余部分是什么样的。如果他对midi2notes_map['c'] 有别的用处，第一个肯定更好；如果没有这种可以想象的用途，可能是第二种。（我认为hash(('c', 60)) 和一个查找会比两个更简单的哈希和查找更快……但是将整个哈希表放入缓存行也可能会有所不同……）
这是一种非常有用且多样化的方法，虽然它现在比我的应用程序需要的要多一些，但如果我将来扩展它（如我计划的那样），这将是巨大的有帮助。

【解决方案3】：

与enumerate():

>>> for x,y in enumerate(midi2notes['c']):
...     if y == "60":
...         print x
... 
5

【讨论】：

不错的一个。可以改成生成器。