Python：使用dict理解生成器计算列表中的出现次数

Python：使用dict理解/生成器计算列表中的出现次数

您无法使用dict-
comprehension有效地（至少在内存方面）做到这一点，因为这样一来，您就必须在另一本词典中跟踪当前计数，即更多的内存消耗。这是使用dict-
comprehension（完全不推荐使用:-）的方法：

>>> words = list('asdsadDASDFASCSAASAS')>>> dct = {}>>> {w: 1 if w not in dct and not dct.update({w: 1})       else dct[w] + 1       if not dct.update({w: dct[w] + 1}) else 1 for w in words}>>> dct{'a': 2, 'A': 5, 's': 2, 'd': 2, 'F': 1, 'C': 1, 'S': 5, 'D': 2}

另一种方法是先对单词列表进行排序，然后使用对其进行分组

itertools.groupby

，然后计算每个组的长度。如果需要，可以将dict-
comprehension转换为生成器，但是是的，这将需要首先读取内存中的所有单词：

from itertools import groupbywords.sort()dct = {k: sum(1 for _ in g) for k, g in groupby(words)}

请注意，其中 最快的一项 是

collections.defaultdict

：

d = defaultdict(int)for w in words: d[w] += 1

时序比较：

>>> from string import ascii_letters, digits>>> %timeit words = list(ascii_letters+digits)*10**4; words.sort(); {k: sum(1 for _ in g) for k, g in groupby(words)}10 loops, best of 3: 131 ms per loop>>> %timeit words = list(ascii_letters+digits)*10**4; Counter(words)10 loops, best of 3: 169 ms per loop>>> %timeit words = list(ascii_letters+digits)*10**4; dct = {}; {w: 1 if w not in dct and not dct.update({w: 1}) else dct[w] + 1 if not dct.update({w: dct[w] + 1}) else 1 for w in words}1 loops, best of 3: 315 ms per loop>>> %%timeit... words = list(ascii_letters+digits)*10**4... d = defaultdict(int)... for w in words: d[w] += 1... 10 loops, best of 3: 57.1 ms per loop>>> %%timeitwords = list(ascii_letters+digits)*10**4d = {}for w in words: d[w] = d.get(w, 0) + 1... 10 loops, best of 3: 108 ms per loop#Increase input size>>> %timeit words = list(ascii_letters+digits)*10**5; words.sort(); {k: sum(1 for _ in g) for k, g in groupby(words)}1 loops, best of 3: 1.44 s per loop>>> %timeit words = list(ascii_letters+digits)*10**5; Counter(words)1 loops, best of 3: 1.7 s per loop>>> %timeit words = list(ascii_letters+digits)*10**5; dct = {}; {w: 1 if w not in dct and not dct.update({w: 1}) else dct[w] + 1 if not dct.update({w: dct[w] + 1}) else 1 for w in words}1 loops, best of 3: 3.19 s per loop>>> %%timeitwords = list(ascii_letters+digits)*10**5d = defaultdict(int)for w in words: d[w] += 1... 1 loops, best of 3: 571 ms per loop>>> %%timeitwords = list(ascii_letters+digits)*10**5d = {}for w in words: d[w] = d.get(w, 0) + 1... 1 loops, best of 3: 1.1 s per loop