排序python 3.7+字典的最快方法

One-time setup with a dict of 1000 floats:>>> import random>>> random.seed(123)>>> d = {random.random(): random.random() for i in range(1000)}按键对数字进行排序(最好到最坏):Sorting numbers by key (best to worst):>>> %timeit {k: d[k] for k in sorted(d)}# 296 µs ± 2.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: d[k] for k in sorted(d.keys())}# 306 µs ± 9.25 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit dict(sorted(d.items(), key=itemgetter(0)))# 345 µs ± 4.15 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: v for k,v in sorted(d.items(), key=itemgetter(0))}# 359 µs ± 2.42 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit dict(sorted(d.items(), key=lambda kv: kv[0]))# 391 µs ± 8.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit dict(sorted(d.items()))# 409 µs ± 9.33 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: v for k,v in sorted(d.items())}# 420 µs ± 5.39 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: v for k,v in sorted(d.items(), key=lambda kv: kv[0])}# 432 µs ± 39.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)按值对数字进行排序(最好到最坏):Sorting numbers by value (best to worst):>>> %timeit {k: v for k,v in sorted(d.items(), key=itemgetter(1))}# 355 µs ± 2.24 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit dict(sorted(d.items(), key=itemgetter(1)))# 375 µs ± 31.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: v for k,v in sorted(d.items(), key=lambda kv: kv[1])}# 393 µs ± 1.89 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit dict(sorted(d.items(), key=lambda kv: kv[1]))# 402 µs ± 9.74 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: d[k] for k in sorted(d, key=d.get)}# 404 µs ± 3.55 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: d[k] for k in sorted(d, key=d.__getitem__)}# 404 µs ± 20.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)>>> %timeit {k: d[k] for k in sorted(d, key=lambda k: d[k])}# 480 µs ± 12 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)具有大量字符串的一次性设置:One-time setup with a large dict of strings:>>> import random>>> from pathlib import Path>>> from operator import itemgetter>>> random.seed(456)>>> words = Path('/usr/share/dict/words').read_text().splitlines()>>> random.shuffle(words)>>> keys = words.copy()>>> random.shuffle(words)>>> values = words.copy()>>> d = dict(zip(keys, values))>>> list(d.items())[:5][('ragman', 'polemoscope'), ('fenite', 'anaesthetically'), ('pycnidiophore', 'Colubridae'), ('propagate', 'premiss'), ('postponable', 'Eriglossa')]>>> len(d)235886通过键对字符串的字典进行排序:Sorting a dict of strings by key:>>> %timeit {k: d[k] for k in sorted(d)}# 387 ms ± 1.98 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: d[k] for k in sorted(d.keys())}# 387 ms ± 2.87 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit dict(sorted(d.items(), key=itemgetter(0)))# 461 ms ± 1.61 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit dict(sorted(d.items(), key=lambda kv: kv[0]))# 466 ms ± 2.62 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: v for k,v in sorted(d.items(), key=itemgetter(0))}# 488 ms ± 10.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: v for k,v in sorted(d.items(), key=lambda kv: kv[0])}# 536 ms ± 16.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit dict(sorted(d.items()))# 661 ms ± 9.09 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: v for k,v in sorted(d.items())}# 687 ms ± 5.38 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)按值对字符串的字典排序:Sorting a dict of strings by value:>>> %timeit {k: v for k,v in sorted(d.items(), key=itemgetter(1))}# 468 ms ± 5.74 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit dict(sorted(d.items(), key=itemgetter(1)))# 473 ms ± 2.52 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit dict(sorted(d.items(), key=lambda kv: kv[1]))# 492 ms ± 9.06 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: v for k,v in sorted(d.items(), key=lambda kv: kv[1])}# 496 ms ± 1.87 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: d[k] for k in sorted(d, key=d.__getitem__)}# 533 ms ± 5.33 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: d[k] for k in sorted(d, key=d.get)}# 544 ms ± 6.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)>>> %timeit {k: d[k] for k in sorted(d, key=lambda k: d[k])}# 566 ms ± 5.77 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 注意:真实世界的数据通常包含已排序序列的较长序列，Timsort算法可以利用这些序列.如果对字典进行排序是您的捷径，那么建议您在得出关于最佳方法的任何结论之前，使用自己的典型数据在自己的平台上进行基准测试.我在每个时间结果上都添加了一个注释字符(#)，以便IPython用户可以复制/粘贴整个代码块，以便在自己的平台上重新运行所有测试. Note: Real-world data often contains long runs of already-sorted sequences, which Timsort algorithm can exploit. If sorting a dict lies on your fast path, then it's recommended to benchmark on your own platform with your own typical data before drawing any conclusions about the best approach. I have prepended a comment character (#) on each timeit result so that IPython users can copy/paste the entire code block to re-run all the tests on their own platform. 这篇关于排序python 3.7+字典的最快方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！