一个基于布尔值选择两个值的简短表达式？

huangapple
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英文:

A short expression to pick from two values based on bool?

问题

  1. 如果一个列表中有两个值：

choices = [1,2]

  1. 而且想要根据一个名为 `ascending` 的 `bool` 参数将这些值分配给两个变量，可以这样做：

if ascending:
a, b = choices[1], choices[0]
else:
a, b = choices[0], choices[1]

  1. 可以使用以下一行代码来表达：

a, b = choices[ascending], choices[not ascending]

  1. 或者

a, b = choices[::(-1)**ascending]

  2. 我更喜欢最后一种方法，但它可能不够清晰。还有其他更简洁但更可读的方式来实现相同的效果吗？
英文:

If one has two values in a list:

  1. choices = [1,2]

and one wants to assign these values to two variables based on a bool parameter (let's call it ascending) like this:

  1. if ascending:
  2.   a,b = choices[1], choices[0]
  3. else:
  4.   a,b = choices[0], choices[1]

one can express this in one line like this:

  1. a,b  = choices[ascending], choices[not ascending]

or

  1. a,b = choices[::(-1)**ascending]

I like the last one better, but it's probably not extremely transparent. What are other concise but more readable ways to do the same?

答案1

得分: 2

如果原始值始终按升序排列，也许这更能表达您的意图：

  1. a, b = sorted(choices, reverse=ascending)

否则，更简单通常更好（参见Kelly的回答），但我可以想出一种方法使它更加晦涩：

例如：

  1. a, b = (iter, reversed)[ascending](choices)

或者，如果其中一种选项比另一种更有可能，您可以基于那个选项进行赋值，如果是另一种情况，则交换变量：

  1. a, b = choices
  2. if ascending: a, b = b, a
英文:

If the original values are always in ascending order, perhaps this would be more expressive of your intent:

  1. a,b = sorted(choices, reverse=ascending)

Otherwise simpler is generally better (see Kelly's answer) but I could think of ways to make it even more obscure:

For example:

  1. a,b = (iter,reversed)[ascending](choices)

Alternatively, if one option is more likely than the other you could do the assignment based on that one and swap the variables if it's the other case:

  1. a,b = choices
  2. if ascending: a,b = b,a

答案2

得分: 1

以下是代码部分的中文翻译：

  1. from timeit import timeit
  2. from statistics import mean, stdev
  4. codes = '''
  6. if ascending:
  7.   a, b = choices[1], choices[0]
  8. else:
  9.   a, b = choices[0], choices[1]
  11. a, b = choices[ascending], choices[not ascending]
  13. a, b = choices[::(-1)**ascending]
  15. b, a = choices[::ascending or -1]
  17. a, b = choices[::-1 if ascending else 1]
  19. if ascending:
  20.   b, a = choices
  21. else:
  22.   a, b = choices
  24. a, b = choices[::-ascending or 1]
  26. a, b = choices[::-1] if ascending else choices
  28. a, b = reversed(choices) if ascending else choices
  30. '''.strip().split('\n\n')
  32. choices = ['foo', 'bar']
  33. for ascending in False, True:
  34.   print(f'{ascending=}')
  35.   for code in codes:
  36.     a = b = None
  37.     exec(code)
  38.     print(a, b)
  39.   print()
  41. times = {c: ([], []) for c in codes}
  42. def stats(c):
  43.   false, true = (
  44.     (mean(ts), stdev(ts))
  45.     for ts in times[c]
  46.     for ts in [[t*1e9 for t in sorted(ts)[:5]]]
  47.   )
  48.   return false[0] + true[0], false, true
  50. for _ in range(25):
  51.   for ascending in False, True:
  52.     setup = f"choices = ['foo', 'bar']; {ascending = }"
  53.     for c in codes:
  54.       number = 10**5
  55.       t = timeit(c, setup, number=number) / number
  56.       times[c][ascending].append(t)
  58. print('asc = False    asc = True')
  59. for c in sorted(codes, key=stats):
  60.   _, false, true = stats(c)
  61.   # print('{:5.1f} ± {:4.1f}   {:5.1f} ± {:4.1f}  '.format(*false, *true), ' '.join(c.split()))
  62.   print('{:5.1f} ± {:4.1f}   {:5.1f} ± {:4.1f}  '.format(*false, *true), c.strip().replace('\n', '\n' + ' '*30))

注意：这是代码的中文翻译，不包括代码的执行结果。

英文:

Some ways...

  1. b,a = choices[::ascending or -1]
  2. a,b = choices[::-ascending or 1]
  3. a,b = choices[::-1] if ascending else choices
  4. a,b = choices.pop(ascending), *choices

The last one modifies the list.

Or with multiple lines but kinda clean and I suspect fastest:

  1. if ascending:
  2. b,a = choices
  3. else:
  4. a,b = choices

Benchmarks (times in nanoseconds):

  1.  asc = False    asc = True
  2. ------------   ------------
  3. 30.2 ±  0.3    30.2 ±  0.3   if ascending:
  4. b,a = choices
  5. else:
  6. a,b = choices
  7. 60.2 ±  1.3    60.8 ±  2.1   a,b = choices[ascending], choices[not ascending]
  8. 62.8 ±  0.6    63.5 ±  0.9   if ascending:
  9. a,b = choices[1], choices[0]
  10. else:
  11. a,b = choices[0], choices[1]
  12. 30.3 ±  0.3   125.0 ±  3.7   a,b = choices[::-1] if ascending else choices
  13. 30.1 ±  0.1   178.2 ±  4.5   a,b = reversed(choices) if ascending else choices
  14. 124.5 ±  3.5   126.1 ±  5.8   b,a = choices[::ascending or -1]
  15. 133.6 ±  7.0   123.8 ±  2.6   a,b = choices[::-1 if ascending else 1]
  16. 143.4 ±  6.4   139.9 ±  6.1   a,b = choices[::-ascending or 1]
  17. 169.4 ± 18.5   441.2 ± 20.4   a,b = choices[::(-1)**ascending]

Code (Attempt This Online!):

  1. from timeit import timeit
  2. from statistics import mean, stdev
  4. codes = '''
  6. if ascending:
  7.   a,b = choices[1], choices[0]
  8. else:
  9.   a,b = choices[0], choices[1]
  11. a,b = choices[ascending], choices[not ascending]
  13. a,b = choices[::(-1)**ascending]
  15. b,a = choices[::ascending or -1]
  17. a,b = choices[::-1 if ascending else 1]
  19. if ascending:
  20.   b,a = choices
  21. else:
  22.   a,b = choices
  24. a,b = choices[::-ascending or 1]
  26. a,b = choices[::-1] if ascending else choices
  28. a,b = reversed(choices) if ascending else choices
  30. '''.strip().split('\n\n')
  32. choices = ['foo', 'bar']
  33. for ascending in False, True:
  34.   print(f'{ascending=}')
  35.   for code in codes:
  36.     a = b = None
  37.     exec(code)
  38.     print(a, b)
  39.   print()
  41. times = {c: ([], []) for c in codes}
  42. def stats(c):
  43.   false, true = (
  44.     (mean(ts), stdev(ts))
  45.     for ts in times[c]
  46.     for ts in [[t*1e9 for t in sorted(ts)[:5]]]
  47.   )
  48.   return false[0] + true[0], false, true
  50. for _ in range(25):
  51.   for ascending in False, True:
  52.     setup = f"choices = ['foo', 'bar']; {ascending = }"
  53.     for c in codes:
  54.       number = 10**5
  55.       t = timeit(c, setup, number=number) / number
  56.       times[c][ascending].append(t)
  58. print('asc = False    asc = True')
  59. for c in sorted(codes, key=stats):
  60.   _, false, true = stats(c)
  61.   # print('{:5.1f} ± {:4.1f}   {:5.1f} ± {:4.1f}  '.format(*false, *true), ' '.join(c.split()))
  62.   print('{:5.1f} ± {:4.1f}   {:5.1f} ± {:4.1f}  '.format(*false, *true), c.strip().replace('\n', '\n' + ' '*30))

huangapple
  • 本文由 发表于 2023年4月7日 00:21:31
  • 转载请务必保留本文链接：https://go.coder-hub.com/75951678.html
  • python
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