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求两个值的最大值和最小值的最佳方法

python

Colin · 技术社区 · 14 年前

我有一个传递两个值的函数,然后在这些值的范围内迭代。这些值可以按任意顺序传递,所以我需要先找出哪个值最低。我的函数是这样写的:

def myFunc(x, y):
    if x > y:
        min_val, max_val = y, x
    else:
        min_val, max_val = x, y
    for i in range(min_val, max_val):
    ...

def myFunc(x, y):
   min_val, max_val = sorted([x, y])
   for i in range(min_val, max_val):
   ...

这有多糟?有没有更好的办法,那还是一条线?

6 回复 | 直到 14 年前

John La Rooy 14 年前

def myFunc(x, y):
    for i in range(*sorted((x, y))):
        ...

不过这要快些

def myFunc(x, y):
    for i in range(x,y) if x<y else range(y,x):
        ...

def f1(x, y):
    for i in range(min(x, y), max(x, y)):
        pass

def f2(x, y):
    for i in range(*sorted((x, y))):
        pass

def f3(x, y):
    for i in range(x, y) if x<y else range(y, x):
        pass

def f4(x, y):
    if x>y:
        x,y = y,x
    for i in range(x, y):
        pass

def f5(x, y):
    mn,mx = ((x, y), (y, x))[x>y]
    for i in range(x,y):
        pass

基准 (无论顺序如何,f3都是最快的)

$ python -m timeit -s"import minmax as mm" "mm.f1(1,2)"
1000000 loops, best of 3: 1.93 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f2(1,2)"
100000 loops, best of 3: 2.4 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f3(1,2)"
1000000 loops, best of 3: 1.16 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f4(1,2)"
100000 loops, best of 3: 1.2 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f5(1,2)"
1000000 loops, best of 3: 1.58 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f1(2,1)"
100000 loops, best of 3: 1.88 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f2(2,1)"
100000 loops, best of 3: 2.39 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f3(2,1)"
1000000 loops, best of 3: 1.18 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f4(2,1)"
1000000 loops, best of 3: 1.25 usec per loop
$ python -m timeit -s"import minmax as mm" "mm.f5(2,1)"
1000000 loops, best of 3: 1.44 usec per loop

miku 14 年前

min 和 max

def myFunc(x, y):
    min_val, max_val = min(x, y), max(x, y)

min-max 一个简单的版本 if . 由于函数调用开销, 最小值最大值 拿 2.5x 再简单不过了 ;参见 http://gist.github.com/571049

Don O'Donnell 14 年前

x 和 y lo hi ),我同意(为了速度和清晰度):

def myfunc(x, y):
    lo, hi = (x, y) if x < y else (y, x)

>>> timeit.repeat("myfunc(10, 5)", "from __main__ import myfunc")
[1.2527812156004074, 1.185214249195269, 1.1886092749118689]
>>> timeit.repeat("foo(10, 5)", "from __main__ import foo")
[1.0397177348022524, 0.9580022495574667, 0.9673979369035806]
>>> timeit.repeat("f3(10, 5)", "from __main__ import f3")
[2.47303065772212, 2.4192818561823515, 2.4132735135754046]

John Kugelman Michael Hodel 14 年前

sorted 一个。聪明但不太聪明。这里有一些其他的选择。

def myFunc(min, max):
    if min > max: min, max = max, min

def myFunc(x, y):
    min, max = min(x, y), max(x, y)

def myFunc(x, y):
    min, max = [f(x, y) for f in (min, max)]

AndiDog 14 年前

一些建议

def myfunc(minVal, maxVal):
    if minVal > maxVal: minVal, maxVal = maxVal, minVal

def myfunc2(a, b):
    minVal, maxVal = ((a, b), (b, a))[a > b] # :-P

在这种情况下,使用sorted,min/max内置函数或上面的第二个解决方案似乎有些过分。

记住这一点 range(min, max) min 到 max - 1 !

Ian 14 年前

最佳答案如下:

def foo(lo, hi):
    if hi < lo: lo,hi = hi,lo

它很清楚,很有意义,而且不会在一堆额外的胶水中模糊意义。它很短。在实践中,它几乎可以肯定与任何其他选择一样快,而且它依赖于最少的聪明。