python - 有没有一种更简单的方法来为给定 if 条件的变量分配值 - Python？

Question

除了以下蛮力方法之外，是否有更简单的方法可以在给定 if 条件的情况下将值分配给变量？

方法一：

a, b, c, d = 0.03,0.4,0.055,0.7
x = 0.2

if a < x:
  a = x
if b < x:
  b = x
if c < x:
  c = x
if d < x:
  d = x

Answer 1

也许：

a, b, c, d = max(a, x), max(b, x), max(c, x), max(d, x)

但是如果你有很多变量以完全相同的方式处理，alist可能会更好。

values = [0.03,0.4,0.055,0.7]
x = 0.2

values = [max(v, x) for v in values]

Answer 2

绝对考虑使用numpy.where哪个是最有效的方法来处理任何大小的数组和维度：

#your example:
a,b,c,d = 0.03,0.4,0.055,0.7
x = 0.2

#solution
values = numpy.asarray([a, b, c, d])
a,b,c,d = numpy.where(values<x, x, values)

#efficiency becomes clear when
values = numpy.random.rand(1000,100,10)     #any size and number of dimensions
values = numpy.where(values<x, x, values)   #just works fine and efficient

#further developments would be possible, e.g., multiple conditions
values = numpy.where((values>=0.3)&(values<0.7), 0.5, values)

Answer 3

也许，更像 Haskell（zipWith）

from itertools import izip, starmap, repeat

a, b, c, d = starmap(max, izip(repeat(0.2), (0.03, 0.4, 0.055, 0.7)))

一些基本时间（darwin 12.2.0，py 2.7.3）：

In [0]: %timeit a,b,c,d = starmap(max, izip(repeat(0.2), (0.03, 0.4, 0.055, 0.7)))
1000000 loops, best of 3: 1.87 us per loop

In [1]: %timeit a,b,c,d = map(max, izip(repeat(0.2), (0.03, 0.4, 0.055, 0.7)))
100000 loops, best of 3: 3.99 us per loop

In [2]: %timeit a,b,c,d = [max(0.2, v) for v in [0.03,0.4,0.055,0.7]]
100000 loops, best of 3: 1.95 us per loop

In [3]: %timeit a,b,c,d = [max(0.2, v) for v in (0.03,0.4,0.055,0.7)]
1000000 loops, best of 3: 1.62 us per loop

结论：

• 元组的迭代速度比列表快？！？

• 星图胜过地图，即使 max(values) 比 max(*values) 快？！？

Answer 4

尝试：

a = x if a < x else a

b = x if b < x else b

c = x if c < x else c

d = x if d < x else d