有没有一种更简单的方法来分配值给一个变量如果条件 - Python？

Question

除了下面的蛮力方法，是否有一个更简单的方法来分配值给变量给定条件？

方法1：

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) 

，但如果你有很多的变量，以完全相同的方式被处理list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

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 

Answer 3

绝对考虑使用numpy.where这是最有效的方式做你想要处理现实与阵列和尺寸的任何尺寸：

#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 4

也许，更多的哈斯克尔样（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timeint（达尔文12.2.0，py 2.7.3）：

个

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：

• 元组是更快地迭代比列表？！？

• 即使max（值）比max（* values）更快？