除了下面的蛮力方法之外，有没有更简单的方法来给给定 if 条件的变量分配值？
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
【问题讨论】：
标签： python if-statement variable-assignment
也许：
a, b, c, d = max(a, x), max(b, x), max(c, x), max(d, x)
但如果您有很多变量以完全相同的方式处理，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]
【讨论】：
绝对考虑使用numpy.where，这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)
【讨论】：
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)))
一些基本的timeits（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？！？
starmap 胜过地图，尽管 max(values) 比 max(*values) 快 ?!?
【讨论】：
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
【讨论】：