python - 有没有比使用 for 循环更干净的方法

Question

我想知道是否有任何其他方法可以循环和操作位于单独数组中的数据。

import numpy as np

a = np.arange(2)
b = np.arange(5)
c = np.arange(5)

l1 = []
for x in a:
    l2 = []
    for y in b: 
        l3 = []
        y = x + 1  
        for z in c:
            z = x + y
            t = (x,y,z)
            l3.append(t)
        l2.append(l3)
    l1.append(l2) 
print l1

Answer 1

此代码完全符合您的要求。

def method(lst, range1, range2):
    for i in lst:
        yield [[(i, i+1, 1+(i*2))]*range2]*range1

甚至可以变成生成器表达式:

def gen_method(lst, r1, r2):
    return ([[(i, i+1, 1+(i*2))]*r2]*r1 for i in lst)

如果喜欢，请自己测试。

---

我的测试:

a = range(2)
b = range(5)
c = range(5)

def your_method(a, b, c):
    l1 = []
    for x in a:
        l2 = []
        for y in b: 
            l3 = []
            y = x + 1  
            for z in c:
                z = x + y
                t = (x,y,z)
                l3.append(t)
            l2.append(l3)
        l1.append(l2)
    return l1

def my_method(lst, range1, range2):
    for i in lst:
        yield [[(i, i+1, 1+(i*2))]*range2]*range1

yours = your_method(a, b, c)
mine = list(my_method(a, len(b), len(c)))

print yours
print '==='
print mine
print '==='
print yours == mine

>>> 
[[[(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)]], [[(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)]]]
===
[[[(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)], [(0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1), (0, 1, 1)]], [[(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)], [(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)]]]
===
True