python - 这个嵌套列表理解有什么问题？

Question

>>> c = 'A/B,C/D,E/F'
>>> [a for b in c.split(',') for (a,_) in b.split('/')]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 1, in <listcomp>
ValueError: need more than 1 value to unpack

预期结果是['A', 'C', 'E']。

这是我期望的方式，但显然它在 Python 中是从头到尾:

>>> [a for (a, _) in b.split('/') for b in c.split(',')]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'b' is not defined

Answer 1

您失败的原因是 b.split('/') 没有产生二元组。 双重列表推导意味着您希望将笛卡尔积视为平面流而不是矩阵。即:

>>> [x+'/'+y for y in 'ab' for x in '012']
['0/a', '1/a', '2/a', '0/b', '1/b', '2/b']
    # desire output 0,1,2
    # not output 0,1,2,0,1,2

你不是在寻找 6 个答案，你在寻找 3 个。你想要的是:

>>> [frac.split('/')[0] for frac in c.split(',')]
['A', 'C', 'E']

即使您使用嵌套列表理解，您也会得到笛卡尔积 (3x2=6) 并意识到您有重复信息(您不需要 x2):

>>> [[x+'/'+y for y in 'ab'] for x in '012']
[['0/a', '0/b'], ['1/a', '1/b'], ['2/a', '2/b']]
    # desire output 0,1,2
    # not [0,0],[1,1],[2,2]

---

以下是等效的做事方式。不过，在此比较中，我略微掩盖了生成器和列表之间的主要区别。

列表形式的笛卡尔积:

((a,b,c) for a in A for b in B for c in C)
            #SAME AS#
((a,b,c) for (a,b,c) in itertools.product(A,B,C))
            #SAME AS#
for a in A:
    for b in B:
        for c in C:
            yield (a,b,c)

矩阵形式的笛卡尔积:

[[[(a,b,c) for a in A] for b in B] for c in C]
            #SAME AS#
def fC(c):
    def fB(b,c):
        def fA(a,b,c):
            return (a,b,c)   
        yield [f(a,b,c) for a in A]
    yield [fB(b,c) for b in B]
[fC(c) for c in C]
            #SAME AS#
Cs = []
for c in C:
    Bs = []
    for b in B:
        As = []
        for a in A:
            As += [a]
        Bs += [As]
    Cs += [Bs]
return Cs

函数重复应用到列表

({'z':z} for x in ({'y':y} for y in ({'x':x} for x in 'abc')))
              #SAME AS#
for x in 'abc':
    x2 = {'x':x}
    y2 = {'y':x2}
    z2 = {'z':y2}
    yield z2
              #SAME AS#
def f(x):
    return {'z':{'y':{'x':x}}}
return [f(x) for x in 'abc']     # or map(f,'abc')