在Python中对齐两个列表

Question

我有两个列表，一个是表示真值的，另一个是包含实验结果及其相关分数的。

truth = [6, 8, 7, 10]
experiment = [(6, 5), (4, 3), (2, 4), (11, 6), (7, 4)]

我想把我的实验结果和真值对齐，确保对齐的真值最大。

        [6, 8, 7, 10]
        [6, missing, 7, missing]

现在我想给那些没有对齐的子序列中的缺失值分配一个值。

在这里，我们从(4, 3), (2, 4), (11, 6)中选择了11，因为它的分数最高。最后一个缺失的值被赋值为0，因为在(7, 4)之后没有其他元素。

truth = [6, 8, 7, 10]
exp  =  [6, 11, 7, 0]   # Zero because 7 is the last element of experiment subsequence.

我在查找difflib库，但没太明白。 我该怎么做呢？

Answer 1

假设你永远不会有两个或更多缺失值挨在一起，你可以使用：

truth = [6, 8, 7, 10]
experiment = [(6, 5), (4, 3), (2, 4), (11, 6), (7, 4)]

exp_list = [x[0] for x in experiment] #[6, 4, 2, 11, 7]

# Part one.
exp = [t if t in exp_list else None for t in truth] #[6, None, 7, None]

# Part two.
for i, t in enumerate(exp):
    if t == None:
        # Get index of previous and next truth values for missing values.
        i_prev = exp_list.index(exp[i-1]) if i else 0
        i_next = exp_list.index(exp[i+1]) if i < len(exp)-1 else len(exp_list)
        # Pick the largest value between them or 0 if no values are there.
        values_between = exp_list[slice(i_prev + 1, i_next)]
        exp[i] = max(values_between) if values_between else 0

print exp #[6, 11, 7, 0]

Answer 2

对于第一部分，你可以这样做：

truth = [6, 8, 7, 10]
experiment = [(6, 5), (4, 3), (2, 4), (11, 6), (7, 4)]

def prepare(exp):
    keys = [i for i, j in exp]
    def select(v):
        return v if v in keys else None
    return select

select = prepare(experiment)

exp = [select(i) for i in truth]

# exp will be [6, None, 7, None]

Answer 3

第一部分可以通过一个简单的列表推导式来完成：

truth = [6, 8, 7, 10]
experiment = [(6, 5), (4, 3), (2, 4), (11, 6), (7, 4)]
res1 = [x if x in dict(experiment) else None for x in truth]

...或者如果你更喜欢用lambda函数的话：

truth = [6, 8, 7, 10]
experiment = [(6, 5), (4, 3), (2, 4), (11, 6), (7, 4)]
res1 = map(lambda x: x if x in dict(experiment) else None, truth)