Question

我有两个清单。例如：

a = [1,None,0,1,None,None]
b=[1,0,None,None,0,None]

a[i]和b[i]是相应的值。它们将是等价的（例如，其中i = 0），一个将包含一个值而另一个将是无（例如，i = 1）或两者都将是None（例如，i = 5）。我希望整理这些列表，以生成列表c：

c = [1,0,0,1,0,None]

目前我这样做：

def combinePredictions (a,b):
    #Naive method to combine lists. Assumes that the lists either have the same value or no value. Not for production use. Only use for testing.
    i = 0
    c = []
    for each in range(len(a)):
        if a[i] == b[i]:
            c.append(a[i])
        elif a[i] == None:
            c.append(b[i])
        elif b[i] == None:
            c.append(a[i])
        i = i +1
    return c

如何更有效地完成这项工作？

Answer 1

以下是我将如何迭代重构此代码您不需要i和each的单独值。如果您在一个范围内进行迭代，那么手动递增索引并没有多大意义。

def combinePredictions (a,b):
    c = []
    for i in range(len(a)):
        if a[i] == b[i]:
            c.append(a[i])
        elif a[i] == None:
            c.append(b[i])
        elif b[i] == None:
            c.append(a[i])
    return c

你可以减少条件数，因为你唯一没有附加[i]的时候就是没有条件：

def combinePredictions (a,b):
    c = []
    for i in range(len(a)):
        if a[i] is None:
            c.append(b[i])
        else:
            c.append(a[i])
    return c

您可以使用zip并行迭代两个列表，而无需参考其索引：

def combinePredictions (a,b):
    c = []
    for x,y, in zip(a,b):
        if x is None:
            c.append(y)
        else:
            c.append(x)
    return c

这可能已经足够了，但是如果我们对减少行数（可以说是以可读性为代价）感兴趣，那就让我们走得更远。
您可以将该条件设置为表达式并将其内联到append调用中：

def combinePredictions (a,b):
    c = []
    for x,y, in zip(a,b):
        c.append(y if x is None else x)
    return c

您可以将该循环变为列表理解：

def combinePredictions (a,b):
    return [y if x is None else x for x,y in zip(a,b)]

Answer 2

尝试列表理解：

a = [1,None,0,1,None,None]
b = [1,0,None,None,0,None]

print([1 if x == 1 or y == 1 else None if x == None and y == None else 0 for x, y in zip(a, b)])
# [1, 0, 0, 1, 0, None]

Answer 3

a = [1, None, 0, 1, None, None]
b = [1, 0, None, None, 0, None]


zipped = [x if y is None else y for x, y in zip(a, b)]

for i in zipped:
    print(i)

Answer 4

我不知道这是否更有效，但你绝对可以让它变得更加抒情和可读：

y_true = [1, None, 0, None]
y_pred = [1, 0, 0, None]
collated = []

for true, predicted in zip(y_true, y_pred):
    if predicted is None:
        collated.append(true)
    else:
        collated.append(predicted)

print(collated)  # [1, 0, 0, None]

根据值整理列表

4 个答案: