Question

我有两个数据集，found和allowed。我想检查一下found中的任何观察是否都没有出现在allowed中，这样我就可以确保将它们标记为不被允许。

found = [["A", "B", "C"], [10, 20, 30], ["X", "Y", "Z"]]
allowed = [["A", "B", "C", "D"], [20, 30, 40], ["W", "X", "Y"]]

如果我运行以下内容，我会得到正确答案，即[(1, 10), (2, 'Z')]

new_values = []
for x in range(0, len(allowed)):
    for val in found[x]:
        if val not in allowed[x]:
            new_values.append((x, val))

但是，如果我运行以下操作，这需要更少的行，我认为更容易，它会返回new_values = [(2, 'Z')]

for x in range(0, len(allowed)):
    new_values = [(x, val) for val in found[x] if val not in allowed[x]]

(1, 10)发生了什么事？如果我运行10 in allowed[1]，我会收到False，这让我认为它应该出现在new_values中！

Answer 1

您正在for循环中重新分配new_values：

for x in range(0, len(allowed)):
    new_values = [(x, val) for val in found[x] if val not in allowed[x]]

试试这个：

new_values = []
for x in range(0, len(allowed)):
    new_values.extend([(x, val) for val in found[x] if val not in allowed[x]])

或者，使用列表推导更简短：

new_values = [(x, val) for x in range(0, len(allowed)) for val in found[x] if val not in allowed[x]]

Answer 2

new_values = [(x, val) for x, allowable in enumerate(allowed)
                        for val in found[x] if val not in allowable]

如果你还需要值的索引，通常认为使用enumerate迭代序列会更加Pythonic。

Answer 3

在嵌套式解析中插入enumerate可以缩短它：

new_values = [(i, x) for i, f in enumerate(found) for x in f if x not in allowed[i]]