我有两个数据集，found和allowed。我想检查中的任何观察值found是否未出现在中allowed，以便确保将其标记为不允许。
found = [["A", "B", "C"], [10, 20, 30], ["X", "Y", "Z"]]
allowed = [["A", "B", "C", "D"], [20, 30, 40], ["W", "X", "Y"]]
如果运行以下命令，则会得到正确的答案，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，如果我运行以下命令，它需要更少的行，并且我认为更容易，它将返回 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]]
发生了什么事(1, 10)？如果我跑步，10 in allowed[1]我会False认为它应该出现在其中new_values！
您正在new_valuesfor循环中重新分配：
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，甚至更短一些，使用列表推导：
new_values = [(x, val) for x in range(0, len(allowed)) for val in found[x] if val not in allowed[x]]
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