我想引用两个列表并通过比较生成一个输出列表。
occupied = [8,9,10]
broken = [1,2,3]
output = ['occupied' if x in occupied else x in broken for x in range(1,11)]
desired_output = ['broken', 'broken', 'broken', 'broken', False, False, False, 'occupied', 'occupied', 'occupied']
一口气就能实现以上目标吗？
目前我正在做两次迭代
['broken' if x2==True else x2 for x2 in ['occupied' if x in occupied else x in broken for x in range(1,11)] ]
我正在寻找这样的东西
['occupied' if x in occupied else 'broken' if x in broken for x in range(1,11)]
但这是不正确的语法
您几乎已经拥有了，只需else在解决方案中添加一个子句（第二个if表达式）：
output = ["broken" if x in broken else 'occupied' if x in occupied else False for x in range(1,11)]
# Out[5]: ['broken', 'broken', 'broken', False, False, False, False, 'occupied', 'occupied', 'occupied']
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我来说两句