python - 循环遍历列表，找到大于前一个最小值的最小值

Question

一个简单的问题。

我有一个列表列表，我希望找到每个列表中大于前一个最小值的最小值。

列表是单独排序的，对于第一个列表，我只是寻找最小值。

例如:

my_list = [[3, 5, 8, 14, 18, 28, 30],
           [4, 6, 7, 9, 13, 15, 17, 19, 22, 26],
           [2, 11, 21, 24, 32],
           [1, 10, 12, 16, 20, 23, 25, 27, 29, 31, 33, 34]]

预期输出为:

[3, 4, 11, 12]

Answer 1

它应该非常可读:

minimums = []
minimum = None
for sublist in my_list:
    try:
        minimum = min(x for x in sublist if minimum is None or x > minimum)
        minimums.append(minimum)
    except ValueError: # No numbers greater than minimum
        # Error handling