python - 如何根据元素的某些方面将 Python 列表分成两个列表

Question

我有一个这样的列表:

[[8, "Plot", "Sunday"], [1, "unPlot", "Monday"], [12, "Plot", "Monday"], [10, "Plot", "Tuesday"], [4, "unPlot", "Tuesday"], [14, "Plot", "Wednesday"], [6, "unPlot", "Wednesday"], [1, "unPlot", "Thursday"], [19, "Plot", "Thursday"], [28, "Plot", "Friday"], [10, "unPlot", "Friday"], [3, "unPlot", "Saturday"]]

我想根据 Plot 和 unPlot 值将它分成两个列表，结果:

list1=[[8, "Plot", "Sunday"], [12, "Plot", "Monday"], ...]
list2=[[1, "unPlot", "Monday"], [4, "unPlot", "Tuesday"], ...]

Answer 1

尝试基本的列表理解:

>>> [ x for x in l if x[1] == "Plot" ]
[[8, 'Plot', 'Sunday'], [12, 'Plot', 'Monday'], [10, 'Plot', 'Tuesday'], [14, 'Plot', 'Wednesday'], [19, 'Plot', 'Thursday'], [28, 'Plot', 'Friday']]
>>> [ x for x in l if x[1] == "unPlot" ]
[[1, 'unPlot', 'Monday'], [4, 'unPlot', 'Tuesday'], [6, 'unPlot', 'Wednesday'], [1, 'unPlot', 'Thursday'], [10, 'unPlot', 'Friday'], [3, 'unPlot', 'Saturday']]

如果您喜欢函数式编程，也可以使用 filter:

>>> filter(lambda x: x[1] == "Plot", l)
[[8, 'Plot', 'Sunday'], [12, 'Plot', 'Monday'], [10, 'Plot', 'Tuesday'], [14, 'Plot', 'Wednesday'], [19, 'Plot', 'Thursday'], [28, 'Plot', 'Friday']]
>>> filter(lambda x: x[1] == "unPlot", l)
[[1, 'unPlot', 'Monday'], [4, 'unPlot', 'Tuesday'], [6, 'unPlot', 'Wednesday'], [1, 'unPlot', 'Thursday'], [10, 'unPlot', 'Friday'], [3, 'unPlot', 'Saturday']]

我个人觉得列表理解更清晰。这当然是最“pythonic”的方式。