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Editorial Team
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Asked: 2026-05-28T20:48:20+00:00

I have a list like this: [[8, Plot, Sunday], [1, unPlot, Monday], [12, Plot,

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I have a list like this:

[[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"]]

I want to separate it into two lists according the Plot and unPlot values, resulting:

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

  1. Editorial Team

    Try with basic list comprehension:

    >>> [ 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']]

    Or with filter if you fancy functional programming:

    >>> 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']]

    I personally find list comprehensions much clearer. It’s certainly the most “pythonic” way.

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