Identify groups of continuous numbers in a list

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误落风尘
误落风尘 2020-11-22 01:12

I\'d like to identify groups of continuous numbers in a list, so that:

myfunc([2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20])

Returns:

         


        
13条回答
  •  失恋的感觉
    2020-11-22 01:52

    EDIT 2: To answer the OP new requirement

    ranges = []
    for key, group in groupby(enumerate(data), lambda (index, item): index - item):
        group = map(itemgetter(1), group)
        if len(group) > 1:
            ranges.append(xrange(group[0], group[-1]))
        else:
            ranges.append(group[0])
    

    Output:

    [xrange(2, 5), xrange(12, 17), 20]
    

    You can replace xrange with range or any other custom class.


    Python docs have a very neat recipe for this:

    from operator import itemgetter
    from itertools import groupby
    data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
    for k, g in groupby(enumerate(data), lambda (i,x):i-x):
        print map(itemgetter(1), g)
    

    Output:

    [2, 3, 4, 5]
    [12, 13, 14, 15, 16, 17]
    

    If you want to get the exact same output, you can do this:

    ranges = []
    for k, g in groupby(enumerate(data), lambda (i,x):i-x):
        group = map(itemgetter(1), g)
        ranges.append((group[0], group[-1]))
    

    output:

    [(2, 5), (12, 17)]
    

    EDIT: The example is already explained in the documentation but maybe I should explain it more:

    The key to the solution is differencing with a range so that consecutive numbers all appear in same group.

    If the data was: [2, 3, 4, 5, 12, 13, 14, 15, 16, 17] Then groupby(enumerate(data), lambda (i,x):i-x) is equivalent of the following:

    groupby(
        [(0, 2), (1, 3), (2, 4), (3, 5), (4, 12),
        (5, 13), (6, 14), (7, 15), (8, 16), (9, 17)],
        lambda (i,x):i-x
    )
    

    The lambda function subtracts the element index from the element value. So when you apply the lambda on each item. You'll get the following keys for groupby:

    [-2, -2, -2, -2, -8, -8, -8, -8, -8, -8]
    

    groupby groups elements by equal key value, so the first 4 elements will be grouped together and so forth.

    I hope this makes it more readable.

    python 3 version may be helpful for beginners

    import the libraries required first

    from itertools import groupby
    from operator import itemgetter
    
    ranges =[]
    
    for k,g in groupby(enumerate(data),lambda x:x[0]-x[1]):
        group = (map(itemgetter(1),g))
        group = list(map(int,group))
        ranges.append((group[0],group[-1]))
    

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