Better way to swap elements in a list?

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臣服心动 2020-12-23 13:32

I have a bunch of lists that look like this one:

l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

I want to swap elements as follows:

fi         


        
14条回答
  •  無奈伤痛
    2020-12-23 14:16

    No need for complicated logic, simply rearrange the list with slicing and step:

    In [1]: l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    
    In [2]: l[::2], l[1::2] = l[1::2], l[::2]
    
    In [3]: l
    Out[3]: [2, 1, 4, 3, 6, 5, 8, 7, 10, 9]
    

     TLDR;

    Edited with explanation

    I believe most viewers are already familiar with list slicing and multiple assignment. In case you don't I will try my best to explain what's going on (hope I do not make it worse).

    To understand list slicing, here already has an excellent answer and explanation of list slice notation. Simply put:

    a[start:end] # items start through end-1
    a[start:]    # items start through the rest of the array
    a[:end]      # items from the beginning through end-1
    a[:]         # a copy of the whole array
    
    There is also the step value, which can be used with any of the above:
    
    a[start:end:step] # start through not past end, by step
    

    Let's look at OP's requirements:

     [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]  # list l
      ^  ^  ^  ^  ^  ^  ^  ^  ^  ^
      0  1  2  3  4  5  6  7  8  9    # respective index of the elements
    l[0]  l[2]  l[4]  l[6]  l[8]      # first tier : start=0, step=2
       l[1]  l[3]  l[5]  l[7]  l[9]   # second tier: start=1, step=2
    -----------------------------------------------------------------------
    l[1]  l[3]  l[5]  l[7]  l[9]
       l[0]  l[2]  l[4]  l[6]  l[8]   # desired output
    

    First tier will be: l[::2] = [1, 3, 5, 7, 9] Second tier will be: l[1::2] = [2, 4, 6, 8, 10]

    As we want to re-assign first = second & second = first, we can use multiple assignment, and update the original list in place:

    first , second  = second , first
    

    that is:

    l[::2], l[1::2] = l[1::2], l[::2]
    

    As a side note, to get a new list but not altering original l, we can assign a new list from l, and perform above, that is:

    n = l[:]  # assign n as a copy of l (without [:], n still points to l)
    n[::2], n[1::2] = n[1::2], n[::2]
    

    Hopefully I do not confuse any of you with this added explanation. If it does, please help update mine and make it better :-)

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