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levon
V2EX  ›  问与答

技术问题,多个数组拼合问题请教

  •  
  •   levon · 2014-11-16 00:13:42 +08:00 · 2764 次点击
    这是一个创建于 3664 天前的主题,其中的信息可能已经有所发展或是发生改变。
    有若干的数组,比如:
    var a = ["A","B","C"];
    var b =["1”,"2","3","4","5"]
    var c =["a","b"]

    要得到 a.count() * b.count() * c.count() 个结果,结果是:
    "A1a","A1b","A2a","A2b","A3a","A3b".......

    数组个数不固定,每个数组的元素个数也是不固定,请问要如何循环?
    语言不限,希望能给实现方法或思路,Help
    15 条回复    2014-11-16 10:01:10 +08:00
    Daniel65536
        1
    Daniel65536  
       2014-11-16 00:27:26 +08:00   ❤️ 1
    python蛋疼写法:

    for x in a:
    for y in b:
    for z in c:
    print('{}{}{}'.format(x, y, z))

    强迫症表示你bc的等号后面没加空格,b的第一项双引号是中文双引号,不爽 +_+
    Daniel65536
        2
    Daniel65536  
       2014-11-16 00:28:09 +08:00
    空格被发帖机制自动消除了……自行脑补缩进吧……
    levon
        3
    levon  
    OP
       2014-11-16 00:30:42 +08:00
    谢谢回复,问题是我手动直接在框里输入的,有点输入错误,抱歉

    但感觉没解决问题,,数组不是固定就3个的,有可能1个,2个,3个,若干个
    levon
        4
    levon  
    OP
       2014-11-16 00:34:06 +08:00
    好吧,还有个集合是:
    var list =[a, b, c];
    a b c就是那三个数组,list是已知的
    arbipher
        5
    arbipher  
       2014-11-16 01:00:04 +08:00   ❤️ 1
    这就是笛卡尔积吧。

    Python写法
    from itertools import product
    result = product(a, b, c)

    result的每个值就是每个列表各取一个元素组成的元组,再拼成字符串就行了。
    Daniel65536
        6
    Daniel65536  
       2014-11-16 01:03:28 +08:00
    a = ["A","B","C"];
    b = ["1","2","3","4","5"]
    c = ["a","b"]
    x = (a, b, c)

    def product(pools):
    result = [[]]
    for pool in pools:
    result = [x+[y] for x in result for y in pool]
    for prod in result:
    yield tuple(prod)

    for i in product(x):
    print(i)
    jiang42
        7
    jiang42  
       2014-11-16 01:05:24 +08:00   ❤️ 1
    def foo(xss, acc):
    if len(xss) == 0:
    return acc
    new_acc = []
    if len(acc) == 0:
    new_acc = [x for x in xss[0]]
    else:
    new_acc = [y+x for y in acc for x in xss[0]]
    return foo(xss[1:], new_acc)
    一个比较仓促的递归实现
    icedx
        8
    icedx  
       2014-11-16 01:32:32 +08:00 via Android   ❤️ 1
    yukirock
        9
    yukirock  
       2014-11-16 02:14:58 +08:00   ❤️ 1
    這東西顯然是用 list comprehension 搞啊。

    用 GHCi:

    Prelude> var a = ["A","B","C"]
    Prelude> var b = ["1;,"2","3","4","5"]
    Prelude> var c = ["a","b"]
    Prelude> [va ++ vb ++ vc | va <- a, vb <- b, vc <- c]
    ["A1a","A1b","A2a","A2b","A3a","A3b","A4a","A4b","A5a","A5b","B1a","B1b","B2a","B2b","B3a","B3b","B4a","B4b","B5a","B5b","C1a","C1b","C2a","C2b","C3a","C3b","C4a","C4b","C5a","C5b"]

    Python 也有類似的東西。
    yukirock
        10
    yukirock  
       2014-11-16 02:16:36 +08:00
    修正

    Prelude> var b = ["1","2","3","4","5"]
    yukirock
        11
    yukirock  
       2014-11-16 02:19:24 +08:00
    ……居然連着打錯這麼多,我還是重新來吧。

    這樣的問題,顯然是通過 list comprehension 來搞。

    用 GHCi:

    Prelude> let a = ["A","B","C"]
    Prelude> let b = ["1","2","3","4","5"]
    Prelude> let c = ["a","b"]
    Prelude> [va ++ vb ++ vc | va <- a, vb <- b, vc <- c]
    ["A1a","A1b","A2a","A2b","A3a","A3b","A4a","A4b","A5a","A5b","B1a","B1b","B2a","B2b","B3a",
    "B3b","B4a","B4b","B5a","B5b","C1a","C1b","C2a","C2b","C3a","C3b","C4a","C4b","C5a","C5b"]

    Python 也有這樣的東西。

    @Livid 煩請代爲刪除 #9,#10 的 typo,非常感謝。
    msg7086
        12
    msg7086  
       2014-11-16 04:27:52 +08:00   ❤️ 1
    Ruby

    [1] pry(main)> a = %w(A B C)
    => ["A", "B", "C"]
    [2] pry(main)> b = %w(1 2 3 4 5)
    => ["1", "2", "3", "4", "5"]
    [3] pry(main)> c = %w(a b)
    => ["a", "b"]
    [4] pry(main)> a.product(b).map(&:join).product(c).map(&:join)
    => ["A1a",
    "A1b",
    "A2a",
    "A2b",
    "A3a",
    "A3b",
    "A4a",
    "A4b",
    "A5a",
    "A5b",
    "B1a",
    "B1b",
    "B2a",
    "B2b",
    "B3a",
    "B3b",
    "B4a",
    "B4b",
    "B5a",
    "B5b",
    "C1a",
    "C1b",
    "C2a",
    "C2b",
    "C3a",
    "C3b",
    "C4a",
    "C4b",
    "C5a",
    "C5b"]
    msg7086
        13
    msg7086  
       2014-11-16 04:32:07 +08:00
    如果要任意数组求积的话可以:

    anyarray = [a, b, c]
    [''].product(*anyarray).map(&:join)
    staticor
        14
    staticor  
       2014-11-16 09:13:48 +08:00   ❤️ 1
    用自己的语言搜索 笛卡尔积 :) 然后可再自己实现或是用自带Module
    levon
        15
    levon  
    OP
       2014-11-16 10:01:10 +08:00
    感谢了,术语笛卡尔积,由于不是科班出身,虽然听过这个词汇,但不知道具体是什么。

    我已经搜索过了,上面好多回复都是可以的。我自己还搜索到一个
    http://stackoverflow.com/questions/533905/get-the-cartesian-product-of-a-series-of-lists-in-python

    Thanks all.
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