技术问题，多个数组拼合问题请教

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这是一个创建于 3669 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

数组

var

c.count

15 条回复 • 2014-11-16 10:01:10 +08:00

Daniel65536

2014-11-16 00:27:26 +08:00

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

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

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

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

2014-11-16 01:00:04 +08:00

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

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

2014-11-16 01:05:24 +08:00

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

2014-11-16 01:32:32 +08:00 via Android

别打我
https://pbs.twimg.com/media/B2f84MMCAAAWMYd.jpg

yukirock

2014-11-16 02:14:58 +08:00

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

2014-11-16 02:16:36 +08:00

修正

Prelude> var b = ["1","2","3","4","5"]

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

2014-11-16 04:27:52 +08:00

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

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

2014-11-16 09:13:48 +08:00

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

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.