python的自由组合问题，我绞尽乳汁了。各位大佬来LOOKLOOK

Z时代
2024-01-10
分类：技术分享

题目来源及自己的思路

{

"binance": [ "ETH/BTC", "NEO/BTC", "NEO/ETH"， "BNT/ETH", "BCC/BTC", "GAS/BTC", "BTC/USDT", "ETH/USDT" ], "huo": [ "ETH/BTC", "NEO/BTC", "BNB/ETH", "NEO/ETH" ], "bitmex": [ "ETH/BTC", "LTC/BTC", "BNB/BTC"

]

}
需要组成三角循环体
{ "ETH/BTC"："binance","NEO/BTC"："binance","NEO/ETH"："binance"}
{ "ETH/BTC"："binance","NEO/BTC"："huo","NEO/ETH"："binance"}
{ "ETH/BTC"："bitmex","NEO/BTC"："binance","NEO/ETH"："binance"}
{ "ETH/BTC"："bitmex","NEO/BTC"："binance","NEO/ETH"："binance"}
等等所有可能的组合
每个字典里面的"ETH/BTC==NEO/BTC==NEO/ETH"就是三角套利的结构，我试过两种方案，第一种是把所有的"/"分割开，提取所有的单独数字货币名字做成列表，在把列表进行组合成三角模型去配对后面交易所的名字上去。不过匹配不完全，pass
第二种是也是做成所有可能列表去求差集，陷入死循环体，出不了结果。
各位大佬有什么好的方法

相关代码

All_bitlist = []
for exchange in bitlist:

exchange = list(data.values())
for Trading_Pair_list in exchange:
    for Trading_Pair in Trading_Pair_list:        All_bitlist.extend(Trading_Pair.split("/", 2))

All_bitlist = list(set(All_bitlist))

Newlist = []
for Tbits in combinations(All_bitlist, 3):

Tbits = list(Tbits)
Newlist2 = []
for Trading_Pair in permutations(Tbits, 2):
    x, y = Trading_Pair
    Trading_Pair = x + "/" + y
    Newlist2.append(Trading_Pair)
Newlist.append(Newlist2)