我正在尝试使用Python中的Bellman-Ford图算法来满足我的需要。在
我已经从一个json文件中计算出了解析部分。在
这是我在github上找到的贝尔曼福特代码: https://github.com/rosshochwert/arbitrage/blob/master/arbitrage.py
下面是我改编的代码:
import math, urllib2, json, re
def download():
graph = {}
page = urllib2.urlopen("https://bittrex.com/api/v1.1/public/getmarketsummaries")
jsrates = json.loads(page.read())
result_list = jsrates["result"]
for result_index, result in enumerate(result_list):
ask = result["Ask"]
market = result["MarketName"]
pattern = re.compile("([A-Z0-9]*)-([A-Z0-9]*)")
matches = pattern.match(market)
if (float(ask != 0)):
conversion_rate = -math.log(float(ask))
if matches:
from_rate = matches.group(1).encode('ascii','ignore')
to_rate = matches.group(2).encode('ascii','ignore')
if from_rate != to_rate:
if from_rate not in graph:
graph[from_rate] = {}
graph[from_rate][to_rate] = float(conversion_rate)
return graph
# Step 1: For each node prepare the destination and predecessor
def initialize(graph, source):
d = {} # Stands for destination
p = {} # Stands for predecessor
for node in graph:
d[node] = float('Inf') # We start admiting that the rest of nodes are very very far
p[node] = None
d[source] = 0 # For the source we know how to reach
return d, p
def relax(node, neighbour, graph, d, p):
# If the distance between the node and the neighbour is lower than the one I have now
if d[neighbour] > d[node] + graph[node][neighbour]:
# Record this lower distance
d[neighbour] = d[node] + graph[node][neighbour]
p[neighbour] = node
def retrace_negative_loop(p, start):
arbitrageLoop = [start]
next_node = start
while True:
next_node = p[next_node]
if next_node not in arbitrageLoop:
arbitrageLoop.append(next_node)
else:
arbitrageLoop.append(next_node)
arbitrageLoop = arbitrageLoop[arbitrageLoop.index(next_node):]
return arbitrageLoop
def bellman_ford(graph, source):
d, p = initialize(graph, source)
for i in range(len(graph)-1): #Run this until is converges
for u in graph:
for v in graph[u]: #For each neighbour of u
relax(u, v, graph, d, p) #Lets relax it
# Step 3: check for negative-weight cycles
for u in graph:
for v in graph[u]:
if d[v] < d[u] + graph[u][v]:
return(retrace_negative_loop(p, source))
return None
paths = []
graph = download()
print graph
for ask in graph:
path = bellman_ford(graph, ask)
if path not in paths and not None:
paths.append(path)
for path in paths:
if path == None:
print("No opportunity here :(")
else:
money = 100
print "Starting with %(money)i in %(currency)s" % {"money":money,"currency":path[0]}
for i,value in enumerate(path):
if i+1 < len(path):
start = path[i]
end = path[i+1]
rate = math.exp(-graph[start][end])
money *= rate
print "%(start)s to %(end)s at %(rate)f = %(money)f" % {"start":start,"end":end,"rate":rate,"money":money}
print "\n"
错误:
^{pr2}$当我打印图表时,我得到了所有需要的东西。它是“LTC”,因为它是列表中的第一个。我试着执行和过滤LTC,它给了我同样的错误,名字出现在图表上:
Traceback (most recent call last):
File "belltestbit.py", line 78, in <module>
path = bellman_ford(graph, ask)
File "belltestbit.py", line 61, in bellman_ford
relax(u, v, graph, d, p) #Lets relax it
File "belltestbit.py", line 38, in relax
if d[neighbour] > d[node] + graph[node][neighbour]:
KeyError: 'NEO'
我不知道该怎么解决这个问题。在
谢谢大家。在
附言:似乎有一个答案被删除了,我是新手,所以我不知道发生了什么。我编辑了这篇文章,因为答案帮助我前进了:)
免责声明:请注意,尽管您可以通过这种方式发现“低效率”,但您实际利用它们赚钱的机会非常低。很可能你真的会损失一些钱。从我在测试期间看到的数据来看,这些“低效率”来自于这样一个事实:汇率在几分钟内的波动性比买卖价差更大。因此,你所看到的低效率可能只是一个过时的数据,你实际上无法迅速执行所有必要的订单,使汇率稳定到足以挣钱。因此,请注意,如果您试图将此应用程序用于任何超出您的好奇心的东西,您可能会损失您的钱。在
现在我们来谈谈商业: 数据的格式与原始代码的格式不同。典型的数据如下:
您感兴趣的是}。你需要理解这些^{} and ^{} 是什么意思。粗略地说,},那么(或者更确切地说是不久前)有一个买家愿意用}。类似地,},那么有一个买家愿意用
MarketName
、Bid
和{Ask
值意味着,如果你想以ETH
的价格出售{0.04986673 BTC
的汇率为1 ETH
购买你的{Bid
值意味着,如果你想以BTC
的价格出售{0.04978511 BTC
的汇率来购买你的ETH
。请注意,此结构意味着您将不会有一个带有"MarketName": "ETH-BTC"
的记录,因为它不提供其他数据。在所以知道你可以用适当的距离填充你的
^{pr2}$graph
,这是相应速率的对数。另外,我相信您的代码中还有另一个错误:由于retrace_negative_loop
的参数p
实际上是前置节点的字典,retrace_negative_loop
以相反的顺序返回负循环。因为你的图是有方向的,同一个循环可能在一个方向上是正的,在另一个方向上是负的。在另外,检查
if path not in paths and not None:
可能还不够,因为它不能过滤我们的path
仅仅是彼此的旋转,但我也没有费心去修复它。在相关问题 更多 >
编程相关推荐