python repeat timestamp error

Author: jklwonder, Created: 2018-03-11 01:03:00, Updated:

Traceback (most recent call last): File “<string>”, line 74, in main File “<string>”, line 24, in Buy_Sold File “<string>”, line 2127, in GetTicker File “<string>”, line 1666, in __delay File “<string>”, line 1685, in rolling O0i1II1Iiii1I11

The error message shown below is roughly the same as calling thegetticker data, which was not a problem on the N calls, so you can't call it if you don't know.


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jklwonderHi, can you add your QQ to communicate?

The Little DreamHello, this problem may require reading the code, can you paste the code and see if it works?

jklwonderThe error message is: Traceback (most recent call last): File "", line 27, in main File "", line 14, in ComparePrice File "", line 2127, in GetTicker File "", line 1665, in __delay File "", line 978, in Sleep O0i1II1Iiii1I11

jklwonder'' backtest Start: 2018-02-03 00:00:00 This is the latest version of the game. End: 2018-02-10 12:00:00 This is the first time I've seen this. period: 1m The following is a list of exchanges: [{"eid":"Bitfinex","currency":"BTC","balance":100000,"stocks":30},{"eid":"OKCoin_EN","currency:"BTC","balance":100000,"stocks":30}] '" Import traceback Import time import numpy as np Import pandas as pd Import os def ComparePrice ((B0P1Diff,B1P0Diff,j): #compare the price difference between two exchanges GetTicker is a free exchange. GetTicker (() # if e0 is higher than e1 B0P1Diff[j]=price0['Buy']/(price1['Sell']) is the difference between the price of the product and the price of the product. B1P0Diff[j]=price0['Sell']/(price1['Buy']) is the difference between the price of a product and the price of the product. def main (: B0P1Diff={} B1P0Diff={} So i is equal to 0. Amount = 0.01 # Amount of coins bought and sold at a time While (True): try: ComparePrice ((B0P1Diff,B1P0Diff,i) is a free app that allows you to compare prices. So i is equal to i plus 1. except Exception: Log ((traceback.format_exc)) is the name of the file. Log (i) exit ((0) B0P1Array=np.array(list(B0P1Diff.values())) B1P0Array=np.array(list(B1P0Diff.values())) ratioB0P1=np.mean (([B0P1Array]) # The price difference between the two exchanges found in the Num iteration ratio B1P0=np.mean (([B1P0Array]) Log (mean B0P1 and ratio B0P1) Log (mean B1P0 and ratio B1P0) # exchange[2].Buy ((11600,1) What do you think? What do you think? def onerror: Log ((Error))