绘制（随时间段的离散总和）与（时间段）产生不连续的图形答案

【问题标题】：Plotting (discrete sum over time period) vs. (time period) yields graph with discontinuities绘制（随时间段的离散总和）与（时间段）产生不连续的图形
【发布时间】：2019-05-10 17:39:04
【问题描述】：

我有一些与买卖比特币有关的清单。一个是（买入或卖出的）价格，另一个是相关日期。当我绘制在不同时间长度内从我的买卖中赚取（或损失）的总钱与那些不同长度的时间时，结果是“不稳定的” - 这不是我的预期。而且我认为我的逻辑可能是错误的

我的原始输入列表如下所示：

dates=['2013-05-12 00:00:00', '2013-05-13 00:00:00', '2013-05-14 00:00:00', ....]

prices=[114.713, 117.18, 114.5, 114.156,...]

#simple moving average of prices calced over a short period
sma_short_list = [None, None, None, None, 115.2098, 116.8872, 118.2272, 119.42739999999999, 121.11219999999999, 122.59219999999998....]

#simple moving average of prices calced over a longer period
sma_long_list = [...None, None, None, None, 115.2098, 116.8872, 118.2272, 119.42739999999999, 121.11219999999999, 122.59219999999998....]

基于移动平均交叉（基于https://stackoverflow.com/a/14884058/2089889 计算），我将在交叉发生的日期/价格买入或卖出比特币。

我想绘制（到今天为止，这种方法可以让我赚多少钱）与（几天前我开始这种方法）的对比图

但是

我遇到的问题是生成的图表非常不稳定。首先，我认为这是因为我买的比卖的多（反之亦然），所以我试图解释这一点。但它仍然波涛汹涌。注意下面的代码在循环中调用for days_ago in reversed(range(0,approach_started_days_ago)):，所以每次执行下面的代码时，它应该吐出如果我开始这种方法会赚多少钱days_ago （我称之为 bank），起伏的情节是 days_ago 与 bank

dates = data_dict[file]['dates']
prices = data_dict[file]['prices']
sma_short_list = data_dict[file]['sma'][str(sma_short)]
sma_long_list = data_dict[file]['sma'][str(sma_long)]

prev_diff=0
bank = 0.0
buy_amt, sell_amt = 0.0,0.0
buys,sells, amt, first_tx_amt, last_tx_amt=0,0,0, 0, 0
start, finish = len(dates)-days_ago,len(dates)
for j in range(start, finish):
    diff = sma_short_list[j]-sma_long_list[j]
    amt=prices[j]

    #If a crossover of the moving averages occured
    if diff*prev_diff<0:
        if first_tx_amt==0:
            first_tx_amt = amt
        #BUY
        if diff>=0 and prev_diff<=0:
            buys+=1
            bank = bank - amt
            #buy_amt = buy_amt+amt
            #print('BUY ON %s (PRICE %s)'%(dates[j], prices[j]))
        #SELL
        elif diff<=0 and prev_diff>=0:
            sells+=1
            bank = bank + amt
            #sell_amt = sell_amt + amt
            #print('SELL ON %s (PRICE %s)'%(dates[j], prices[j]))
    prev_diff=diff

last_tx_amt=amt
#if buys > sells, subtract last
if buys > sells:
    bank = bank + amt
elif sells < buys:
    bank = bank - amt

#THIS IS RELATED TO SOME OTHER APPROACH I TRIED
#a = (buy_amt) / buys if buys else 0
#b = (sell_amt) / sells if sells else 0
#diff_of_sum_of_avg_tx_amts = a - b

start_date = datetime.now()-timedelta(days=days_ago)

return bank, start_date

【问题讨论】：

标签： python numpy numerical-methods

【解决方案1】：

我推断我在“银行”中的金额将是我已售出的金额 - 我已购买的金额

但是，如果第一个交叉是卖出，我不想计算（我将假设我做的第一笔交易将是买入。

如果我最后一笔交易是买入（对我的银行不利），我会将今天的价格计入我的“银行”

if last_tx_type=='buy':
    sell_amt=sell_amt+prices[len(prices)-1] #add the current amount to the sell amount if the last purchase you made is a buy
if sell_first==True:
    sell_amt = sell_amt - first_tx_amt #if the first thing you did was sell, you do not want to add this to money made b/c it was with apriori money

bank = sell_amt-buy_amt

【讨论】：