如何获得其所属列表的加权平均值'；的权重受Python 3.6中的变量限制

我希望这个标题有意义。我想得到的是鞋子的加权平均价格，这些鞋子的价格不同，数量也不同。例如，我有：

list_prices = [12,12.7,13.5,14.3]
list_amounts = [85,100,30,54]
BuyAmount = x

我想知道我的加权平均价格，以及如果我买x数量的鞋子，我每只鞋子支付的最高价格（假设我想先买最便宜的）

这就是我现在拥有的（我使用numpy）：

这段代码可以工作，但可能过于复杂和扩展。特别是因为我想能够处理20多个项目的金额和价格清单

---

更好的方法是什么？

这里有一个通用的矢量化解决方案，使用

cumsum

替换切片求和，并使用

argmax

获取适当的索引，用于设置那些

IF case

操作的切片限制-

# Use cumsum to replace sliced summations - Basically all those 
# `list_amounts[0]`, `sum(list_amounts[0: 2]))`, `sum(list_amounts[0: 3])`, etc.
c = np.cumsum(list_amounts)

# Use argmax to decide the slicing limits for the intended slicing operations.
# So, this would replace the last number in the slices - 
# list_prices[0: 2], list_prices[0: 3], etc.
idx = (c >= BuyAmount).argmax()

# Use the slicing limit to get the slice off list_prices needed as the first
# input to numpy.average
l = list_prices[:idx+1]

# This step gets us the weights. Now, in the weights we have two parts. E.g.
# for the third-IF we have : 
# [list_amounts[0],list_amounts[1],BuyAmount - (sum(list_amounts[0: 2]))]
# Here, we would slice off list_amounts limited by `idx`.
# The second part is sliced summation limited by `idx` again.
w = np.r_[list_amounts[:idx], BuyAmount - c[idx-1]]

# Finally, plug-in the two inputs to np.average and get avgprice output.
avgprice = np.average(l,weights=w)

# Get idx element off list_prices as the highprice output.
highprice = list_prices[idx]

我们可以进一步优化以删除连接步骤（使用

np.r

）并获得

avgprice

，如下所示-

slice1_sum = np.multiply(list_prices[:idx], list_amounts[:idx]).sum()
        # or np.dot(list_prices[:idx], list_amounts[:idx])
slice2_sum = list_prices[idx]*(BuyAmount - c[idx-1])
weight_sum = np.sum(list_amounts[:idx]) + BuyAmount - c[idx-1]
avgprice = (slice1_sum+slice2_sum)/weight_sum

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您的确是对的，您的代码缺乏灵活性。但在我看来，你是从一个有效的角度来看待这个问题，但还不够全面

换句话说，您的解决方案实现了这一想法：“让我先检查一下——给定每个价格的可用数量（我将其完美地排列在一个数组中）——我必须从哪些不同的卖家那里购买，然后进行所有计算。”

一个更灵活的想法可以是：“让我开始尽可能多地从更便宜的商品中购买。当我的订单完成时，我会停止，并一步一步地计算。”。这意味着您要构建一个迭代代码，一步一步累积总花费量，并在完成计算后计算每件产品的平均价格和最高价格（即在您的订单列表中最后访问的价格）

要将此想法转化为代码：

list_prices = [12,12.7,13.5,14.3]
list_amounts = [85,100,30,54]
BuyAmount = x

remaining = BuyAmount
spent_total = 0
current_seller = -1 # since we increment it right away 

while(remaining): # inherently means remaining > 0
    current_seller += 1
    # in case we cannot fulfill the order
    if current_seller >= len(list_prices):
        # since we need it later we have to restore the value
        current_seller -= 1
        break
    # we want either as many as available or just enough to complete 
    # BuyAmount
    buying = min([list_amounts[current_seller], remaining])
    # update remaining
    remaining -= buying
    # update total
    spent_total += buying * list_prices[current_seller]

# if we got here we have no more remaining or no more stock to buy

# average price
avgprice = spent_total / (BuyAmount - remaining) 

# max price - since the list is ordered -
highprice = list_prices[current_seller]

print(avgprice)
print(highprice)

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大多数情况下，使用numpy比普通python效率更高，但代码看起来往往晦涩难懂：您能添加一些注释和解释吗？@FabioVeronese补充道。非常感谢您的帮助。我不得不做很多研究，因为我不熟悉cumsum、argmax和np\r。cumsum真的大大缩短了代码的长度哈哈，这正是你真正需要的。您如何找到amountindex，其中的金额填充了订单，这也是非常方便的，并且最终您拥有了一个非常高效的代码！我有一个问题，为什么删除连接步骤的优化是一件好事？再次感谢您的帮助，学到了很多@连接需要额外的内存来存储连接的数组。通过将其替换为后一部分所示的原位操作，我们节省了内存，并有望提高性能效率。非常感谢您的回复。很高兴听到我的思路是正确的。这就是说，我甚至没有想过在一个遥远的地方编程，它只需要看看相关的金额。我像这样实现了它，它工作了！我还学到了很多关于你的代码以及它比我的代码更高效的方法：）再次感谢

list_prices = [12,12.7,13.5,14.3]
list_amounts = [85,100,30,54]
BuyAmount = x

remaining = BuyAmount
spent_total = 0
current_seller = -1 # since we increment it right away 

while(remaining): # inherently means remaining > 0
    current_seller += 1
    # in case we cannot fulfill the order
    if current_seller >= len(list_prices):
        # since we need it later we have to restore the value
        current_seller -= 1
        break
    # we want either as many as available or just enough to complete 
    # BuyAmount
    buying = min([list_amounts[current_seller], remaining])
    # update remaining
    remaining -= buying
    # update total
    spent_total += buying * list_prices[current_seller]

# if we got here we have no more remaining or no more stock to buy

# average price
avgprice = spent_total / (BuyAmount - remaining) 

# max price - since the list is ordered -
highprice = list_prices[current_seller]

print(avgprice)
print(highprice)