向量化numpy中的複雜賦值邏輯

Question

我在模擬中有一些複雜的賦值邏輯，我想優化性能。當前的邏輯被實現爲一組嵌套for循環遍歷各種numpy數組。我想這個矢量化分配邏輯，但一直沒能弄清楚這是否可能

import numpy as np 
from itertools import izip 

def reverse_enumerate(l): 
    return izip(xrange(len(l)-1, -1, -1), reversed(l)) 

materials = np.array([[1, 0, 1, 1], 
        [1, 1, 0, 0], 
        [0, 1, 1, 1], 
        [1, 0, 0, 1]]) 

vectors = np.array([[1, 1, 0, 0], 
        [0, 0, 1, 1]]) 

prices = np.array([10, 20, 30, 40]) 
demands = np.array([1, 1, 1, 1]) 

supply_by_vector = np.zeros(len(vectors)).astype(int) 

#go through each material and assign it to the first vector that the material covers 
for m_indx, material in enumerate(materials): 
    #find the first vector where the material covers the SKU 
    for v_indx, vector in enumerate(vectors): 
     if (vector <= material).all(): 
      supply_by_vector[v_indx] = supply_by_vector[v_indx] + 1 
      break 

original_supply_by_vector = np.copy(supply_by_vector) 
profit_by_vector = np.zeros(len(vectors)) 
remaining_ask_by_sku = np.copy(demands) 

#calculate profit by assigning material from vectors to SKUs to satisfy demand 
#go through vectors in reverse order (so lowest priority vectors are used up first) 
profit = 0.0 
for v_indx, vector in reverse_enumerate(vectors): 
    for sku_indx, price in enumerate(prices): 
     available = supply_by_vector[v_indx] 
     if available == 0: 
      continue 

     ask = remaining_ask_by_sku[sku_indx] 
     if ask <= 0: 
      continue 

     if vector[sku_indx]: 
      assign = ask if available > ask else available 
      remaining_ask_by_sku[sku_indx] = remaining_ask_by_sku[sku_indx] - assign 
      supply_by_vector[v_indx] = supply_by_vector[v_indx] - assign 

      profit_by_vector[v_indx] = profit_by_vector[v_indx] + assign*price 
      profit = profit + assign * price 

print 'total profit:', profit 
print 'unfulfilled demand:', remaining_ask_by_sku 
print 'original supply:', original_supply_by_vector 

結果：

total profit: 80.0 
unfulfilled demand: [0 1 0 0] 
original supply: [1 2] 

Answer 1

似乎有最裏面的內迭代之間的依賴關係在嵌套循環的第二部分/組中嵌套循環，對我來說，如果不是無法進行矢量化，看起來很困難。這樣，此信息基本上是試圖代替進行矢量兩個嵌套循環的第一組的部分解決方案，這是 -

supply_by_vector = np.zeros(len(vectors)).astype(int) 
for m_indx, material in enumerate(materials): 
    #find the first vector where the material covers the SKU 
    for v_indx, vector in enumerate(vectors): 
     if (vector <= material).all(): 
      supply_by_vector[v_indx] = supply_by_vector[v_indx] + 1 
      break 

即整個部分可以通過的向量化一行代碼被替換，像​​這樣 -

supply_by_vector = ((vectors[:,None] <= materials).all(2)).sum(1)