Question

我正在寻找最有效的方法来进行以下计算：

我有三个矩阵，像这样：

[[Brand1, operationCost], [Brand2, operationCost],...]

[[Brand1, maintenanceCost],[Brand2, maintenanceCost]...]

[[Brand1, replacementCost],[Brand2, replacementCost]...]

我需要计算每个品牌的总成本，运营+维护+更换。可以确定的是，相同的标记不在所有矩阵中。并获得另一个这样的矩阵：

[[Brand1, totalCost],[Brand2, totalCost]...]

Answer 1

Numpy应该解决你的问题：

示例：

import numpy as np
c = np.array([[1, 2, 3], [1, 2, 3]])
c.sum(0)
Out[5]: array([2, 4, 6])

如果你想让你的品牌与变量结合起来我会使用pandas：

示例：

import pandas as pd
In[9]: df = pd.DataFrame({'brand1': [1, 2, 3], 'brand2': [1, 2, 3]})
In[10]: df
Out[10]: 
    brand1  brand2
0       1       1
1       2       2
2       3       3

In[11]: df.sum()
Out[11]: 
brand1    6
brand2    6

Answer 2

由于您似乎不使用python词典，因此应该可以使用：

[Object]0: 
    Object
    barcode: "abc"
    name: "barcoded"
    sellPrice: "122"
    unit: "200"
    __proto__:Object
    length: 1
    __proto__: Array[0]

修改：

但是，如果列表的长度或品牌的顺序发生变化，则您无法使用上述代码。所以最好的解决方案是使用词典：

operation = [[Brand1, operationCost], [Brand2, operationCost],...]
maintenance = [[Brand1, maintenanceCost], [Brand2, maintenanceCost],...]
replacement = [[Brand1, replacementCost], [Brand2, replacementCost],...]

total = [ [ope[0], ope[1]+mai[1]+rep[1]] for ope,mai,rep in zip(operation,maintenance,replacement) ]

或者对于2.7版本之前的python：

# Your matrix as dictionaries
operation = {Brand1: operationCost, Brand2: operationCost, ...}
maintenance = {Brand1: maintenanceCost, Brand2: maintenanceCost, ...}
replacement = {Brand1: replacementCost, Brand2: replacementCost, ...}
# Get all brands in a container
all_brands = set(operation.keys()+maintenance.keys()+replacement.keys())
# Return 0 as default value if a brand is not in the dictionary
f = lambda x, dic: dic[x] if x in dic else 0
# Compute the total cost of each brand
total = {brand: f(brand,operation)+f(brand,maintenance)+f(brand,replacement) for brand in all_brands}

Answer 3

这个解决方案是纯Python（它不依赖于第三方依赖）并且即使列表的长度不同也应该有效：

oc = [['Brand1', <operationCost>], 
      ['Brand2', <operationCost>],
      ...,
      ]
mc = [['Brand1', <maintenanceCost>], 
      ['Brand2', <maintenanceCost>],
      ...,
      ]
rc = [['Brand1', <replacementCost>], 
      ['Brand2', <replacementCost>],
      ...,
      ]
total = {}
for lst in [oc, mc, rc]:
    for brand, cost in lst:
        total[brand] = total.get(brand, 0) + cost

Python和2D列表列重合

3 个答案: