Question

我正在寻找建立出口的快速方法。

将三个数组作为输入数据-两个常规数组和一个字典数组：

companies = ["company1", "company2", "company3", "company4", "company5", "company6", "company7", "company8", "company9"]
products = ["product1", "product2", "product3", "product4", "product5"]
mappings = [{"company": "company1", "product": "product1"},
            {"company": "company1", "product": "product2"},
            {"company": "company1", "product": "product3"},
            {"company": "company3", "product": "product6"},
            {"company": "company3", "product": "product9"},
            {"company": "company4", "product": "product2"}
            ]

“映射”数组可能包含0到20,000条记录。我需要一种快速的方法来建立以下导出：

mappings_export = [{"company": "company1", "product": "product1", "is_mapped": 1},
                   {"company": "company1", "product": "product2", "is_mapped": 1},
                   {"company": "company1", "product": "product3", "is_mapped": 1},
                   {"company": "company1", "product": "product4", "is_mapped": 0},
                   {"company": "company1", "product": "product5", "is_mapped": 0},
                   {"company": "company1", "product": "product6", "is_mapped": 1}
                   ]

我尝试使用这种方式，但是速度很慢：

mappings_export = []
for company in companies:
    for product in products:
        found = filter(lambda x: x["product"] == product and x["company"] == company, mappings)
        if len(found) > 0:
            mapped = 1
        else:
            mapped = 0
        mappings_export.append({"company": company,
                                "product": product,
                                "mapped": mapped})

谢谢！

Answer 1

这是python3（对不起，我没有python 2编译器，所以我无法检查），但它非常基础，也许您可以采用它。之所以速度更快，是因为我创建了一组映射，以避免一直都在运行过滤器。

company_product = set([])

for m in mappings:
    company_product.add((m["company"],m["product"]))

mappings_export = []
for c in companies:
    for p in products:
        mappings_export.append({"company": c,
                                "product": p,
                                "mapped": 1 if (c,p) in company_product else 0})

寻找一种构建所有可能性导出的快速方法（Python 2.5）

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