寻找一种构建所有可能性导出的快速方法Python 2.5

如何解决寻找一种构建所有可能性导出的快速方法Python 2.5

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"},"product": "product3"},{"company": "company3","product": "product6"},"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},"product": "product2","product": "product3","product": "product4","is_mapped": 0},"product": "product5","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！

解决方法

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})