寻找一种快速的方法来构建所有可能性的导出(Python 2.5)
寻找一种快速的方法来构建所有可能性的导出(Python 2.5)
提问于 2020-09-08 14:50:20
我正在寻找一种快速的方式来建立出口。
获得三个数组作为输入数据-两个规则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"}
]“‘mappings”数组可以包含0到20,000条记录。我需要一种快速的方法来构建以下导出：
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}
]我试着使用这种方法，但速度非常慢：
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!
回答 1
Stack Overflow用户
回答已采纳
发布于 2020-09-08 15:03:55
这是python3 (对不起，我没有Python2编译器，所以我无法检查)，但是非常基础，也许你可以采用它。它的速度更快，因为我创建了一组映射，以避免总是运行过滤器。
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})页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/63788509
复制相关文章
相似问题
腾讯云开发者
