我正在寻找一种快速构建导出的方法。
得到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"}
]
"mappings"数组可能包含0到20000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}
]
我尝试过使用这种方式，但速度很慢：
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！
这是python3(对不起，我没有python 2编译器，所以我不能检查(，但它非常基本，也许你可以采用它。它更快，因为我创建了一组映射，以避免一直运行过滤器。
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})