url_list = ['www.scrape.com/file1', 'www.scrape.com/file2', ''www.scrape.com/file3']
category_id_list = ['12345','abcde','ABCDE']
zip_list = ['10075','10017','10028']
我使用三个变量来创建要请求的URL。 顺序为:url_list + zip + categoryid
然后将网址传递给具有刮刮代码的函数
我正在使用3个for循环来遍历这些列表,但这是高度冗余的
for url_ in url_list:
for category_id in category_id_list:
for zip_ in zip_list:
request_url = url_+category_zip_
func(request_url)
这可以完成工作,但是还有更好的方法吗?谢谢!
答案 0 :(得分:2)
您可以使用itertools.product
import itertools
for url in (str.join("",url) for url in itertools.product(url_list,category_id_list,zip_list)):
func(url)
答案 1 :(得分:0)
这可能有点晚了,但这是我做到的方式:
cats = ["a","b","c","d"]
zips = ["25320","53902","59607","53123"]
base = "https://example.com"
for i in range(4):
url = "{}/{}/{}".format(base, cats[i], zips[i])
print(url)
输出:
https://example.com/a/25320
https://example.com/b/53902
https://example.com/c/59607
https://example.com/d/53123
答案 2 :(得分:0)
一种避免编写多个for循环的方法是使用zip。它允许您一次访问每个列表中的第ith个元素。因此,您可以执行以下操作:
url_list = ['www.scrape.com/file1', 'www.scrape.com/file2', 'www.scrape.com/file3']
category_id_list = ['12345','abcde','ABCDE']
zip_list = ['10075','10017','10028']
for url, id, zip in zip(url_list, category_id_list, zip_list):
request_url = url + id + zip
func(request_url)