下载多个具有不同名称的文件Python

如何解决下载多个具有不同名称的文件Python

Python的新手，以前主要使用R。我正在尝试从网页API（气候数据）下载多个文件，请参见链接 https://opendata-download-grid-archive.smhi.se/data/6/201604/MESAN_201604050000+000H00M

我想每年（每月2008-2019年）每个月的每个小时下载一个文件，即将“ 201604050000”部分更改为“ 201604050100”，“ 201604050200”，“ 201604050300”等。其他所有内容看起来都一样，我只需要更改时间/天/月/年就可以下载文件。

这是我获取每个文件的Python代码，但是我对如何下载它感到迷茫。

for a in [2010,2011,2012,2013,2014,2015,2016,2017,2018,2019]:
    for b in [1,2,3,4,5,6,7,8,9,10,11,12]:
        for c in [1,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31]:
            for d in ["00","01","02","03","04","05","06","07","08","09","10","11","12","13","14","15","16","17","18","19","20","21","22","23"]:
                s = "grib_%d_%d_%d_%s" %(a,b,c,d) 
                print(s)

如何下​​载所有这些文件？我知道它会填满我的记忆，所以我很高兴能从一天开始下载24个文件开始。其他人遇到过这个问题吗？

最后，当我管理完文件后，我也想删除它们（这样我就不会分配太多内存）。

解决方法

不是完整的答案，但有一些建议：

使用更好的变量名

使用range()代替较长的常量列表（如果适用）

for year in range(2010,2019 + 1):
    for month in range(1,12 + 1):
        for day in range(1,31 + 1):  # beware,not every month has 31 days
            for hour in ["00","01","02","03","04","05","06","07","08","09","10","11","12","13","14","15","16","17","18","19","20","21","22","23"]:
                s = "grib_%d_%d_%d_%s" %(year,month,day,hour) 
                print(s)