概述
items = (functions.getItems(item,date) print items test = sum(abs(l[-1]) for l in items) total = open('total' +str(datetime.today- datetime.timedelta(1)),'a')
我想要timedelta(1)循环遍历每个日期,以便输出文件的格式为total2012-01-01的第一天,循环,直到创建文件total2012-06-09. Item的date参数也是MM-DD-YYYY的格式
我以为我可以这样做:
sd = 01-01-2012 ed = 06-09-2012 delta = datetime.timedelta(days=1) diff = 0 while sd != ed # do functions # (have output files (datetime.today - datetime.delta(diff)) diff +=1 sd+=delta
所以本质上我只是想弄清楚如何循环使功能从01-01-2012开始,结束于06-10-2012,不包括周末.我无法弄清楚如何排除周末,以及如何让它按正确的顺序循环
谢谢
start = datetime(2012,1,1) end = datetime(2012,10,6) delta = timedelta(days=1) d = start diff = 0 weekend = set([5,6]) while d <= end: if d.weekday() not in weekend: diff += 1 d += delta
总结
以上是编程之家为你收集整理的python – 循环通过日期,除了周末全部内容,希望文章能够帮你解决python – 循环通过日期,除了周末所遇到的程序开发问题。
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