Question

我是高级PHP和Perl老兄，但Python对我来说是新手。我很开心学习它！我编写了下面的代码，但是我无法摆脱那些具有“高级”Python技能的人更好地编写的感觉。你真正的Python人是一个有趣的挑战吗？

请注意：我希望代码可读。 Python被认为是可读的 - 我们不是在这里写人Perl！（例如：我喜欢'工作日'是字符串而不是整数，非常清楚）

import datetime
today = datetime.datetime.now()
weekday = today.strftime("%a")
hourmin = int(today.strftime("%H%M"))
print "today here is: " + today.strftime("%c") # for debug
if weekday == "Sat" or \
   (weekday == "Sun" and hourmin < 2000) or \
   (weekday == "Fri" and hourmin > 1630) or \
   (hourmin >= 1630 and hourmin < 2000) :
    print "bad time"
else:
    print "good time"

Answer 1

您可以将day字符串映射到需要hourmin的lambda，并确定它是否错误。例如：

# establish the "rules"    
bad_time = {
    'Sat': lambda h: True,    # always bad time!
    'Sun': lambda h: h < 2000,
    'Fri': lambda h: h > 1630,
}

# ... get your `weekday` and `hourmin` values

is_bad = bad_time.get(weekday, lambda h: (1630 <= h < 2000))(hourmin)
print 'bad time' if is_bad else 'good time'

编辑：关注kindall的建议。

Answer 2

在python中，如果你有不匹配的parens，你可以继续下一行，不需要反斜杠。您还可以在一个布尔表达式中进行两次比较。

if (weekday == "Sat" or
    (weekday == "Sun" and hourmin < 2000) or
    (weekday == "Fri" and hourmin > 1630) or
    1630 <= hourmin < 2000):
    print "bad time"
else:
    print "good time"

Answer 3

我建议使用一个类，让您轻松确定时间是否在预定义的范围内，然后将此类的实例放在字典中，以记录每天的时间是“坏”。也许使用几个预定义的范围以方便和以后更改。

class hmrange(object):
    def __init__(self, start, end):
        self.start, self.end = start, end
    def __contains__(self, hm):
        return self.start <= hm < self.end

alldayhours  = hmrange(0000, 2400)
weekdayhours = hmrange(1630, 2000)

badhours = {
    'Sun': hmrange(0000, 2000)
    'Mon': weekdayhours
    'Tue': weekdayhours
    'Wed': weekdayhours
    'Thu': weekdayhours
    'Fri': hmrange(1630, 2400)
    'Sat': alldayhours
}

badtime =  hourmin in badhours[weekday]

Answer 4

import datetime
today = datetime.datetime.now()
weekday = today.strftime("%a")
hourmin = int(today.strftime("%H%M"))
print "today here is: " + today.strftime("%c") # for debug
days = ["Sun", "Fri"]
times = [2000, 1630]
if weekday == "Sat" or \
    weekday in days and hourmin < times[days.index(weekday)] or \
    2000 > hourmin >= 1630:
    print "bad time"
else:
    print "good time"

Pythonians：有没有更好的方法在Python中编写代码？

4 个答案: