Question

我试图将我的工作日作为字符串转换为python中的int，这就是我所拥有的

$(".acol").each(function(i) {
  console.log($(this).html()); 
});

但是我得到了这个错误

from datetime import date, datetime, time, time delta
from pytz import timezone

now = datetime.now(timezone('UTC'))
dt_tz = now.astimezone(timezone('US/Eastern'))
dt = dt_tz.replace(tzinfo=None)
weekday = 'Friday'
weekday_as_int = dt.strptime(weekday, "%A")

为什么不将ValueError: time data 'Friday' does not match format '%A'更改为int？

Answer 1

整个工作日的正确格式为%A：

import time
weekday_as_int = time.strptime('friday', "%A").tm_wday

>>> weekday_as_int.tm_wday
4

星期一的计数从零开始：

>>> time.strptime('Monday', "%A").tm_wday
0

一些时间：

@Rustem的版本：

%%timeit
days = dict(zip(calendar.day_name, range(7))); 
days['Friday']
10000 loops, best of 3: 104 µs per loop

此版本：

%timeit time.strptime('Friday', "%A").tm_wday
10000 loops, best of 3: 19.7 µs per loop

看起来strptime的速度提高了五倍。

Answer 2

最快的方式是：

import calendar
days = dict(zip(calendar.day_name, range(7))); 
days['Friday']

是的，strptime很慢，对于这种微不足道的操作来说太过分了。

与@Mike版本相比

In [7]: timeit time.strptime('Friday', "%A").tm_wday
The slowest run took 337.46 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 7.14 µs per loop

In [8]: timeit days['Friday']
The slowest run took 22.05 times longer than the fastest. This could mean that an intermediate result is being cached
10000000 loops, best of 3: 54.1 ns per loop

工作日作为字符串编号

2 个答案: