Question

我将如何处理这个过程？假设有人输入“3：5”，我应该能够得到输出“3:05 AM”。或者，如果有人输入“00:00”，我应该得到“12:00 AM”。如果时间格式只是输入为“hhmm”，我会意识到过多的方法，但输入的范围可以是“h：m”，“hh：m”，“hh：m”和“hh：毫米”。所以这是非常情境化的。

def timeConvert():
    miliTime = int(input("Enter a time in hh:mm (military) format: "))
    miliTime.split(":")
    if len(miliTime) == 3:
        hours = miliTime[0]
        minutes = miliTime[2]
        if hours < 0:
            print("Hours can't be less than 0.")
    elif len(miliTime) == 4:
        if miliTime[0:2] >= 10:
            hours = militime[0:2]
            minutes = militime[3]
            if minutes < 0:
                print("Minutes can't be less than 0.")
            if minutes < 10:
                minutes = 0 + minutes
        else:
            hours = miliTime[0]
            minutes = militime[2:]
            if minutes >= 60:
                print("Too big of a number for minutes.")
    else:
        hours = miliTime[0:2]
        minutes = miliTime[3:]
    setting = AM
    if hours > 12:
        setting = PM
        hours -= 12
    print(hours + ":" + minutes + setting)

timeConvert()

Answer 1

使用datetime模块的strptime()将字符串作为日期时间对象读取，然后使用strftime()将其转换回字符串，例如 -

>>> datetime.datetime.strptime('3:5','%H:%M').strftime('%I:%M %p')
'03:05 AM'
>>> datetime.datetime.strptime('18:30','%H:%M').strftime('%I:%M %p')
'06:30 PM'
>>> datetime.datetime.strptime('00:00','%H:%M').strftime('%I:%M %p')
'12:00 AM'

解释的格式 -

％H - 小时（24小时fomrat）

％M - 分钟

％I - 小时（12小时格式）

％p - AM或PM

功能看起来像 -

def timeConvert():
    import datetime
    ds = input("Enter a time in hh:mm (military) format: ")
    newds = datetime.datetime.strptime(ds, ,'%H:%M').strftime('%I:%M %p')
    print(newds)

Answer 2

将输入分割为:，并在最后使用格式化输出：

def timeConvert():
    miliTime = input("Enter a time in hh:mm (military) format: ")
    hours, minutes = miliTime.split(":")
    hours, minutes = int(hours), int(minutes)
    setting = "AM"
    if hours > 12:
        setting = "PM"
        hours -= 12
    print(("%02d:%02d" + setting) % (hours, minutes))

Answer 3

您可以简单地将2列合并为一个字符串函数。我假设您正在使用24小时制或军事时间（精确到100分钟）

import pandas as pd

df = pd.read_clipboard()

for i in df.columns:

    df[i] = df[i].apply(str)


df['datetime'] = pd.to_datetime(df['date'] + ' ' + df['time'].apply(lambda x: x.zfill(4)))


df
Out[11]: 
      ccy      date time    open    high     low   close            datetime
0  EURUSD  20190211  100  1.1318  1.1319  1.1317  1.1319 2019-02-11 01:00:00
1  EURUSD  20190211  200   1.132  1.1322  1.1319  1.1319 2019-02-11 02:00:00
2  EURUSD  20190211  300  1.1318  1.1319  1.1318  1.1319 2019-02-11 03:00:00
3  EURUSD  20190211  400  1.1319  1.1319  1.1318  1.1318 2019-02-11 04:00:00
4  EURUSD  20190211  500  1.1318  1.1318  1.1318  1.1318 2019-02-11 05:00:00

Answer 4

military_time = input("Enter a military time (0000 to 2359): ")
hours = int(military_time[0:2])
minutes = int(military_time[2:4])
if hours >= 12:
    meridian = "pm"
    hours %= 12
else:
    meridian = "am"
if hours == 0:
    hours = 12
print("The regular time is {0}:{1} {2}.".format(hours, minutes, meridian))

0-23小时军用时钟到标准时间（hh：mm）

4 个答案: