Question

我是编程的新手，我正在Pyschools上尝试这个问题。

任何人都可以帮助我吗？

编写一个将时间转换为24小时格式的函数。实例

>>> time24hr('12:34am')
'0034hr'
>>> time24hr('12:15pm')
'1215hr'

这个问题属于条件限制。

Answer 1

从末尾删除“am”或“pm”并将其保存在某个地方
拆分“：”，以便分开小时和分钟
如果时间是“am”，则打印小时，除非小时为“12”，在这种情况下打印“00”
否则（如果时间是“pm”），打印小时+12，除非小时为“12”，在这种情况下打印“12”
打印分钟

Answer 2

def time24hr(tstr):
    time_list = tstr[:-2].split(':')
    time_int = [int(x) for x in time_list]
    am_pm = tstr[-2:]
    hours = time_int[0]
    minutes = time_int[1]
    if am_pm == 'am':
        if hours < 12:
            hours = hours + 12          
            return "%02d%02dhr" % (hours,minutes)
        else:
            return "00%2dhr" % (minutes)        
    else:
        if hours < 12:
            hours = hours + 12
            return "%02d%02dhr" % (hours,minutes)
        else:
            return "%02d%02dhr" % (hours,minutes)

Answer 3

def time24hr(tstr):
    newTime = ""
    if "am" in tstr:
        for i in tstr:
            if i not in ":am":
                newTime += i
        if "12" in newTime[:2]:
            newTime = newTime.replace("12", "00")
            return newTime + "hr"
        elif len(newTime) == 3:
            newTime = "0" + newTime + "hr"
            return newTime
    elif "pm" in tstr:
        for i in tstr:
            if i not in "pm":
                newTime += i
        if "10" in newTime[:2]:
            newTime = newTime.replace("10", "22")
            newTime = newTime.replace(":", "")
        elif "11" in newTime[:2]:
            newTime = newTime.replace("11", "23")
            newTime = newTime.replace(":", "")
        elif "12" in newTime[:2]:
            newTime = newTime.replace(":", "")
        elif "1" in newTime[:1]:
            newTime = newTime.replace("1", "13")
            newTime = newTime.replace(":", "")
        elif "2" in newTime[:1]:
            newTime = newTime.replace("2", "14")
            newTime = newTime.replace(":", "")
        elif "3" in newTime[:1]:
            newTime = newTime.replace("3", "15")
            newTime = newTime.replace(":", "")
        elif "4" in newTime[:1]:
            newTime = newTime.replace("4", "16")
            newTime = newTime.replace(":", "")
        elif "5" in newTime[:1]:
            newTime = newTime.replace("5", "17")
            newTime = newTime.replace(":", "")
        elif "6" in newTime[:1]:
            newTime = newTime.replace("6", "18")
            newTime = newTime.replace(":", "")
        elif "7" in newTime[:1]:
            newTime = newTime.replace("7", "19")
            newTime = newTime.replace(":", "")
        elif "8" in newTime[:1]:
            newTime = newTime.replace("8", "20")
            newTime = newTime.replace(":", "")
        elif "9" in newTime[:1]:
            newTime = newTime.replace("9", "21")
            newTime = newTime.replace(":", "")
        return newTime + "hr"

Answer 4

对我来说，它可以工作（但不太漂亮：'C）：

def time24hr(tstr):
  a=list(tstr)
  if len(a)==6:
    a.insert(0,'0')
  if a[5]=='a' and a[0]=='1' and a[1]=='2':
    del a[2]
    del a[4:6]
    a.append('h')
    a.append('r')
    a[0]='0'
    a[1]='0'
    w=''.join(a)
    return w
  elif a[5]=='p':
    del a[2]
    del a[4:6]
    a.append('h')
    a.append('r')
    x=a[0]+a[1]
    print x
    y=int(x)
    print y
    if y<12:
      z=y+12
      g=str(z)
      a[0:2]=g
      w=''.join(a)
      return w
    else:
      w=''.join(a)
      return w
  else:
    del a[2]
    del a[4:6]
    a.append('h')
    a.append('r')
    w=''.join(a)
    return w

Answer 5

内联评论

def time24hr(s):
    c = s.split(':') # Split string to list ["hh", "mmAP"] AP is "am" or "pm"
    h = int(c[0]) # convert hours string to integer
    m = int(c[1][0:-2]) # remove "AP" from "mmAP" and convert to integer. This will handle "mAP" input correctly as well.
    h = h % 12 if s[-2:] != 'pm' else 12 if h == 12 else h + 12 # convert hours to 24hr format, handling edge cases.
    return "%02d%02dhr" % (h, m) # Print hours and minutes as %02d - two digits, potentially filling with zeroes

Answer 6

这对我来说更好。我根据接受的答案对此进行了改进。

我没有使用循环将列表转换为整数，而是从实际列表中调用值并将变量声明为int类型。简单！
我仍然没有理由在上午时间小于12小时的情况下增加小时+ 12 – 12时和24时时间格式与上午时间格式相同，但12时除外。因此，我只是从早上返回该函数的初始值，当它刚好是12时，我返回小时％12以拥有凌晨00点。

def time24hr(tstr):
    time_list = tstr[:-2].split(':')
    
    t_list_int = []
    for e in time_list:
        t_list_int.append(e)
   
    minutes = int(time_list[1])
    hours = int(time_list[0])
    am_pm = tstr[-2:]
    
    if am_pm == 'am':
        if hours == 12:
            return "%02d%02dhr" % (hours%12,minutes)
        elif hours <12:
            return "%02d%02dhr" %(hours, minutes)
    else:
        if hours == 12:
            return "%02d%02dhr" %(hours, minutes)
        elif hours <12:
            return "%02d%02dhr" %(hours+12, minutes)

print(time24hr('9:34pm'))
print(time24hr('12:00am'))

Answer 7

def time24hr(tstr):
    time=tstr.replace(':','')
    if 'am' in time:
        tim=int(time.replace('am',''))
    elif 'pm' in time:
        tim=int(time.replace('pm',''))
    mini=tim%100
    hour=tim//100 
    if mini<10:
        mini='0'+str(mini)
    else:
        mini=str(mini)
    if 'am' in time:
        if hour<10:
            hour='0'+str(hour)
        elif hour==12:
            hour='0'+str(hour-12)
            time1=time.replace('am','')
            time1=str(hour)+str(mini)+'hr'
    elif 'pm' in time:
        if hour<12:
            hour=12+hour
        elif hour==12:
            hour=hour
    time1=time.replace('pm','')
    time1=str(hour)+str(mini)+'hr'
    return time1

Answer 8

def time24hr(tstr): 
  a = tstr
  b = int(a[:2])+12
  if a[-2:] =='am' and a[:2]=='12':
      return '\'00'+a[3:-2]+'hr\''
  elif a[-2:]=='pm' and a[:2]=='12':
      return '\'12'+a[3:-2]+'hr\''
  else:
      if a[-2:] =='pm':
          return  "\'"+str(b)+a[3:-2]+'hr\''
      else:
        return "'"+a[:2]+a[3:-2]+'hr\''

print time24hr('12:15pm')
# '1215hr'
print  time24hr('12:15pm')
print time24hr('12:34am')
# '0034hr'
print time24hr('08:34am')
# '0834hr'
print time24hr('08:34pm')
'2034hr'

Pyschools主题3 Q 9

8 个答案: