strptime如何解析：01/Jul/1995:00:00:010400

Question

我已经尝试了大量的选择，并得到了一些类似的黑客一起解析，但我很好奇如何用strptime做到这一点？在

item = "01/Jul/1995:00:00:01-0400"

checkdate = datetime.strptime(item,"%Y-%m-%dT:%H:%M%S%z")
checkdate = datetime.strptime(item,"%Y/%m/%dT:%H:%M:%S%z")
checkdate = datetime.strptime(item,"%Y-%b-%d:%H:%M%S%z")
checkdate = datetime.strptime(item,"%Y-%b-%dT:%H:%M:%S%z")
checkdate = datetime.strptime(item,"%Y/%m/%d:%H:%M:%S%z")

我每次尝试得到的结果是：

^{pr2}$

正确的strptime格式是什么？在

编辑： 所以你是对的，我做了一个小测试

def split_date (stringdate):    
     datepart = [] 
     monthDict = {'Jan':'01','Feb':'02','Mar':'03','Apr':'04','May':'05',
 'Jun':'06','Jul':'07','Aug':'08','Sep':'09','Oct':'10','Nov':'11','Dec':'12'}
    split1 = [part for part in stringdate.split('/')]
    day = split1[0]
    month = split1[1]
    month = monthDict.get(month)
    split2 = [part for part in split1[2].split(":")]
    year = split2[0]
    hour = split2[1]
    minute = split2[2]
    split3 = [part for part in split2[3].split('-')]
    second = split3[0]
    timezone = split3[1]
    return datetime(int(year), int(month), int(day), int(hour), int(minute), int(second), int(timezone)

datetime_received_split = []
datetime_received_strp = []
s = time.time()
for date in data.time_received:
    try: 
        datetime_received_split.append(split_date(date))
    except:
        split_fail.append(date)
e = time.time()
print ('split took {} s '.format(e-s))
s = time.time() 
for date in data.time_received:
    try: 
             datetime_received_strp.append(datetime.strptime(item,"%d/%b/%Y:%H:%M:%S-     %f"))
    except: 
        strp_fail.append(date)
e = time.time() 
print ('strp took {} s'.format(e-s))

我发现手动拆分实际上要快得多？在

Answer 1

我修正了你的日期转换。最棒的是，python2.7和3.x都支持%f

from datetime import datetime

item = "01/Jul/1995:00:00:01-0400"
checkdate = datetime.strptime(item,"%d/%b/%Y:%H:%M:%S-%f")

print(checkdate)