将字符串分割成给定大小的组 [英] split string into groups of given size

问题描述

我的情况是我有一个时间戳的项目列表，但实际的时间是一起挤压在一起。像这样：

  time =165420＃16:54:20（或，4:54 PM和20秒） 
  

我想要一个脚本，可以将时间戳转换成日期-时间。我有一年，一个月，一天已经没有问题。但是对于时间的局面，我绝望地找到一个优雅简单的解决方案。我决定先强制它继续执行我的代码，最后是：

  import datetime 
 
 dt_timestamp = extract_dt（2012-11-09-165420）＃example 
 
 def extract_dt（timestamp）：
 ds = timestamp.split（' - '）#split into组（0,1,2是日期，3是时间）
d = [int（i）for ds in [ds] [] 3]] [int（ds [3] [0：2]）， int（ds [3] [2：4]），int（ds [3] [4：6]）] 
 return datetime.datetime（* d）
  / pre> 
 
 
在尝试解决我的情况时，我看到很多变体，有很多答案。
 
 
 
我发现此解决方案 ：
 
 
  def nsplit（s，n）：
 return [s [k：k + n] for k in xrange（0，len（s），n）] 
  
哪个对我很好：
 >>> s =165420
>>> n = 2 
 >>> [s] [k：k + n] for x in xrange（0，len（s），n）] 
 ['16'，'54'，'20'] 
  
但我确信，必须将一个更好，更优雅的方式将字符串分割成相同大小的组，如<我提供了一个href =http://bytes.com/topic/python/answers/157385-splitting-string-into-groups-three-characters =nofollow>链接（但是由于链接海报有一个错字）：
 
 
 TheFunction（Hello World，3）
 
 
 
返回：
 
 
 
 ['Hel'，'lo'，'Wor'，'ld'] 
 
 
如果可能的话，我更喜欢它不是一个功能，我想要一个在线的方法。它应该尽可能高效。我总是可以从我发现的函数中提取列表的理解，但是我觉得它的效率不是很高，而且看起来并不正确。
 
 
 
答案可能是很简单，我只是不知道。再次感谢
解决方案
我不知道比您在问题中提到的更优雅的分块字符串解决方案，但是  datetime.datetime.strptime（） 解决了您的实际问题：
 >>> from datetime import datetime 
>>>> datetime.strptime（2012-11-09-165420，％Y-％m-％d-％H％M％S）
 datetime.datetime（2012，11，9，16，54， 20）
  
 
My situation is I have a list of time-stamped items, but the actual time-of-day is all squished together. Like this:
time = "165420" # 16:54:20 (or, 4:54 PM and 20 seconds.)
I wanted a script to go over the lines and be able to convert the time-stamp into a date-time. I have the year, month, day already that was no problem. But for the time situation I was desperate to find an elegant and simple solution. I decided to brute force it at first to continue with my code, and ended up with:
import datetime

dt_timestamp = extract_dt("2012-11-09-165420") # example

def extract_dt(timestamp):
    ds = timestamp.split('-') #split into groups (0,1,2 are the date, 3 is the time.
    d = [int(i) for i in ds[:3]] + [int(ds[3][0:2]), int(ds[3][2:4]), int(ds[3][4:6])]
    return datetime.datetime(*d)
While trying to solve my situation I've seen many variations on this question, with many answers.

I have found this solution:
def nsplit(s, n):
    return [s[k:k+n] for k in xrange(0, len(s), n)]
Which works nicely for me:
>>> s = "165420"
>>> n = 2
>>> [s[k:k+n] for k in xrange(0, len(s), n)]
['16', '54', '20']
But I am pretty sure there MUST be a better, more elegant way of splitting a string into groups of equal size, as was mentioned in the link I provided (but fixed because the link poster had a typo):

  
TheFunction("Hello World",3)
  
  
Returns:
  
  
['Hel','lo ','Wor','ld']
I would prefer, if possible, that it not be a function, I want a way to do it in-line. And it should be as efficient as possible. I can always extract the list-comprehension from the function I found, but I have a feeling its not very efficient, and it just doesn't look right.

The answer is probably something very simple I just didn't know about. Thanks again.
解决方案
I don't know of a more elegant solution to chunking strings than those you mention in your question, but datetime.datetime.strptime() solves your practical problem:
>>> from datetime import datetime
>>> datetime.strptime("2012-11-09-165420", "%Y-%m-%d-%H%M%S")
datetime.datetime(2012, 11, 9, 16, 54, 20)


                        
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