从python的csv文件的当前日期时间减去日期时间

Question

我试图计算创建和更新文件之前的时间差，以分钟计算丢失数据的数量。但我无法正确减去它们。我的代码如下。

在文件中的数据是这样的：

DATE,temp1 
2017-08-14 17:10:01,0.44 
2017-08-14 17:15:01,0.62 
2017-08-14 17:20:01,0.99 
2017-08-14 17:25:01,0.85 
2017-08-14 17:30:01,0.13 
2017-08-14 17:35:01,0.24 
2017-08-14 17:40:02,0.95 
2017-08-14 17:45:01,0.65 

而且代码：

#!/usr/bin/env python 
# -*-coding:utf-8 -* 
import os 
import sys 
import time 

from time import mktime, strftime, localtime, sleep 
from datetime import datetime, timedelta 

def missing_data(expect, missed): 
. 
. 
. 
. 

def main(): 
    fmt = '%Y-%m-%d %H:%M:%S' 
    with open('/home/kamil/Desktop/sensor1/temp.dat', 'rb') as f: 
     read = f.readlines()[1:] 
     for line in read: 
      line = line.strip().split(',') 
      # print line 
      start_date = line[0] # type is str so i convert it to datetime 
      start_date = datetime.strptime(start_date, fmt) # type is datetime.datetime and here I need the first timestamp in the file only 
      current_time = datetime.now() 
      step_size = 5 
      differ_time = current_time - start_date 
      minutes_values = differ_time.total_seconds()/60 
      print 'minutes_values:::::::::::::::::::', minutes_values 
      missed = int(minutes_values/step_size) 
      expected = ......... 
      missing_data(expected, missed)  


if __name__ == '__main__': 
    main() 

如果我打印minutes_values，它给我的结果是这样的：

40.755 
35.7553 
30.755 
25.755 
20.755 
15.755 
10.755 
5.755 
0.755 

但我只是需要最后的值，例如40.755在这种情况下。我是编码新手。有人可以帮助我，在这方面如何获得单一的价值。

Answer 1

首先，创建一个列表来保存minutes_values，然后使用内建的max（）来获得最大值。

def main(): 
    fmt = '%Y-%m-%d %H:%M:%S' 
    with open('/home/kamil/Desktop/sensor1/temp.dat', 'rb') as f: 
     read = f.readlines()[1:] 
     minutes_values = [] 
     for line in read: 
      line = line.strip().split(',') 
      # print line 
      start_date = line[0] # type is str so i convert it to datetime 
      start_date = datetime.strptime(start_date, fmt) # type is datetime.datetime and here I need the first timestamp in the file only 
      current_time = datetime.now() 
      step_size = 5 
      differ_time = current_time - start_date 
      minutes_values.append(differ_time.total_seconds()/60) 
      missed = int(minutes_values/step_size) 
      expected = ......... 
      missing_data(expected, missed) 

    print 'minutes_values:::::::::::::::::::', max(minutes_values) 

Answer 2

我也是初学者，但在我看来for循环是不必要的。看起来你只需要循环返回的第一个值（40.755），并且每个连续的值减少5，与步长= 5有关？如果你改变step_size，那么连续的时间改变那个数量？

如果for循环对于其他东西是必需的，则分开代码的两部分：一个计算时间差，一个计算for循环正在做的事情。