從python的csv文件的當前日期時間減去日期時間

Question

我試圖計算創建和更新文件之前的時間差，以分鐘計算丟失數據的數量。但我無法正確減去它們。我的代碼如下。

在文件中的數據是這樣的：

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 

而且代碼：

#!/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，它給我的結果是這樣的：

40.755 
35.7553 
30.755 
25.755 
20.755 
15.755 
10.755 
5.755 
0.755 

但我只是需要最後的值，例如40.755在這種情況下。我是編碼新手。有人可以幫助我，在這方面如何獲得單一的價值。

Answer 1

首先，創建一個列表來保存minutes_values，然後使用內建的max（）來獲得最大值。

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

我也是初學者，但在我看來for循環是不必要的。看起來你只需要循環返回的第一個值（40.755），並且每個連續的值減少5，與步長= 5有關？如果你改變step_size，那麼連續的時間改變那個數量？

如果for循環對於其他東西是必需的，則分開代碼的兩部分：一個計算時間差，一個計算for循環正在做的事情。