How to count accesses per hour from log file entries?

Question

I have a log-file where every line contains IP address, time of access, and the URL accessed. I want to count the accesses per hour.

Time of access data looks like this

[01/Jan/2017:14:15:45 +1000]
[01/Jan/2017:14:15:45 +1000]
[01/Jan/2017:15:16:05 +1000]
[01/Jan/2017:16:16:05 +1000] 

How can I improve it so I don't need to set up the variable and if statement for every hour?

twoPM = 0
thrPM = 0
fouPM = 0
timeStamp = line.split('[')[1].split(']')[0]
formated_timeStamp = datetime.datetime.strptime(timeStamp,'%d/%b/%Y:%H:%M:%S %z').strftime('%H')
if formated_timeStamp == '14':
    twoPM +=1
if formated_timeStamp == '15':
    thrPM +=1
if formated_timeStamp == '16':
    fouPM +=1

Answer 1

1. You can include the brackets into your strptime format description:

   datetime.datetime.strptime(line.strip(),'[%d/%b/%Y:%H:%M:%S %z]')
   

2. You can extract the hour using the .hour attribute of any datetime.datetime object:

   timestamp = datetime.datetime.strptime(…)
   hour = timestamp.hour
   

3. You can count the number of elements using a collections.Counter:

   from collections import Counter
   
   
   def read_logs(filename):
       with open(filename) as log_file:
            for line in log_file:
                timestamp = datetime.datetime.strptime(
                        line.strip(),
                        '[%d/%b/%Y:%H:%M:%S %z]')
                yield timestamp.hour
   
   
   def count_access(log_filename):
       return Counter(read_logs(log_filename))
   
   
   if __name__ == '__main__':
       print(count_access('/path/to/logs/'))