Grouping Messages by Time Intervals

Question

I'm currently trying to group messages that are sent out by 1 second time intervals. I'm currently calculating time latency with this:

def time_deltas(infile): 
entries = (line.split() for line in open(INFILE, "r")) 
ts = {}
for e in entries: 
    if " ".join(e[2:5]) == "T out: [O]": 
        ts[e[8]] = e[0]    
    elif " ".join(e[2:5]) == "T in: [A]":    
        in_ts, ref_id = e[0], e[7] 
        out_ts = ts.pop(ref_id, None) 
        yield (float(out_ts),ref_id[1:-1],(float(in_ts)*1000 - float(out_ts)*1000))

INFILE = 'C:/Users/klee/Documents/test.txt'
import csv 

with open('test.csv', 'w') as f: 
csv.writer(f).writerows(time_deltas(INFILE)) 

HOWEVER I want to calculate the number of "T in: [A]" messages per second that are sent out, and have been trying to work with this to do so:

import datetime
import bisect
import collections

data=[ (datetime.datetime(2010, 2, 26, 12, 8, 17), 5594813L), 
  (datetime.datetime(2010, 2, 26, 12, 7, 31), 5594810L), 
  (datetime.datetime(2010, 2, 26, 12, 6, 4) , 5594807L),
]
interval=datetime.timedelta(seconds=50)
start=datetime.datetime(2010, 2, 26, 12, 6, 4)
grid=[start+n*interval for n in range(10)]
bins=collections.defaultdict(list)
for date,num in data:
idx=bisect.bisect(grid,date)
   bins[idx].append(num)
for idx,nums in bins.iteritems():
print('{0} --- {1}'.format(grid[idx],len(nums)))

which can be found here: Python: group results by time intervals

(I realize the units would be off for what I want, but I'm just looking into the general idea...)

I've been mostly unsuccessful thus far and would appreciate any help.

Also, The data appears as:

082438.577652 - T in: [A] accepted. ordID [F25Q6] timestamp [082438.575880] RefNumber [6018786] State [L]

Answer 1

Assuming you want to group your data by those issued within 1 second intervals on the second, we can make use of the fact that your data is ordered and that int(out_ts) truncates the timestamp to the second which we can use as a grouping key.

Simplest way to do the grouping would be to use itertools.groupby:

from itertools import groupby

data = get_time_deltas(INFILE)  
get_key = lambda x: int(x[0])  # function to get group key from data
bins = [(k, list(g)) for k, g in groupby(data, get_key)]

bins will be a list of tuples where the first value in the tuple is the key (integer, e.g. 082438) and the second value is the a list of data entries that were issued on that second (with timestamp = 082438.*).

Example usage:

# print out the number of messages for each second
for sec, data in bins:
    print('{0} --- {1}'.format(sec, len(data)))

# write (sec, msg_per_sec) out to CSV file
import csv
with open("test.csv", "w") as f:
    csv.writer(f).writerows((s, len(d)) for s, d in bins)

# get average message per second
message_counts = [len(d) for s, d in bins]
avg_msg_per_second = float(sum(message_count)) / len(message_count)

P.S. In this example, a list was used for bins so that the order of data is maintained. If you need random access to the data, consider using an OrderedDict instead.

---

Note that it is relatively straight-forward to adapt the solution to group by multiples of seconds. For example, to group by messages per minute (60 seconds), change the get_key function to:

get_key = lambda x: int(x[0] / 60)  # truncate timestamp to the minute

Answer 2

This is easier if you don't base your grid on time intervals with bisection.

Instead, do this. Transform each interval to a single number.

def map_time_to_interval_number( epoch, times )
    for t in times:
        delta= (t - epoch)
        delta_t= delta.days*60*60*24 + delta.seconds + delta.microseconds/1000000.0
        interval = delta_t / 50
        yield interval, t

counts = defaultdict( int )
epoch = min( data ) 
for interval, time in map_time_to_interval_number( epoch, data ):
    counts[interval] += 1

The interval will be an integer. 0 is the first 50-second interval. 1 is the second 50-second interval. etc.

You can reconstruct the timestamp from the interval knowing that each interval is 50-seconds wide and begins at epoch.