First Solution

Here is how I approach the problem:

import json
import collections

if __name__ == '__main__':    
    # Load file into data
    with open('raw.json') as f:
        data = [json.loads(line) for line in f]

    # Calculate count and total
    time_total = collections.defaultdict(float)
    time_count = collections.defaultdict(int)
    for row in data:
        time_count[row['name']] += 1
        time_total[row['name']] += row['time']

    # Calculate average
    time_average = {}
    for name in time_count:
        time_average[name] = time_total[name] / time_count[name]

    # Report
    for name in sorted(time_count):
        print '{:<10} {:2} {:8.2f} {:8.2f}'.format(
            name,
            time_count[name],
            time_total[name],
            time_average[name])

Discussion

• data is a list of dict with keys such as name, time, ...
• I used three additional dictionaries to keep track of the count, total, and average per machine.
• I assume you want your calculation based on the time value. If not, it is an easy fix.
• The defaultdict is a nice way to tally numbers. If an int value is not already created, it will be created and assign value of 0, very convenient. You should look it up.

Second Solution

Here is a different approach: since your data looks like a table, why not use a database to handle your data. The advantage of this approach is you don't have to do calculations yourself.

import json
import sqlite3

if __name__ == '__main__':
    # Create an in-memory database for calculation
    connection = sqlite3.connect(':memory:')
    cursor = connection.cursor()
    cursor.execute('DROP TABLE IF EXISTS time_table')
    cursor.execute('CREATE TABLE time_table (name text, time real)')
    connection.commit()

    # Load file into database
    with open('raw.json') as f:
        for line in f:
            row = json.loads(line)
            cursor.execute('INSERT INTO time_table VALUES (?,?)', (row['name'], row['time']))
            connection.commit()

    # Report: print the name, count, sum, and average
    cursor.execute('SELECT name, COUNT(time), SUM(time), AVG(time) FROM time_table GROUP BY name')
    print '%-10s %8s %8s %8s' % ('NAME', 'COUNT', 'SUM', 'AVERAGE')
    for row in cursor.fetchall():
        print '%-10s %8d %8.2f %8.2f' % row

    connection.close()

Output

NAME          COUNT      SUM  AVERAGE
machine1          1    12.64    12.64
machine2          1    12.64    12.64
machine3          4    50.77    12.69
machine4          3    38.03    12.68
machine5          5    63.45    12.69

Discussion

• In this solution, I created an in-memory SQLite3 database
• Since we are only interested in the name and time columns, the table only contains those two.
• We got all the statistical functions such as SUM, COUNT, and AVG for free, just by using the database.

Addition to First Solution

To answer the question: Given machine5, how can I get the last value? By that, I assume you want to filter your data down to those containing machine5, then sort them by time and select the last row. For the first solution, append the following block of code and run it:

# Filter data: prints all rows with 'machine5'
print '\nFilter by machine5'
machine5 = [row for row in data if row['name'] == 'machine5']
machine5 = sorted(machine5, key=lambda row: int(row['time']))
pprint(machine5)

# Get the last instance
print '\nLast instance of machine5:'
latest_row = machine5[-1]
pprint(latest_row)

Don't forget to add the following at the beginning of the script:

from pprint import pprint

Output

Filter by machine5
[{u'name': u'machine5', u'time': 12.67007, u'value': 5.068},
 {u'name': u'machine5', u'time': 12.6801, u'value': 2.0868},
 {u'name': u'machine5', u'time': 12.6901, u'value': 12.633},
 {u'name': u'machine5', u'time': 12.69512, u'value': 13.13},
 {u'name': u'machine5', u'time': 12.71517, u'value': 131.633}]

Last instance of machine5:
{u'name': u'machine5', u'time': 12.71517, u'value': 131.633}

Discussion

If you do not want to sort the rows by time, then remove the sorted() line and that will give you the unsorted output.