Text processing/Max licenses in use: Difference between revisions

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=={{header|Python}}==
=={{header|Python}}==
Python 2.4+
<python>
<python>
out = 0
out = 0
maxout = -1
max_out = -1
maxtimes = []
max_times = []


for job in open('mlijobs.txt'):
for job in open('mlijobs.txt'):
out += (1 if "OUT" in job else -1)
if "OUT" in job:
if out > maxout:
out += 1
else:
maxout = out
maxtimes = []
out -= 1
if out == maxout:
if out > max_out:
maxtimes.append(job.split()[3])
max_out = out
max_times = []
if out == max_out:
max_times.append(job.split()[3])


print "Maximum simultaneous license use is", maxout, "at the following times:"
print "Maximum simultaneous license use is", max_out, "at the following times:"
for time in maxtimes:
for time in max_times:
print " ", time
print " ", time</python>
</python>
Example output:
Example output:
<pre>Maximum simultaneous license use is 99 at the following times:
<pre>Maximum simultaneous license use is 99 at the following times:

Revision as of 13:55, 17 November 2008

Task
Text processing/Max licenses in use
You are encouraged to solve this task according to the task description, using any language you may know.

A company currently pays a fixed sum for the use of a particular licensed software package. In determining if it has a good deal it decides to calculate its maximum use of the software from its license management log file.

Assume the softwares file faithfully records a checkout event when a copy of the software starts and a checkin event when the software finishes. An example of checkout and checkin events are:

 License OUT @ 2008/10/03_23:51:05 for job 4974
 ...
 License IN  @ 2008/10/04_00:18:22 for job 4974


Save the 10,000 line log file from here into a local file then write a program to scan the file extracting both the maximum licenses that were out at any time, and the time(s) at which this occurs.

APL

Works with: APL2
Translation of: J
      ⍝  Copy/paste file's contents into TXT (easiest), or TXT ← ⎕NREAD
      I  ←  TXT[;8+⎕IO]
      D  ←  TXT[;⎕IO+14+⍳19]
      lu ←  +\ ¯1 * 'OI' ⍳ I
      mx ←  (⎕IO+⍳⍴lu)/⍨lu= max ← ⌈/ lu
      ⎕  ←  'Maximum simultaneous license use is ' , ' at the following times:' ,⍨ ⍕max ⋄ ⎕←D[mx;]
Maximum simultaneous license use is 99 at the following times:
2008/10/03_08:39:34
2008/10/03_08:40:40

Fortran

Works with: Fortran version 90 and later
PROGRAM MAX_LICENSES
  IMPLICIT NONE

  INTEGER :: out=0, maxout=0, maxcount=0, err
  CHARACTER(50) :: line
  CHARACTER(19) :: maxtime(100)

  OPEN (UNIT=5, FILE="Licenses.txt", STATUS="OLD", IOSTAT=err)
  IF (err > 0) THEN
    WRITE(*,*) "Error opening file Licenses.txt"
    STOP
  END IF

  DO 
    READ(5, "(A)", IOSTAT=err) line
    IF (err == -1) EXIT          ! EOF detected
    IF (line(9:9) == "O") THEN
      out = out + 1
    ELSE IF (line(9:9) == "I") THEN
      out = out - 1
    END IF
    IF (out > maxout ) THEN
      maxout = maxout + 1
      maxcount = 1
      maxtime(maxcount) = line(15:33)
    ELSE IF (out == maxout) THEN
      maxcount = maxcount + 1
      maxtime(maxcount) = line(15:33)
    END IF
  END DO
 
  CLOSE(5)
 
  WRITE(*,"(A,I4,A)") "Maximum simultaneous license use is", maxout, " at the following times:"
  WRITE(*,"(A)") maxtime(1:maxcount)
 
END PROGRAM MAX_LICENSES

Output

Maximum simultaneous license use is  99 at the following times:
2008/10/03_08:39:34                                           
2008/10/03_08:40:40

J

   NB.  Parse data, select columns
   'I D' =: (8 ; 14+i.19) { ::]"1 L:0 ];._2 ] 1!:1 ::(''"_) <'licenses.txt'
   
   NB.  Calculate number of licenses used at any given time
   lu    =:  +/\ _1 ^ 'OI' i. I
   
   NB.  Find the maxima
   mx    =:  (I.@:= >./) lu
  
   NB.  Output results
   (mx { D) ,~ 'Maximum simultaneous license use is ' , ' at the following times:' ,~ ": {. ,mx { lu
Maximum simultaneous license use is 99 at the following times:
2008/10/03_08:39:34                                           
2008/10/03_08:40:40

Python

<python> out = 0 max_out = -1 max_times = []

for job in open('mlijobs.txt'):

   if "OUT" in job:
       out += 1
   else:
       out -= 1
   if out > max_out:
       max_out = out
       max_times = []
   if out == max_out:
       max_times.append(job.split()[3])

print "Maximum simultaneous license use is", max_out, "at the following times:" for time in max_times:

   print " ", time</python>

Example output:

Maximum simultaneous license use is 99 at the following times:
  2008/10/03_08:39:34
  2008/10/03_08:40:40