CODEWITHSUNDEEP
pythontuples
mburkenysdot
mburkenysdot

Reputation: 15

Extract the tuple with the largest item from sublist (find most recent report in directory)

I have a script that creates a directory listing of all pdfs within a certian series of subdirectories. The outputs are tuples that include the year of the file saved as a string as well as an id for the unit that generated the report that looks something like the following:

unit1, 2010
unit2, 2002
unit2, 2005
unit2, 2010 
unit3, 2003

What I'm looking to now do is create a report that finds the most recent report based on the tuple that contains the max value in its second item. Normally, I would do this in Access with a MAX query, however, I am trying to elimate that step since and write the extract all at once. Using my orginal code, my output would consist of the following:

unit1, '2010'
unit2, '2010'
unit3, '2003'

I did some looking around and realize that I need to change my script so that it would generate a list of the tuples that matched every unique id. Using the great answer I found from Split a list of tuples into sub-lists of the same tuple field I was able to get the results split into a group of sublists. This means my output is now the following:

[[(unit1, '2010')],[(unit12, '2010'), (unit2, '2010'), (unit2, '2005'), (unit2, '2002')],[(unit3, '2003']]

My difficulty now is trying to extract the tuple from each sublist that contains the highest value item. I tried the following:

import glob, os, itertools, operator  
dirtup = []
for f in glob.glob('P:\Office*\Technical*\Bureau*\T*\*\YR2*\R*\*\*.pdf'):
    fpath, fname = os.path.split(f)
    fyr = fpath[91:95]
    vcs = 'Volume'
    rname, extname = os.path.splitext(fname)
    rcid = fname[0:7]
    dirtup.append ((f, fyr, rcid, vcs))

dirtup2 = sorted(dirtup, key=operator.itemgetter(2))

for key, group in itertools.groupby(dirtup2, operator.itemgetter(2)):
    maxval = max(x[1] for x in dirtup2)

print [x for x in dirtup2 if x[1] == maxval]

This returns only the tuple that match the max of fyr rather then the max of fyr per each sublist.

Edit

Using mglison's first answer I was able to get the output (tuple that contained second item with max value).

Upvotes: 0

Views: 218

Answers (1)

mgilson
mgilson

Reputation: 310117

You can sort each sublist based on the particular field and take the first element of the sorted sublist.

for key,group in itertools.groupby(dirtup2,operator.itemgetter(2)):
    newlist=sorted(group,key=operator.itemgetter(1),reverse=True)
    tuple_with_max=newlist[0]
    print tuple_with_max

Upvotes: 1

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