dnf_count_files.py

#!/usr/bin/python
# James Shubin, @purpleidea, 2016+, AGPLv3+
# Count number of files in each package, and figure out which has the most
# We took a string based parsing approach to the xml filelists for simplicity
# When I ran this, the max was: kcbench-data-4.0, with 52116 files
# Verify with dnf repoquery --quiet -l kcbench-data-4.0 | wc -l
# To run this script, do something like the following:
# wget http://mirror.its.dal.ca/pub/fedora/linux/releases/23/Everything/x86_64/os/repodata/874f220caf48ccd307c203772c04b8550896c42a25f82b93bd17082d69df80db-filelists.xml.gz
# gunzip 874f220caf48ccd307c203772c04b8550896c42a25f82b93bd17082d69df80db-filelists.xml.gz
# time cat 874f220caf48ccd307c203772c04b8550896c42a25f82b93bd17082d69df80db-filelists.xml | ./dnf_count_files.py > /tmp/output
# head /tmp/output

import sys
import operator

count = {}
name = ""
m = 0
for line in sys.stdin:	# this is a dirty hack, don't parse xml this way
	if line.startswith("<package"):
		if name != "":
			count[name] = z # store

		start = line.find('name="')
		q1 = line.find('"', start)
		q2 = line.find('"', start+6) # name= is the 5+1
		name = line[q1+1:q2]
		z = 0
	if line.startswith("  <file"):
		z = z + 1

sorted_counts = sorted(count.items(), key=operator.itemgetter(1))
for ix in reversed(sorted_counts):
	print("%s\t%d" % (ix[0], ix[1]))