This challenge is interesting in that it is confirming to me how inexperienced my programming actually is. I can get to the solution but the answer shows how limited I am. Here is my solution to 2 (I’d copied and pasted the text from the source to a document 2_message.txt.)<\/p>\n
# find rare characters in the mess below:
maxoccurence = 1 #how many do I think is rare?
# load up the file
s = open(r\"2_message.txt\", \"r\").read()
#find the unique characters
uniqueones = set(s)
#find the rare characters (<= maxoccurence) and their position
rareones = []
for one in uniqueones:
position = []
if s.count(one) <=1:
position.append(one)
position.append(s.index(one))
rareones.append(position)
#rareones is now a multi dimensional list with the character and its position
from operator import itemgetter
solution = \"\"
for final in sorted(rareones, key=itemgetter(1)):
solution += final[0]
print solution <\/pre>\nBut here is the first of the suggested solutions <\/p>\n
import collections
data = ''.join([line.rstrip() for line in open('2_message.txt')])
OCCURRENCES = collections.OrderedDict()
for c in data: OCCURRENCES[c] = OCCURRENCES.get(c, 0) + 1
avgOC = len(data) \/\/ len(OCCURRENCES)
print ''.join([c for c in OCCURRENCES if OCCURRENCES[c] < avgOC]) <\/pre>\nI want to understand this! So I spend as much time working it out as I do solving the first problem.\u00a0 This is a great way of learning. I have had to learn what a Dictionary is in python<\/a> (a sort of associative 2 dimensional array).\u00a0 Link this to the collections library<\/a> and you have got something pretty powerful.
The code appears to<\/p>\n\n
- Load the data in to a massive string, removing the carriage returns.<\/li>\n
- Make a container \"Dictionary\" for the characters <\/li>\n
- Whiz through the massive string using the character as a key and simply adding 1 to the number attached to the key.\u00a0\u00a0<\/li>\n
- Getting the integer (floor) average (\/\/) for each letter if they were distributed evenly.<\/li>\n
- Concatenating all the items in the Dictionary that are less frequent than the average<\/li>\n<\/ol>\n
This is actually not that different to what I did!!!\u00a0 Just much more efficient. <\/p>\n
[c for c in OCCURRENCES if OCCURRENCES[c] < avgOC]<\/pre>\nIs very clever.\u00a0 It is one line that only outputs a string if the occurrences are less than average.\u00a0 I am still wrapping my head around it.<\/p>\n
My Python 3.2.3 Solution<\/h3>\n
# find rare characters in the mess below:
# official version
import collections
data = ''.join([line.rstrip() for line in open('2_message.txt')])
OCCURRENCES = collections.OrderedDict()
for c in data: OCCURRENCES[c] = OCCURRENCES.get(c, 0) + 1
# avgOC = len(data) \/\/ len(OCCURRENCES)
avgOC = 1
print (''.join([c for c in OCCURRENCES if OCCURRENCES[c] == avgOC]) )<\/pre>\nThis just needed the modification to \"Print\" though I did find on python 3 it didn't run because there was no collections library.<\/p>\n","protected":false},"excerpt":{"rendered":"
This challenge is interesting in that it is confirming to me how inexperienced my programming actually is. I can get to the solution but the answer shows how limited I am. Here is my solution to 2 (I’d copied and pasted the text from the source to a document 2_message.txt.) My Python 2.7 Solution # … Continue reading The Python Challenge – 2<\/span>