Agenda
- Announcements
- Structure
- Tuple review
- APT and for loops
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Agenda Announcements Structure Tuple review APT and for loops - - PowerPoint PPT Presentation
Agenda Announcements Structure Tuple review APT and for loops 1/14/2013 CompSci101 Peter Lorensen 1 Variable Type High level Value Modules Functions( ) Functions( ) Datastructures code Variables (String) Datastructures
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High level Modules Functions( ) code
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Functions( ) Variables Datastructures Control structures Functions( ) Variable Type Value Datastructures (String) List Set Tuple Dictionary Control structures If For loop While loop (Exceptions)
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def load_lines(filename): lines = [] f = open(filename) for line in f.readlines(): line = line.strip() lines.append(line) return lines
file.read([size]) Read at most size bytes from the file (less if… file.readlines([sizehint]) Read until EOF using readline() and return a list… file.seek(offset[, whence]) Set the file’s current position, like stdio‘s fseek()… file.write(str) Write a string to the file. There is no return… ......... ........
File Object
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def whosDishonest(club1, club2, club3): guilty = set() # Empty set """ Making a list of clubs converted to sets""" allClubs = [ set(club1), set(club2), set(club3)] for redIndx in range( len( allClubs) ): for greenIndx in range(redIndx, len( allClubs)): common = allClubs[redIndx] & allClubs[greenIndx] guilty = guilty.union( common ) for redIndx in range( len( allClubs) ): for greenIndx in range(redIndx+1, len( allClubs)): common = allClubs[redIndx] & allClubs[greenIndx] guilty = guilty.union( common ) for redIndx in range( len( allClubs) ): for greenIndx in range(redIndx, len( allClubs)-1): common = allClubs[redIndx] & allClubs[greenIndx] guilty = guilty.union( common ) result = list( guilty ) # Converting to list result.sort() # Sorting return result
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2-friends consist of to counts: 1) all your friends (called immediate friends) 2) all of your friends friends, BUT you can NOT count someone twice! Run through all people and for each of them do the following:
a) count the number of their immediate friends. b) for each immediate friend, count his friends. a and b are added. The result is compared to the previous highest found value and if higher then replaces it. The final high score is returned.
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