+- Hack Forums (https://hackforums.net)
+-- Forum: Tech (https://hackforums.net/forumdisplay.php?fid=444)
+--- Forum: Coding (https://hackforums.net/forumdisplay.php?fid=151)
+---- Forum: Software Development (https://hackforums.net/forumdisplay.php?fid=118)
+---- Thread: [Python] Dictionaries (/showthread.php?tid=5587725)
Dictionaries - twistingtide - Mar 29, 2017
1. So I have a dictionary with the set up below
The actor name, the movies they were in, and number to the right of the movies is the actors popularity
dictionary = {1: ['Actor1', {movie1, movie3, movie7}, 89], 2:['Actor35, {movie3, movie7}, 57], 3:['Actor7',{movie7} 64], 4:['Actor9, {movie90} 3 ]}
2. So If someone types in the number 3 (Or any number within the length of my dictionary they go to that actor) into my input they go to Actor7
3. Once they are at actor 7 they look at ALL the movies the actor has been in and compare to all the movies the other actors have been in
4. If any of the actors share a movie append there name to a new list in the order of decreasing popularity(Popularity is the last element in the list)
So it should look like this if they type in 3 because the only actors who share a movie with actor 7 is actor1 and actor35:
final_list = [Actor1,Actor35]
Thanks guys!!!🤗
* - (╯°□°）╯︵ ┻━━┻ - Mar 30, 2017
I'm on my phone, but to begin, you need to put the movies in a list instead of a dictionary (the {movies1...} part).
* - hunter24957 - Mar 30, 2017
NGL, that's the worst layout of a dictionary I've ever seen. I would fix it for you but I'm still not sure what you're going for...
Maybe use something like this:
PHP Code
<?php
{'Actor1': [movie1, movie3, movie7, 89], 'Actor35': [movie3, movie7, 57], 'Actor7': [movie7, 64], 'Actor9': [movie90, 3]}
And your dictionary keys are redundant, you can just use a list...
* - (╯°□°）╯︵ ┻━━┻ - Mar 30, 2017
This feels like homework to me, so I'm going to make it fun! Good ol one liner :D
Code
actors = [
("Actor 1", ["Movie 1", "Movie 3", "Movie 7"], 9),
("Actor 35", ["Movie 3", "Movie 7"], 57),
("Actor 7", ["Movie 7"], 64),
("Actor 9", ["Movie 90"], 3),
]
get_actors = lambda index: list(map(lambda x: x[0], sorted(((actor[0], actor[2]) for actor in actors if actor[0] is not actors[index - 1][0] and list(filter(lambda movie: movie in actors[index - 1][1], actor[1]))), key=lambda x: x[1], reverse=True)))
print(get_actors(3))* - Folks - Mar 30, 2017
(Mar 30, 2017 07:57 AM)(╯°□°）╯︵ ┻━━┻ Wrote: This feels like homework to me, so I'm going to make it fun! Good ol one liner :D
Codeactors = [
("Actor 1", ["Movie 1", "Movie 3", "Movie 7"], 9),
("Actor 35", ["Movie 3", "Movie 7"], 57),
("Actor 7", ["Movie 7"], 64),
("Actor 9", ["Movie 90"], 3),
]
get_actors = lambda index: list(map(lambda x: x[0], sorted(((actor[0], actor[2]) for actor in actors if actor[0] is not actors[index - 1][0] and list(filter(lambda movie: movie in actors[index - 1][1], actor[1]))), key=lambda x: x[1], reverse=True)))
print(get_actors(3))
One liners gave me an aneurism until I finally started to understand how to read through them lol
* - (╯°□°）╯︵ ┻━━┻ - Mar 30, 2017
(Mar 30, 2017 08:16 AM)Folks Wrote: One liners gave me an aneurism until I finally started to understand how to read through them lol
I wrote it out first and then I looked at it and figured I could one line it so I did.