Re: [Tutor] Fastest find in 2 2D lists with else statement
On Fri, Mar 20, 2015 at 08:22:03AM -0400, Kale Good wrote:
Hello all,
I'm new to python and a bit of a weekend-warrior programmer (guitar
teacher by trade), so there are plenty of computer sciencey concepts
that I haven't grasped yet, and I think I'm bumping up against those now.
Read from separate csv files, I have something like
a=[['bass',9],['eagle',36],['human',68]]
b=[['bass','fish'],['eagle','bird'],['dog',land']]
c=[[1,'fish','water'],[2, 'mammal','land'],[3,'bird','air']]
What do the numbers 9, 36, 68 in a list mean?
If I'm reading b correctly, you say a bass is a fish, an eagle is a
bird, and a dog is a land.
What do the numbers 1, 2, 3 in c list mean?
What I want is to return where each animal lives. What I've been able to
glisten from the manuals and stack overflow is:
Let's start by writing in English how you would do this. My guess is:
For each animal:
find out what kind of animal it is;
then look up where that kind of animal lives.
This tells you that the most natural sort of information you want is a
list of animals:
# [...] is used for lists
animals = [bass, eagle, human, spider, salmon]
plus a dict that associates each animal with its kind:
# {...} is used for dicts
kinds = {bass: fish,
eagle: bird,
human: mammal,
spider: arachnid,
salmon: fish,
}
plus a second dict that maps each kind to a location:
locations = {fish: water,
bird: air,
mammal: land,
insect: land,
arachnid: land,
mollusk: water,
}
Then your code becomes trivially easy!
for animal in animals:
kind = kinds[animal]
print(animal, lives in or on, locations[kind])
The dict lookup kinds[animal] takes care of all the searching for you,
no need to write a for loop at all.
To handle the case where the animal or kind is unknown:
for animal in animals:
kind = kinds.get(animal, None)
if kind is None:
# Ask the user what sort of kind of animal it is.
# In Python 2, you MUST use raw_input instead of input!
kind = input(What kind of animal is '%s'? % animal)
kind = kind.strip().lower()
kinds[animal] = kind
location = locations.get(kind, None)
if location is None:
# Ask the user where this kind of animal lives.
# In Python 2, you MUST use raw_input instead of input!
location = input(Where do '%s' animals live? % kind)
location = location.strip().lower()
locations[kind] = location
print(animal, lives in or on, location)
Note that direct lookups into a dict are written with square brackets:
kinds[animal]
but optional lookups with a default value use round brackets
(parentheses):
kinds.get(animal, None) # None is the default.
Also, there's nothing you can write in Python which will be faster than
looking it up in a dict. If you have concerns about looping over lists
repeatedly, you are right to worry about speed. But if you can get the
same data into a dict or two, dict lookups are *blazingly* fast.
Are you able to move your data into the list and dictionary formats
shown above? The secret to programming successfully is that getting the
data into the right format is half the battle. Once you have the data in
the right format, everything else becomes so much simpler.
For example, I might take your a list and convert it like this:
a=[['bass',9],['eagle',36],['human',68]]
animals = [sublist[0] for sublist in a]
That extracts out the strings bass, eagle etc. and puts them into a
list called animals.
Converting your b list to a dict should be even easier:
b=[['bass', 'fish'],['eagle', 'bird'],['dog', 'mammal']]
kinds = dict(b)
Can you work out how to get the locations you need, or do you need more
help? Feel free to ask any questions you have!
--
Steve
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Re: [Tutor] Fastest find in 2 2D lists with else statement
On 20/03/2015 12:22, Kale Good wrote: Hello all, I'm new to python and a bit of a weekend-warrior programmer (guitar teacher by trade), so there are plenty of computer sciencey concepts that I haven't grasped yet, and I think I'm bumping up against those now. Welcome to the club :) Read from separate csv files, I have something like a=[['bass',9],['eagle',36],['human',68]] b=[['bass','fish'],['eagle','bird'],['dog',land']] c=[[1,'fish','water'],[2, 'mammal','land'],[3,'bird','air']] What I want is to return where each animal lives. What I've been able to glisten from the manuals and stack overflow is: for i in range(len(a)): for x in range(len(b)): if a in b[x][1]: clss = (b[x][0]) occurrence w/ else statement for y in range(len(c)): if clss in c[y][1]: place = (c[y][2]) a[i].append(place) Classic newbie stuff that immediately flags up a code smell. You do not need to loop around Python containers using indexes. This is typically written something like. for animal in animals: doSomething(animal) However see my comment below, you probably don't need so many loops if you hold your data in dicts. a)I'd like to have an else statement; if a match isn't found in b, prompt the user for one of the values in c[:][0]. (no need to print the list c; the user will have that available elsewhere) b)Is there a faster way to do this? List a has ~300 elements, while list b has ~45000, so I'm curious if I can make it faster. Take a look at the DictReader here https://docs.python.org/3/library/csv.html as when it comes to looking things up dicts are far faster than lists. Try coding it up and come back to us if you get any problems. We don't bite although I have been known to bark :) -I only need to find the first match, so it seems a next statement would do the trick. However, I can't figure out how to use next for finding in two 2d arrays. -From my understanding, for this use, b needs to be a list. However, each line is unique, so it could be a set if necessary. Thanks in advance. Best, Kale -- My fellow Pythonistas, ask not what our language can do for you, ask what you can do for our language. Mark Lawrence ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
[Tutor] Fastest find in 2 2D lists with else statement
Hello all, I'm new to python and a bit of a weekend-warrior programmer (guitar teacher by trade), so there are plenty of computer sciencey concepts that I haven't grasped yet, and I think I'm bumping up against those now. Read from separate csv files, I have something like a=[['bass',9],['eagle',36],['human',68]] b=[['bass','fish'],['eagle','bird'],['dog',land']] c=[[1,'fish','water'],[2, 'mammal','land'],[3,'bird','air']] What I want is to return where each animal lives. What I've been able to glisten from the manuals and stack overflow is: for i in range(len(a)): for x in range(len(b)): if a in b[x][1]: clss = (b[x][0]) occurrence w/ else statement for y in range(len(c)): if clss in c[y][1]: place = (c[y][2]) a[i].append(place) a)I'd like to have an else statement; if a match isn't found in b, prompt the user for one of the values in c[:][0]. (no need to print the list c; the user will have that available elsewhere) b)Is there a faster way to do this? List a has ~300 elements, while list b has ~45000, so I'm curious if I can make it faster. -I only need to find the first match, so it seems a next statement would do the trick. However, I can't figure out how to use next for finding in two 2d arrays. -From my understanding, for this use, b needs to be a list. However, each line is unique, so it could be a set if necessary. Thanks in advance. Best, Kale -- Kale Good: Guitar Instructor ♫ phillyguitarlessons.com http://phillyguitarlessons.com phone: (215)260-5383 * 4705 Baltimore Ave, Phila, PA 19143 :Mailing Lessons * 1867 Frankford Ave. Phila, PA 19125 :Lessons Google+ https://plus.google.com/b/105422331794047992190/ Facebook http://facebook.com/KaleGoodGuitarStudio Read my article The Seven Secrets to Six String Success http://www.guitarnoise.com/lesson/seven-secrets-to-six-string-success/ at GuitarNoise.com http://guitarnoise.com Leading the Journey from No-Skills-Guitarist to Talented Musician! ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
