Create an infinite nested list of list - python

I have a task to make an list of list following specific rules. List must represent a tree with a root and branches from this root with specific color. Each branch should be represented as a list of its child elements (one black branch generates 3 white; 1 white branch generates 2 black). For example: root=['black'], first branches [['white','white','white']], next iteration should be [[[black,black],[black,black],[black,black]]] and so on.
This infinite list should be stored in global variable. Is it possible?
My code, which generates such list do it only for for a predetermined number of steps.
root = ['b']
def change(root):
for index, item in enumerate(root):
if isinstance(item, list):
change(item)
elif item == 'b':
root[index] = ['w','w','w']
elif item == 'w':
root[index] = ['b','b']
return root
for i in range(3):
tree=change(root)
print(tree)
How can I generate infinite list if it is possible?

If I'm understanding your requirements correctly, this creates the entire infinite tree in memory (but requiring only finite memory because it refers to itself):
black = []
white = [black, black]
black.append([white, white, white])
root = black # or white, if you prefer
Note that the 'b' and 'w' leaf nodes never appear in the tree, as you can never actually reach them.

In the comments on your question, I suggested:
#MartijnPieters Perhaps he [OP] means he needs to write a program that yields a potentially-infinite list of lists that follows a certain pattern.
You replied that that's what you want, so I had a go at it - the code I post involves using a function generator, which you can use to generate a potentially never-ending sequence that follows the pattern you specify (if we were to assume infinite memory).
A short while ago I saw an incredible solution posted by someone to generate the sequence of coprime pairs. It was the first time I'd been made aware of the whole mind-blowing concept of "self-recursive generators". You can use them to do some pretty awesome things, like generate the Kolakoski sequence.
In any case, I had a gut feeling that I could use self-recursive generators to solve your problem. The goods news is that my code works. The bad news is that I have no idea why my code works. Here it is:
from enum import Enum
class Colour(Enum):
BLACK = 'black'
WHITE = 'white'
def generate_tree(curr_colour = Colour.BLACK):
yield [curr_colour]
for sub_tree in generate_tree(Colour.WHITE if curr_colour == Colour.BLACK else Colour.BLACK):
if curr_colour == Colour.WHITE:
tree = [sub_tree, sub_tree]
else:
tree = [sub_tree, sub_tree, sub_tree]
yield tree
if __name__ == '__main__':
generator = generate_tree()
for _ in range(3):
print(next(generator))
Output for n=3
[<Colour.BLACK: 'black'>]
[[<Colour.WHITE: 'white'>], [<Colour.WHITE: 'white'>], [<Colour.WHITE: 'white'>]]
[[[<Colour.BLACK: 'black'>], [<Colour.BLACK: 'black'>]], [[<Colour.BLACK: 'black'>], [<Colour.BLACK: 'black'>]], [[<Colour.BLACK: 'black'>], [<Colour.BLACK: 'black'>]]]
You can use the variant below to have a program that keeps generating the tree indefinitely, waiting at each iteration for user input:
if __name__ == '__main__':
generator = generate_tree()
while True:
print(next(generator))
input()

Related

QTextEdit- Change level of one list item without creating new list

I made a custom QTextEdit class and implemented a nesting feature for QTextList.
The bullet list works fine but the numbered list is difficult to realize.
I want to have nested numbered lists look like this:
Item
1.1 Nested Item
1.2 Nested Item
     1.2.1 Nested Item
...
Item 2
...
The numbering of the individual items works good but when I decrease the list level the numbering begins from 1 because I use createList(). I know that the composition of the numbering would also need to be changed. I could implement a check if it's an increase or decrease. Decreasing the level like this works well for the bullet list because there is no numbering but I haven't found an alternative for the numbered list.
Is there any way to increase/decrease the level of a single list item so that the numbers are changed accordingly?
Credits to this thread from where I adopted most parts of the code:
QTextEdit: Tab key changes bullet list indentation
# incr=1 -> increase level; incr=-1 -> decrease level
def changeIndentation(self, incr=1) -> None:
cursor = self.textCursor()
tlist = cursor.currentList()
if tlist:
lformat = tlist.format()
lstyle = lformat.style()
lformat.setIndent(lformat.indent() + incr)
# Bullet list
if lstyle == QTextListFormat.Style.ListDisc or lstyle == QTextListFormat.Style.ListCircle or lstyle == QTextListFormat.Style.ListSquare:
if lformat.indent() == 1: lformat.setStyle(QTextListFormat.Style.ListDisc)
elif lformat.indent() == 2: lformat.setStyle(QTextListFormat.Style.ListCircle)
elif lformat.indent() >= 3: lformat.setStyle(QTextListFormat.Style.ListSquare)
# Numbered list
elif lformat.style() == QTextListFormat.Style.ListDecimal:
lformat.setNumberPrefix(f"{tlist.format().numberPrefix()}{tlist.itemNumber(cursor.block())}.")
if lformat.indent() > 1:
lformat.setNumberSuffix("")
cursor.createList(lformat)
I tried setIndent() of QTextListFormat as described in the thread I mentioned before but it sets the whole indentation of the list and I only want to change the level of a single item.

AIO Castle Cavalry - My code is too slow, is there a way I can shorten this?

So I am currently preparing for a competition (Australian Informatics Olympiad) and in the training hub, there is a problem in AIO 2018 intermediate called Castle Cavalry. I finished it:
input = open("cavalryin.txt").read()
output = open("cavalryout.txt", "w")
squad = input.split()
total = squad[0]
squad.remove(squad[0])
squad_sizes = squad.copy()
squad_sizes = list(set(squad))
yn = []
for i in range(len(squad_sizes)):
n = squad.count(squad_sizes[i])
if int(squad_sizes[i]) == 1 and int(n) == int(total):
yn.append(1)
elif int(n) == int(squad_sizes[i]):
yn.append(1)
elif int(n) != int(squad_sizes[i]):
yn.append(2)
ynn = list(set(yn))
if len(ynn) == 1 and int(ynn[0]) == 1:
output.write("YES")
else:
output.write("NO")
output.close()
I submitted this code and I didn't pass because it was too slow, at 1.952secs. The time limit is 1.000 secs. I wasn't sure how I would shorten this, as to me it looks fine. PLEASE keep in mind I am still learning, and I am only an amateur. I started coding only this year, so if the answer is quite obvious, sorry for wasting your time 😅.
Thank you for helping me out!
One performance issue is calling int() over and over on the same entity, or on things that are already int:
if int(squad_sizes[i]) == 1 and int(n) == int(total):
elif int(n) == int(squad_sizes[i]):
elif int(n) != int(squad_sizes[i]):
if len(ynn) == 1 and int(ynn[0]) == 1:
But the real problem is your code doesn't work. And making it faster won't change that. Consider the input:
4
2
2
2
2
Your code will output "NO" (with missing newline) despite it being a valid configuration. This is due to your collapsing the squad sizes using set() early in your code. You've thrown away vital information and are only really testing a subset of the data. For comparison, here's my complete rewrite that I believe handles the input correctly:
with open("cavalryin.txt") as input_file:
string = input_file.read()
total, *squad_sizes = map(int, string.split())
success = True
while squad_sizes:
squad_size = squad_sizes.pop()
for _ in range(1, squad_size):
try:
squad_sizes.remove(squad_size) # eliminate n - 1 others like me
except ValueError:
success = False
break
else: # no break
continue
break
with open("cavalryout.txt", "w") as output_file:
print("YES" if success else "NO", file=output_file)
Note that I convert all the input to int early on so I don't have to consider that issue again. I don't know whether this will meet AIO's timing constraints.
I can see some things in there that might be inefficient, but the best way to optimize code is to profile it: run it with a profiler and sample data.
You can easily waste time trying to speed up parts that don't need it without having much effect. Read up on the cProfile module in the standard library to see how to do this and interpret the output. A profiling tutorial is probably too long to reproduce here.
My suggestions, without profiling,
squad.remove(squad[0])
Removing the start of a big list is slow, because the rest of the list has to be copied as it is shifted down. (Removing the end of the list is faster, because lists are typically backed by arrays that are overallocated (more slots than elements) anyway, to make .append()s fast, so it only has to decrease the length and can keep the same array.
It would be better to set this to a dummy value and remove it when you convert it to a set (sets are backed by hash tables, so removals are fast), e.g.
dummy = object()
squad[0] = dummy # len() didn't change. No shifting required.
...
squad_sizes = set(squad)
squad_sizes.remove(dummy) # Fast lookup by hash code.
Since we know these will all be strings, you can just use None instead of a dummy object, but the above technique works even when your list might contain Nones.
squad_sizes = squad.copy()
This line isn't required; it's just doing extra work. The set() already makes a shallow copy.
n = squad.count(squad_sizes[i])
This line might be the real bottleneck. It's effectively a loop inside a loop, so it basically has to scan the whole list for each outer loop. Consider using collections.Counter for this task instead. You generate the count table once outside the loop, and then just look up the numbers for each string.
You can also avoid generating the set altogether if you do this. Just use the Counter object's keys for your set.
Another point unrelated to performance. It's unpythonic to use indexes like [i] when you don't need them. A for loop can get elements from an iterable and assign them to variables in one step:
from collections import Counter
...
count_table = Counter(squad)
for squad_size, n in count_table.items():
...
You can collect all occurences of the preferred number for each knight in a dictionary.
Then test if the number of knights with a given preferred number is divisible by that number.
with open('cavalryin.txt', 'r') as f:
lines = f.readlines()
# convert to int
list_int = [int(a) for a in lines]
#initialise counting dictionary: key: preferred number, item: empty list to collect all knights with preferred number.
collect_dict = {a:[] for a in range(1,1+max(list_int[1:]))}
print(collect_dict)
# loop though list, ignoring first entry.
for a in list_int[1:]:
collect_dict[a].append(a)
# initialise output
out='YES'
for key, item in collect_dict.items():
# check number of items with preference for number is divisilbe
# by that number
if item: # if list has entries:
if (len(item) % key) > 0:
out='NO'
break
with open('cavalryout.txt', 'w') as f:
f.write(out)

Algorithm to return all possible paths in this program to a nested list

So I have a game with a function findViableMoves(base). If i call this function at the start with the parameter base, I get an output [move1, move2 ... moven] which denotes all of the n viable moves that the user can perform give the state base. (there are in fact 2 root moves)
Upon performing a move, say move2 - base gets changed depending on the move, the function gets called again and now we have an output for findViableMoves(base) of [move21,move22 .... move2n].
Depth-first-tree
If you look at this diagram, it's a very similar situation - there's no looping back, it's just a normal tree. I need a program that performs a depth-first search (i think?) on all the possible moves given a starting state of base, and then returns then in a list as such:
[[move1,move11,move111],[move1,move11,move112],....[moven,moven1,moven11],...]
There will be more elements in these lists (14 at most), but I was just wondering if someone could provide any hints over how I can build an algorithm to do this? Efficiency doesn't really matter to me as there isn't too many paths, I just want it done for now.
I'm not 100% clear on what you're after, but if you have a list or similar iterable that is changing while the loop is happening you could try something like the below.
This example allows the list and the loop condition to both remain dynamic during the loop execution.
import random
import sys
import time
changing_list = ['A', 27, 0.12]
def perform_operation(changing_list, counter):
sometimes_add_another_element_threshold = 0.6
if random.random() > sometimes_add_another_element_threshold:
changing_list.append(random.random())
print(changing_list[counter])
def main(z=0):
safety_limit = 100
counter = 0
condition = True
while condition and counter < safety_limit:
perform_operation(changing_list, counter)
counter += 1
condition = counter<len(changing_list)
print("loop finished")
if __name__ == '__main__':
start_time = time.time()
main(int(sys.argv[1])) if len(sys.argv)>1 else main()
print(time.time() - start_time)
which provides output of varying length that looks something like:
A
27
0.12
0.21045788812161237
0.20230442292518247
loop finished
0.0058634281158447266

Creating a list of dictionaries where each dictionary contains another dictionary as values

Im writing an algorithm in Python which plays this game.
The current state of the board of tiles in the game is a dictionary in the form of:
{
<tile_id>: {
'counters': <number of counters on tile or None>,
'player': <player id of the player who holds the tile or None>,
'neighbours': <list of ids of neighbouring tile>
},
...
}
I have another dictionary which stores all of my tiles which are 'full' (i.e. a tile which has one less counter than its number of neighbours and where the player is me) This dictionary, full_tiles, is in the same form as the board dictionary above.
I am now trying to create a list, chains, where each element in the list is a dictionary of my full tiles that are neighbouring at least one other full tile (i.e a chain of full tiles). So this will be a list of all my seperate chains on the board.
Here is my code so far:
for tile_id, tile in full_tiles.items(): #iterates through all full tiles
current_tile = {tile_id : tile} #temporarily stores current tile
if not chains: #if chains list is empty
chains.append(current_tile) #begin list
else: #if list is not empty
for index, chain in enumerate(chains): #iterate though list of chains
if not (current_tile in chain): #if current tile is not in current chain
for tile_id2, tile2 in chain.items(): #iterate through tiles in current chain
for neighbour in tile2["neighbours"]: #iterate through each neighbour of current tile
#neighbour is in the form of tile_id
if neighbour in chain: #if current tile's neighbour is in chain
chain[tile_id] = tile #add tile to chain
It is very difficult for me to test and debug my code and check if it is working correctly as the code can only be run in an application that simulates the game. As you can see, there is quite a lot going on in this block of code with all of the nested loops which are difficult to follow. I cannot seem to think straight at the minute and so I cannot determine if this mess, in all honesty, will function as I hope.
While I am writing this, I have just realised that - on line 7 of this code - I am only checking if the current tile is not in the current chain and so there will be intersecting chains which, of course, will be a mess. Instead of this, I need to first check if the current tile is in not in any of the chains, not just one.
Apart from this error, will my code achieve what I am attempting? Or can you recommend a simpler, neater way to do it? (There has to be!)
Also, let me know if I have not given enough information on how the code is run or if I need to explain anything further, such as the contents of the board dictionary.
Thank you for any help in advance.
EDIT: Unfortunately, I was under a time constraint to complete this project, and as it was my first time ever working with Python, I currently lack the knowledge in the language to optimise my solution using the sources given below. Here is my final extremely ugly and messy solution to this problem which, in the end, worked fine and wasn't terribly inefficient given the small data set that the code works on.
for x in range(0, len(my_hexplode_chains) - 1):
match_found = False
for y in range(x + 1, len(my_hexplode_chains)):
for tile_id_x, tile_x in my_hexplode_chains[x].items(): #compare each chain in list
for tile_id_y, tile_y in my_hexplode_chains[y].items(): #to every other chain
for neighbour in tile_x["neighbours"]: #if tiles in different lists
if neighbour == tile_id_y: #are neighbours
match_found = True
my_hexplode_chains[x].update(my_hexplode_chains[y]) #append one chain to the other
del my_hexplode_chains[y] #delete appended chain
if match_found: #continue loop at next chain
break #very ugly way to do this
if match_found:
break
if match_found:
break
if match_found:
break
How about this optimization?
def find_match (my_hexplode_chains):
x = 0
len_chain = len(my_hexplode_chains)
while x <= len_chain:
y = x + 1
for tile_id_x, tile_x in my_hexplode_chains[x].items():
for tile_id_y, tile_y in my_hexplode_chains[y].items():
if tile_id_y in tile_x["neighbours"]:
my_hexplode_chains[x].update(my_hexplode_chains[y])
del my_hexplode_chains[y]
return True
x += 1
return False
You could pass this function after each move in your game and trace the output.

Python: How to speed up creating of objects?

I'm creating objects derived from a rather large txt file. My code is working properly but takes a long time to run. This is because the elements I'm looking for in the first place are not ordered and not (necessarily) unique. For example I am looking for a digit-code that might be used twice in the file but could be in the first and the last row. My idea was to check how often a certain code is used...
counter=collections.Counter([l[3] for l in self.body])
...and then loop through the counter. Advance: if a code is only used once you don't have to iterate over the whole file. However You are stuck with a lot of iterations which makes the process really slow.
So my question really is: how can I improve my code? Another idea of course is to oder the data first. But that could take quite long as well.
The crucial part is this method:
def get_pc(self):
counter=collections.Counter([l[3] for l in self.body])
# This returns something like this {'187':'2', '199':'1',...}
pcode = []
#loop through entries of counter
for k,v in counter.iteritems():
i = 0
#find post code in body
for l in self.body:
if i == v:
break
# find fist appearence of key
if l[3] == k:
#first encounter...
if i == 0:
#...so create object
self.pc = CodeCana(k,l[2])
pcode.append(self.pc)
i += 1
# make attributes
self.pc.attr((l[0],l[1]),l[4])
if v <= 1:
break
return pcode
I hope the code explains the problem sufficiently. If not, let me know and I will expand the provided information.
You are looping over body way too many times. Collapse this into one loop, and track the CodeCana items in a dictionary instead:
def get_pc(self):
pcs = dict()
pcode = []
for l in self.body:
pc = pcs.get(l[3])
if pc is None:
pc = pcs[l[3]] = CodeCana(l[3], l[2])
pcode.append(pc)
pc.attr((l[0],l[1]),l[4])
return pcode
Counting all items first then trying to limit looping over body by that many times while still looping over all the different types of items defeats the purpose somewhat...
You may want to consider giving the various indices in l names. You can use tuple unpacking:
for foo, bar, baz, egg, ham in self.body:
pc = pcs.get(egg)
if pc is None:
pc = pcs[egg] = CodeCana(egg, baz)
pcode.append(pc)
pc.attr((foo, bar), ham)
but building body out of a namedtuple-based class would help in code documentation and debugging even more.

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