I'm trying to make CV website which has a sudoku generator and solver. The generator and solver are two separate functions in the functions.py file imported into the views.py file.
My sudokus are stored in a list of 9 lists of 9 ints.
from views.py:
from functions import *
def sudoku(request):
grid = make_sudoku() # generates unsolved sudoku list
solved = solve(grid) # generates solved list of unsolved list fed in.
count = [i for i in range(9)]
context = {
'grid': grid,
'solved': solved,
'count': count
}
return render(request, 'main/sudoku.html', context)
if I print grid, I get an unsolved sudoku list. If I print solved, I get a the same list which has been solved. Everything works dandy up until that point, but if I go to sudoku.html and type {{ grid }}, I get a solved sudoku list.
As the tree said to the lumberjack, I'm STUMPED! I'm completely baffled as to why this might happen because at no point in sudoku.html do I refer to grid or solved outside of passing them on to sudoku.js which actually makes the puzzle.
your solve() function probably is not only returning the solved grid but it'd also modify the existing grid and solve it in place, it's not making a copy but solving it on the same grid. read more about passing by value and passing by reference (here grid is being passes by reference)
https://flexiple.com/python/python-pass-by-reference/
additionally, you can replace
count = [i for i in range(9)]
by
count = list(range(9))
Related
I have a list that contains sublists. The sequence of the sublist is fixed, as are the number of elements.
schedule = [['date1', 'action1', beginvalue1, endvalue1],
['date2', 'action2', beginvalue2, endvalue2],
...
]
Say, I have a date and I want find what I have to do on that date, meaning I require to find the contents of the entire sublist, given only the date.
I did the following (which works): I created a intermediate list, with all the first values of the sublists. Based on the index i was able to retrieve its entire contents, as follows:
dt = 'date150' # To just have a value to make underlying code more clear
ls_intermediate = [item[0] for item in schedule]
index = ls_intermediate.index(dt)
print(schedule[index])
It works but it just does not seem the Python way to do this. How can I improve this piece of code?
To be complete: there are no double 'date' entries in the list. Every date is unique and appears only once.
Learning Python, and having quite a journey in front of me...
thank you!
I'm new to coding and have searched as best I can to find out how to solve this before asking.
I'm trying to pull information from poloniex.com REST api, which is in JSon format I believe. I can import the data, and work with it a little bit, but when I try to call and use the elements in the contained dictionaries, I get "'unicode' object not callable". How can I use this information? The end goal with this data is to pull the "BTC: "(volume)" for each coin pair and test if it is <100, and if not, append it to a new list.
The data is presented like this or you can see yourself at https://poloniex.com/public?command=return24hVolume:
{"BTC_LTC":{"BTC":"2.23248854","LTC":"87.10381314"},"BTC_NXT":{"BTC":"0.981616","NXT":"14145"}, ... "totalBTC":"81.89657704","totalLTC":"78.52083806"}
And my code I've been trying to get to work with currently looks like this(I've tried to iterate the information I want a million different ways, so I dunno what example to give for that part, but this is how I am importing the data):
returnvolume = urllib2.urlopen(urllib2.Request('https://poloniex.com/public?command=return24hVolume'))
coinvolume = json.loads(returnvolume.read())
coinvolume = dict(coinvolume)
No matter how I try to use the data I've pulled, I get an error stating:
"unicode' object not callable."
I'd really appreciate a little help, I'm concerned I may be approaching this the wrong way as I haven't been able to get anything to work, or maybe I'm just missing something rudimentary, I'm not sure.
Thank you very much for your time!
Thanks to another user, downshift, I have discovered the answer!
d = {}
for k, v in coinvolume.items():
try:
if float(v['BTC']) > 100:
d[k] = v
except KeyError:
d[k] = v
except TypeError:
if v > 100:
d[k] = k
This creates a new list, d, and adds every coin with a 'BTC' volume > 100 to this new list.
Thanks again downshift, and I hope this helps others as well!
How can i duplicate a list of lists (or any other types) in a way that the resulting lists are new objects and not references to the old ones? As an example i have the following list of lists:
l=[[1,2],[3,4]]
what i want as result is:
l=[[1,2],[3,4],[1,2],[3,4]]
If i do l*=2 the new sub-lists are references to the old sub-lists.
Doing l[0].append("python") will result in
l=[[1,2,'python'],[3,4],[1,2,'python'],[3,4]]
Also creating a new list like:
l2=list(l)
or
l2=l[:]
doesn't solve the problem. I want to have new sub-lists which are independent of their origin and which upon changing have no impact on their old fellows. How can i do this i python?
In general, the best way to copy a nested data structure so that copies get made of all the references (not just the ones at the top level) is to use copy.deepcopy. In your nested list example, you can do:
l.extend(copy.deepcopy(l))
deepcopy will still work even if the data structure contains references to itself, or multiple references to the same object. It usually works for objects stored as attributes on an instances of custom classes too. You can define a __deepcopy__ method if you want to give a class special copying behavior (e.g. if some of its attributes are bookkeeping data that shouldn't be copied).
Here's a version of your nested list example code using instances of a linked list class rather than Python lists. copy.deepcopy does the right thing!
class linked_list(object):
def __init__(self, value, next=None):
self.value = value
self.next = next
def __repr__(self):
if self.next is not None:
return "({!r})->{!r}".format(self.value, self.next)
else:
return "({!r})".format self.value
lst = linked_list(linked_list(1, linked_list(2)),
linked_list(linked_list(3, linked_list(4))))
print(lst) # prints ((1)->(2))->((3)->(4))
lst.next.next = copy.deepcopy(lst)
print(lst) # prints ((1)->(2))->((3)->(4))->((1)->(2))->((3)->(4))
lst.value.value = 5
print(lst) # prints ((5)->(2))->((3)->(4))->((1)->(2))->((3)->(4))
I need to create a structure, in my mind similar to an array of linked lists (where a python list = array and dictionary = linked list). I have a list called blocks, and this is something like what I am looking to make:
blocks[0] = {dictionary},{dictionary},{dictionary},...
blocks[1] = {dictionary},{dictionary},{dictionary},...
etc..
currently I build the blocks as such:
blocks = []
blocks.append[()]
blocks.append[()]
blocks.append[()]
blocks.append[()]
I know that must look ridiculous. I just cannot see in my head what that just made, which is part of my problem. I assign to a block from a different list of dictionary items. Here is a brief overview of how a single block is created...
hold = {}
hold['file']=file
hold['count']=count
hold['mass']=mass_lbs
mg1.append(hold)
##this append can happen several times to mg1
blocks[i].append(mg1[j])
##where i is an index for the block I want to append to, and j is the list index corresponding to whichever dictionary item of mg1 I want to grab.
The reason I want these four main indices in blocks is so that I have shorter code with just the one list instead of block1 block2 block3 block4, which would just make the code way longer than it is now.
Okay, going off of what was discussed in the comments, you're looking for a simple way to create a structure that is a list of four items where each item is a list of dictionaries, and all the dictionaries in one of those lists have the same keys but not necessarily the same values. However, if you know exactly what keys each dictionary will have and that never changes, then it might be worth it to consider making them classes that wrap dictionaries and have each of the four lists be a list of objects. This would be easier to keep in your head, and a bit more Pythonic in my opinion. You also gain the advantage of ensuring that the keys in the dictionary are static, plus you can define helper methods. And by emulating the methods of a container type, you can still use dictionary syntax.
class BlockA:
def __init__(self):
self.dictionary = {'file':None, 'count':None, 'mass':None }
def __len__(self):
return len(self.dictionary)
def __getitem__(self, key):
return self.dictionary[key]
def __setitem__(self, key, value):
if key in self.dictionary:
self.dictionary[key] = value
else:
raise KeyError
def __repr__(self):
return str(self.dictionary)
block1 = BlockA()
block1['file'] = "test"
block2 = BlockA()
block2['file'] = "other test"
Now, you've got a guarantee that all instances of your first block object will have the same keys and no additional keys. You can make similar classes for your other blocks, or some general class, or some mix of the two using inheritance. Now to make your data structure:
blocks = [ [block1, block2], [], [], [] ]
print(blocks) # Or "print blocks" if you're not using Python 3.x
blocks[0][0]['file'] = "some new file"
print(blocks)
It might also be worthwhile to have a class for this blocks container, with specific methods for adding blocks of each type and accessing blocks of each type. That way you wouldn't trip yourself up with accidentally adding the wrong kind of block to one of the four lists or similar issues. But depending on how much you'll be using this structure, that could be overkill.
I have what I believe to be an embarrassingly simple problem, but three hours of googling and checking stackoverflow have not helped.
Let's say I have a very simple piece of code:
def secret_formula(started):
jelly_beans = started*500
jars = jelly_beans/1000
crates = jars/100
return jelly_beans,jars,crates
start_point = 10000
print("We'd have {} beans, {} jars, and {} crates.".format(secret_formula(start_point)))
What happens is I get the "IndexError: tuple index out of range". So I just print the secret_formula function to see what it looks like, and it looks like this:
(5000000, 5000.0, 50.0)
Basically, it is treating the output as one 'thing' (I am still very new, sorry if my language is not correct). My question is, why does it treat it like this and how do I make it pass the three outputs (jelly_beans, jars, and crates) so that it formats the string properly?
Thanks!
The format function of the string take a variable number of argument. The secret_formula function is returning a tuple. You want to convert that to a list of arguments. This is done using the following syntax:
print("We'd have {} beans, {} jars, and {} crates.".format(*secret_formula(start_point)))
The important par is the * character. It tell that you want to convert the following iterable into a list of argument to pass to the function.