I have a dictionary that looks something like that
{A:{'score': 0, 'throw1': [3, 2, 5, 6], 'throw2': [1, 5, 5, 1]},
'B': {'score': 0, 'throw1': [2, 2, 3, 6], 'throw2': [6, 4, 2, 2]}}
A and B are players in this game and the throw1 and throw2 are their dice rolls. Each player has 4 attempts.
My question is how do i extract both throw1 and throw2 from the dictionary and sum their respective attempts together for each player. For instance, Player A threw 3 and 1 for their first attempt on both throws. I want the answer to be 4
you can use the zip that returns an iterator of tuples
data = {'A': {'score': 0, 'throw1': [3, 2, 5, 6], 'throw2': [1, 5, 5, 1]},
'B': {'score': 0, 'throw1': [2, 2, 3, 6], 'throw2': [6, 4, 2, 2]}}
player_A_1_results = data['A']['throw1']
player_A_2_results = data['A']['throw2']
for f, s in zip(player_A_1_results, player_A_2_results):
print(f + s)
You could iterate each players throw summary like below
data = {'A': {'score': 0, 'throw1': [3, 2, 5, 6], 'throw2': [1, 5, 5, 1]},
'B': {'score': 0, 'throw1': [2, 2, 3, 6], 'throw2': [6, 4, 2, 2]}}
for player, info in data.items():
print("Throw summary of player "+player)
for t1, t2 in zip(info["throw1"], info["throw2"]):
print(t1+t2)
`So,I am given a list-
group = [2,1,3,4]
Each index of this list represents a group.
So group 0 = 2
group 1 = 1
group 2 = 3
group 3 = 4
I am given another list called :
l =[[[0, 0, 3, 3, 3, 3], [0, 0, 1, 3, 3, 3, 3]], [[0, 1]], [[2, 2, 2, 3, 3, 3, 3], [2, 2, 2, 3, 3, 3, 3], [2, 2, 2, 3, 3, 3, 3]], [[0, 0, 2, 2, 2, 3, 3, 3], [0, 0, 2, 2, 2, 3, 3, 3], [0, 0, 2, 2, 2, 3, 3, 3, 3], [0, 0, 2, 2, 2, 3, 3, 3, 3]]]
the output I want is:
dict = {0:[0,3],1;[1],2:[2,3],3:[0,2]}
If the number of times the element appears in each sublist of l ie if both l[0][0] and l[0][1] have 0's appear 2 times, it is added to the index 0 of the dict. Since both l[0][0] and l[0][1] have 3 appear 4 times(this is because group[3] is 4), it is added to the index 0.
now in l[1][0] and 0 appears just once(instead of twice) so its not added to index 1. However 1 just appears once so it is added to index 1.Thanks!
What I have tried so far:
def tryin(l,groups):
for i in range(len(l)):
count = 0
for j in range(len(l[i])):
if j in (l[i][j]):
count+=1
if count == groups[i]:
print(i,j)
try this code :
input:
group = [2,1,3,4]
l =[[[0, 0, 3, 3, 3, 3], [0, 0, 1, 3, 3, 3, 3]], [[0, 1]], [[2, 2, 2, 3, 3, 3, 3], [2, 2, 2, 3, 3, 3, 3], [2, 2, 2, 3, 3, 3, 3]], [[0, 0, 2, 2, 2, 3, 3, 3], [0, 0, 2, 2, 2, 3, 3, 3], [0, 0, 2, 2, 2, 3, 3, 3, 3], [0, 0, 2, 2, 2, 3, 3, 3, 3]]]
def IntersecOfSets(list_):
result = set(list_[0])
for s in list_[1:]:
result.intersection_update(s)
return result
def nb_occ(l,group):
d = {}
for i in l:
l2=[]
for j in i:
l1 = []
for x in group:
if j.count(group.index(x)) >= x :
y=group.index(x)
l1.append(y)
l2.append(l1)
if len(l2)>1:
d[str(l.index(i))]= IntersecOfSets(l2)
else:
d[str(l.index(i))]= l2[0]
return d
print(nb_occ(l,group))
output:
{'2': {2, 3}, '1': [1], '0': {0, 3}, '3': {0, 2}}
I'm trying to find all palindromic sequences of length k that sum to n. I have a specific example (k=6):
def brute(n):
J=[]
for a in range(1,n):
for b in range(1,n):
for c in range(1,n):
if (a+b+c)*2==n:
J.append((a,b,c,c,b,a))
return(J)
The output gives me something like:
[(1, 1, 6, 6, 1, 1),
(1, 2, 5, 5, 2, 1),
(1, 3, 4, 4, 3, 1),
(1, 4, 3, 3, 4, 1),
(1, 5, 2, 2, 5, 1),
(1, 6, 1, 1, 6, 1),
(2, 1, 5, 5, 1, 2),
(2, 2, 4, 4, 2, 2),
(2, 3, 3, 3, 3, 2),
(2, 4, 2, 2, 4, 2),
(2, 5, 1, 1, 5, 2),
(3, 1, 4, 4, 1, 3),
(3, 2, 3, 3, 2, 3),
(3, 3, 2, 2, 3, 3),
(3, 4, 1, 1, 4, 3),
(4, 1, 3, 3, 1, 4),
(4, 2, 2, 2, 2, 4),
(4, 3, 1, 1, 3, 4),
(5, 1, 2, 2, 1, 5),
(5, 2, 1, 1, 2, 5),
(6, 1, 1, 1, 1, 6)]
The issue is that I have no idea how to generalize this to any values of n and k. I hear that dictionaries would be helpful. Did I mention I was new to python? any help would be appreciated
thanks
The idea is that we simply count from 0 to 10**k, and consider each of these "integers" as a palindrome sequence. We left pad with 0 where necessary. So, for k==6, 0 -> [0, 0, 0, 0, 0, 0], 1 -> [0, 0, 0, 0, 0, 1], etc. This enumerates over all possible combinations. If it's a palindrome, we also check that it adds up to n.
Below is some code that (should) give a correct result for arbitrary n and k, but is not terribly efficient. I'll leave optimizing up to you (if it's necessary), and give some tips on how to do it.
Here's the code:
def find_all_palindromic_sequences(n, k):
result = []
for i in range(10**k):
paly = gen_palindrome(i, k, n)
if paly is not None:
result.append(paly)
return result
def gen_palindrome(i, k, n):
i_padded = str(i).zfill(k)
i_digits = [int(digit) for digit in i_padded]
if i_digits == i_digits[::-1] and sum(i_digits) == n:
return i_digits
to test it, we can do:
for paly in find_all_palindromic_sequences(n=16, k=6):
print(paly)
this outputs:
[0, 0, 8, 8, 0, 0]
[0, 1, 7, 7, 1, 0]
[0, 2, 6, 6, 2, 0]
[0, 3, 5, 5, 3, 0]
[0, 4, 4, 4, 4, 0]
[0, 5, 3, 3, 5, 0]
[0, 6, 2, 2, 6, 0]
[0, 7, 1, 1, 7, 0]
[0, 8, 0, 0, 8, 0]
[1, 0, 7, 7, 0, 1]
[1, 1, 6, 6, 1, 1]
[1, 2, 5, 5, 2, 1]
[1, 3, 4, 4, 3, 1]
[1, 4, 3, 3, 4, 1]
[1, 5, 2, 2, 5, 1]
[1, 6, 1, 1, 6, 1]
[1, 7, 0, 0, 7, 1]
[2, 0, 6, 6, 0, 2]
[2, 1, 5, 5, 1, 2]
[2, 2, 4, 4, 2, 2]
[2, 3, 3, 3, 3, 2]
[2, 4, 2, 2, 4, 2]
[2, 5, 1, 1, 5, 2]
[2, 6, 0, 0, 6, 2]
[3, 0, 5, 5, 0, 3]
[3, 1, 4, 4, 1, 3]
[3, 2, 3, 3, 2, 3]
[3, 3, 2, 2, 3, 3]
[3, 4, 1, 1, 4, 3]
[3, 5, 0, 0, 5, 3]
[4, 0, 4, 4, 0, 4]
[4, 1, 3, 3, 1, 4]
[4, 2, 2, 2, 2, 4]
[4, 3, 1, 1, 3, 4]
[4, 4, 0, 0, 4, 4]
[5, 0, 3, 3, 0, 5]
[5, 1, 2, 2, 1, 5]
[5, 2, 1, 1, 2, 5]
[5, 3, 0, 0, 3, 5]
[6, 0, 2, 2, 0, 6]
[6, 1, 1, 1, 1, 6]
[6, 2, 0, 0, 2, 6]
[7, 0, 1, 1, 0, 7]
[7, 1, 0, 0, 1, 7]
[8, 0, 0, 0, 0, 8]
Which looks similar to your result, plus the results that contain 0.
Ideas for making it faster (this will slow down a lot as k becomes large):
This is an embarrassingly parallel problem, consider multithreading/multiprocessing.
The palindrome check of i_digits == i_digits[::-1] isn't as efficient as it could be (both in terms of memory and CPU). Having a pointer at the start and end, and traversing characters one by one till the pointers cross would be better.
There are some conditional optimizations you can do on certain values of n. For instance, if n is 0, it doesn't matter how large k is, the only palindrome will be [0, 0, 0, ..., 0, 0]. As another example, if n is 8, we obviously don't have to generate any permutations with 9 in them. Or, if n is 20, and k is 6, then we can't have 3 9's in our permutation. Generalizing this pattern will pay off big assuming n is reasonably small. It works the other way, too, actually. If n is large, then there is a limit to the number of 0s and 1s that can be in each permutation.
There is probably a better way of generating palindromes than testing every single integer. For example, if we know that integer X is a palindrome sequence, then X+1 will not be. It's pretty easy to show this: the first and last digits can't match for X+1 since we know they must have matched for X. You might be able to show that X+2 and X+3 cannot be palindromes either, etc. If you can generalize where you must test for a new palindrome, this will be key. A number theorist could help more in this regard.
HTH.