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I'm writing a sum up game where two players will take turns picking a random number in the range (1,9), no repeated number allowed. So I'm struggling at
If at any point exactly three of the player's numbers sum to 15, then that player has won.
If the first player picks [7, 2, 3, 5], he will win because 7+3+5 = 15
So my question is why doesn't the program stop when first_player has inputs == 15
I want to avoid importing any libs.
Instead of generating all permutations at each step, maintain a map of what each permutation sums to, then add two branches to each branch at each move.
Think of each entry as a set of bits, ie with each permutation you either include a given entry or not, eg if the numbers are [7, 3, 2] you might store [1, 0, 1] for the combination of the 7 and the 2.
You can make a hashmap of 101->9 etc and when someone adds a 3 to it you add an entry for 1010->9 and 1011->12. As soon as you see the target you know the game is over.
So the evolution of [7, 3, 2] would be
0->0
1->7
00->0
01->3
10->7
11->10
000->0
001->2
010->3
011->5
100->7
101->9
110->10
111->12
A more efficient way would be to find only those numbers whose sum is equal to the target that is 15.
entry = [7, 5, 1, 3]
def is_sum_15(nums):
res = []
search_numbers(nums, 3, 15, 0, [], res)
return len(res) != 0
def search_numbers(nums, k, n, index, path, res):
if k < 0 or n < 0:
return
if k == 0 and n == 0:
res.append(path)
for i in range(index, len(nums)):
search_numbers(nums, k-1, n-nums[i], i+1, path+[nums[i]], res)
print(is_sum_15(entry)) # True
An inefficient but easy way is to use itertools.permutations:
>>> entry = [7, 2, 3, 5]
>>> import itertools
>>> [sum(triplet) for triplet in itertools.permutations(entry, r=3) if sum(tr]
[12, 14, 12, 15, 14, 15, 12, 14, 12, 10, 14, 10, 12, 15, 12, 10, 15, 10, 14, 15, 14, 10, 15, 10]
>>> any(sum(triplet) == 15 for triplet in itertools.permutations(entry, r=3))
True
It's inefficient, because you would be trying all permutations every time entry gets expanded with a new number.
I working on a problem where I'm supposed to generate ten random but unique numbers that range from 1 to 15 inclusive. The thing is, I'm supposed to write everything in one line and to also get this output:
[2, 4, 6, 7, 8, 9, 11, 12, 13, 15]
Below I have some code I wrote but, it's not getting the output I want. What am I doing wrong and can I perhaps see a solution with a break so I know how to do this going down the road?
import random
print(sorted(random.sample(range(1,16),15)))
Output:
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
The output I want is:
[2,4,6,7,8,9,11,12,13,15]
How do I get this in one line of code?
>>> help(random.sample)
sample(population, k): method of random.Random instance
Chooses k unique random elements from a population sequence or set.
I'm supposed to write everything in one line and to also get this output:
[2, 4, 6, 7, 8, 9, 11, 12, 13, 15]
>>> sorted(__import__('random').Random(4225).sample(range(1, 16), 10))
[2, 4, 6, 7, 8, 9, 11, 12, 13, 15]
If you want to generate ten numbers in range 1-15, change
print(sorted(random.sample(range(1,16),15)))
to
print(sorted(random.sample(range(1,16),10)))
# From the documentation :
# random.sample(population, k)
import random
population = range(16)
how_may_sample = 10
random.sample(population, how_many_sample)
# Now in one line
random.sample(range(16), 10)
I want to print the top 10 distinct elements from a list:
top=10
test=[1,1,1,2,3,4,5,6,7,8,9,10,11,12,13]
for i in range(0,top):
if test[i]==1:
top=top+1
else:
print(test[i])
It is printing:
2,3,4,5,6,7,8
I am expecting:
2,3,4,5,6,7,8,9,10,11
What I am missing?
Using numpy
import numpy as np
top=10
test=[1,1,1,2,3,4,5,6,7,8,9,10,11,12,13]
test=np.unique(np.array(test))
test[test!=1][:top]
Output
array([ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
Since you code only executes the loop for 10 times and the first 3 are used to ignore 1, so only the following 3 is printed, which is exactly happened here.
If you want to print the top 10 distinct value, I recommand you to do this:
# The code of unique is taken from [remove duplicates in list](https://stackoverflow.com/questions/7961363/removing-duplicates-in-lists)
def unique(l):
return list(set(l))
def print_top_unique(List, top):
ulist = unique(List)
for i in range(0, top):
print(ulist[i])
print_top_unique([1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], 10)
My Solution
test = [1,1,1,2,3,4,5,6,7,8,9,10,11,12,13]
uniqueList = [num for num in set(test)] #creates a list of unique characters [1,2,3,4,5,6,7,8,9,10,11,12,13]
for num in range(0,11):
if uniqueList[num] != 1: #skips one, since you wanted to start with two
print(uniqueList[num])
I'm traversing a two-dimensional list (my representation of a matrix) in an unusual order: counterclockwise around the outside starting with the top-left element.
I need to do this more than once, but each time I do it, I'd like to do something different with the values I encounter. The first time, I want to note down the values so that I can modify them. (I can't modify them in place.) The second time, I want to traverse the outside of the matrix and modify the values of the matrix as I go, perhaps getting my new values from some generator.
Is there a way I can abstract this traversal to a function and still achieve my goals? I was thinking that this traverse-edge function could take a function and a matrix and apply the function to each element on the edge of the matrix. However, the problems with this are two-fold. If I do this, I don't think I can modify the matrix that's given as an argument, and I can't yield the values one by one because yield isn't a function.
Edit: I want to rotate a matrix counterclockwise (not 90 degrees) where one rotation moves, for example, the top-left element down one spot. To accomplish this, I'm rotating one "level" (or shell) of the matrix at a time. So if I'm rotating the outermost level, I want to traverse it once to build a list which I can shift to the left, then I want to traverse the outermost level again to assign it those new values which I calculated.
Just create 4 loops, one for each side of the array, that counts through the values of the index that changes for that side. For example, the first side, whose x index is always 0, could vary the y from 0 to n-2 (from the top-left corner to just shy of the bottom-left); repeat for the other sides.
I think there are two approaches you can take to solving your problem.
The first option is to create a function that returns an iterable of indexes into the matrix. Then you'd write your various passes over the matrix with for loops:
for i, j in matrix_border_index_gen(len(matrix), len(matrix[0])): # pass in dimensions
# do something with matrix[i][j]
The other option is to write a function that works more like map that applies a given function to each appropriate value of the matrix in turn. If you sometimes need to replace the current values with new ones, I'd suggest doing that all the time (the times when you don't want to replace the value, you can just have your function return the previous value):
def func(value):
# do stuff with value from matrix
return new_value # new_value can be the same value, if you don't want to change it
matrix_border_map(func, matrix) # replace each value on border of matrix with func(value)
I have added a few lines of python 3 code here. It has the mirror function and a spiral iterator (not sure, if that's what you meant). No doc strings (sorry). It is readable though. Change print statement for python 2.
EDIT : FIXED A BUG
class Matrix():
def __init__(self, rows=5, cols=5):
self.cells = [[None for c in range(cols)] for r in range(rows)]
def transpose(self):
self.cells = list(map(list, zip(*self.cells)))
def mirror(self):
for row in self.cells:
row.reverse()
def invert(self):
self.cells.reverse()
def rotate(self, clockwise=True):
self.transpose()
self.mirror() if clockwise else self.invert()
def iter_spiral(self, grid=None):
grid = grid or self.cells
next_grid = []
for cell in reversed(grid[0]):
yield cell
for row in grid[1:-1]:
yield row[0]
next_grid.append(row[1:-1])
if len(grid) > 1:
for cell in grid[-1]:
yield cell
for row in reversed(grid[1:-1]):
yield row[-1]
if next_grid:
for cell in self.iter_spiral(grid=next_grid):
yield cell
def show(self):
for row in self.cells:
print(row)
def test_matrix():
m = Matrix()
m.cells = [[1,2,3,4],
[5,6,7,8],
[9,10,11,12],
[13,14,15,16]]
print("We expect the spiral to be:", "4, 3, 2, 1, 5, 9, 13, 14, 15, 16, 12, 8, 7, 6, 10, 11", sep='\n')
print("What the iterator yields:")
for cell in m.iter_spiral():
print(cell, end=', ')
print("\nThe matrix looks like this:")
m.show()
print("Now this is how it looks rotated 90 deg clockwise")
m.rotate()
m.show()
print("Now we'll rotate it back")
m.rotate(clockwise=False)
m.show()
print("Now we'll transpose it")
m.transpose()
m.show()
print("Inverting the above")
m.invert()
m.show()
print("Mirroring the above")
m.mirror()
m.show()
if __name__ == '__main__':
test_matrix()
This is the output:
We expect the spiral to be:
4, 3, 2, 1, 5, 9, 13, 14, 15, 16, 12, 8, 7, 6, 10, 11
What the iterator yields:
4, 3, 2, 1, 5, 9, 13, 14, 15, 16, 12, 8, 7, 6, 10, 11,
The matrix looks like this:
[1, 2, 3, 4]
[5, 6, 7, 8]
[9, 10, 11, 12]
[13, 14, 15, 16]
Now this is how it looks rotated 90 deg clockwise
[13, 9, 5, 1]
[14, 10, 6, 2]
[15, 11, 7, 3]
[16, 12, 8, 4]
Now we'll rotate it back
[1, 2, 3, 4]
[5, 6, 7, 8]
[9, 10, 11, 12]
[13, 14, 15, 16]
Now we'll transpose it
[1, 5, 9, 13]
[2, 6, 10, 14]
[3, 7, 11, 15]
[4, 8, 12, 16]
Inverting the above
[4, 8, 12, 16]
[3, 7, 11, 15]
[2, 6, 10, 14]
[1, 5, 9, 13]
Mirroring the above
[16, 12, 8, 4]
[15, 11, 7, 3]
[14, 10, 6, 2]
[13, 9, 5, 1]
I would go with generator functions. They can be used to create iterators over which we can iterate. An Example of a generator function -
def genfunc():
i = 0
while i < 10:
yield i
i = i + 1
>>> for x in genfunc():
... print(x)
...
0
1
2
3
4
5
6
7
8
9
When calling the generator function, it returns a generator object -
>>> genfunc()
<generator object genfunc at 0x00553AD0>
It does not start going over the function at that point. When you start iterating over the generator object, calling for its first element, it starts going over the function, untill it reaches the first yield statement, and at that point it returns the value (in above case, it returns value of i) . And it also saves the state of the function at that point (that is it saves at what point the execution was when the value was yielded, what were the values for the variables in the local namespace, etc).
Then when it tries to get the next value, again execution starts from where it stopped last time, till it again yield another value. And this continues on.
Hey im new to python. How do you get a portion of a list by the relative value of its sorting key.
example...
list = [11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10]
list.sort()
newList = list.split("all numbers that are over 13")
assert newList == [14,15,16]
>>> l = [11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10]
>>> sorted(x for x in l if x > 13)
[14, 15, 16]
or with filter (would be a little bit slower if you have big list, because of lambda)
>>> sorted(filter(lambda x: x > 13, l))
[14, 15, 16]
Use [item for item in newList if item > 13].
There is a decent chance this could be replaced with the generator expression (item for item in newList if item > 13), which filters lazily rather than storing the whole list in memory.
You might also be interested in changing the code just a bit to something like
all_numbers = [11, 12, 13, 14, 15, 16, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
filtered_sorted_numbers = sorted(number for number in all_numbers if number > 13)
which performs the sorting—a worst case O(n log n) operation—on only the filtered values.