I have to do a function that accepts an array, where the first element is the number of columns and rows of a matrix, the rest is the matrix itself. I have to return the absolute difference between the sums of its diagonals.
My problem is that the interpreter returns this traceback:
File "Solution.py", line 23, in diagonalDifference
while counter != n + 1:
TypeError: can only concatenate list (not "int") to list
I don't why it intrepet '''n''' like a list, it's the first element of the'''lis''' list.
Here is the code
def diagonalDifference(arr):
sumleftdiagonal = 0
sumrightdiagonal = 0
counter = 0
lis = [i for i in arr]
n = lis[0]
lis.remove(lis[0])
while counter != n + 1:
sumleftdiagonal += lis[0 + counter]
sumrightdiagonal += [n - counter]
counter += 1
for i in lis[:n+1]:
lis.remove(i)
return abs(sumleftdiagonal - sumrightdiagonal)
Solution
Your code does not work because you are essentially evaluating the difference of the first row of the square matrix from itself.
# this picks up members of first row (left to right)
sumleftdiagonal += lis[0 + counter]
# this picks up members of first row (right to left)
sumrightdiagonal += [n - counter]
counter += 1
You can verify that sumleftdiagonal = [ 1, 2, 3, 4, 5] and sumrightdiagonal = [5, 4, 3, 2, 1] for the dummy data I used.
Purely pythonic solution.
Method-1: just python
# using the dummy data below: sum(diag_left - diag_right) = 0
n = arr[0]
sum([arr[1+n*i+i] - arr[1+(n-1)*(i+1)] for i in range(n)])
Output:
0
Using numpy
If you can use numpy then there are at least two methods.
Method-2: using numpy
import numpy as np
def diagdiff(arr):
n, arr = arr[0], arr[1:]
arr = np.array(arr).reshape((n,n))
return sum(np.diag(arr) - np.diag(np.fliplr(arr)))
# Dummy Data
a = np.arange(25).reshape((5,5)) + 1
print(f'square-array: \n\n{a}\n')
arr = [int(np.sqrt(a.size))] + a.flatten().tolist()
dd = diagdiff(arr)
print(f'diagonal-difference: {dd}')
Output:
square-array:
[[ 1 2 3 4 5]
[ 6 7 8 9 10]
[11 12 13 14 15]
[16 17 18 19 20]
[21 22 23 24 25]]
diagonal-difference: 0
Method-3: using numpy
n, a = arr[0], np.array(arr[1:])
a = a.reshape((n,n))
sum([a[i, 0+i] - a[i, -(1+i)] for i in range(n)])
Output:
0
sumrightdiagonal += [n - counter]
On this line you are trying to append a list containing n - counter, I assume you mean lis[n - counter] here.
+++
We are given the following table on the number of homeless people in each of 3 cities, where n the number of homes to be built.
City n=0 n=1 n=2
A 15 12 10
B 9 8 4
C 8 6 5
We are after a housing strategy that minimises the total numbers of
homeless people across all 3 cities for a given number of homes.
What would be the optimal allocation of n homes to homeless, and how many homeless would that result in total?
The solution is to employ dynamic programming & be in Python.
It appears like an unbounded knapsack problem with capacity n but what is hard to find are the values & weights to be used as parameters of the algorithm.
Please share your views.
+++
+++
I have managed to solve the task as a bounded knapsack case with multiple values as follows.
def assign_home(cpct, tbl):
ctNum = len(tbl[0])
wghts = [1] * ctNum
vls = get_vls(tbl)
hms = [0] * ctNum
HMSBND = 2
hmlsVal = [sum(row[0] for row in tbl) for hmIndx in range(min(cpct + 1, ctNum * HMSBND + 1))]
for hmIndx in range(min(wghts), min(cpct + 1, ctNum * HMSBND + 1)):
chsn = -1
for ct, wght in enumerate(wghts):
if wght <= hmIndx and hms[ct] < HMSBND and hmlsVal[hmIndx] > hmlsVal[hmIndx - wght] - vls[ct][hms[ct]]:
hmlsVal[hmIndx] = hmlsVal[hmIndx - wght] - vls[ct][hms[ct]]
chsn = ct
if chsn != -1:
hms[chsn] += 1
return hmlsVal[min(cpct, ctNum * HMSBND)], hms
def get_vls(tbl):
[A, B, C] = tbl
Adf = [A[0] - A[1], A[1] - A[2]]
Bdf = [B[0] - B[1], B[1] - B[2]]
Cdf = [C[0] - C[1], C[1] - C[2]]
return [Adf, Bdf, Cdf]
tbl=[[15, 12, 10], [9, 8, 4], [8, 6, 5]]
print(assign_home(5, tbl))
# (20, [2, 2, 1])
+++
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I tried to write code to solve the standard Integer Partition problem (Wikipedia). The code I wrote was a mess. I need an elegant solution to solve the problem, because I want to improve my coding style. This is not a homework question.
A smaller and faster than Nolen's function:
def partitions(n, I=1):
yield (n,)
for i in range(I, n//2 + 1):
for p in partitions(n-i, i):
yield (i,) + p
Let's compare them:
In [10]: %timeit -n 10 r0 = nolen(20)
1.37 s ± 28.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [11]: %timeit -n 10 r1 = list(partitions(20))
979 µs ± 82.9 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [13]: sorted(map(sorted, r0)) == sorted(map(sorted, r1))
Out[14]: True
Looks like it's 1370 times faster for n = 20.
Anyway, it's still far from accel_asc:
def accel_asc(n):
a = [0 for i in range(n + 1)]
k = 1
y = n - 1
while k != 0:
x = a[k - 1] + 1
k -= 1
while 2 * x <= y:
a[k] = x
y -= x
k += 1
l = k + 1
while x <= y:
a[k] = x
a[l] = y
yield a[:k + 2]
x += 1
y -= 1
a[k] = x + y
y = x + y - 1
yield a[:k + 1]
It's not only slower, but requires much more memory (but apparently is much easier to remember):
In [18]: %timeit -n 5 r2 = list(accel_asc(50))
114 ms ± 1.04 ms per loop (mean ± std. dev. of 7 runs, 5 loops each)
In [19]: %timeit -n 5 r3 = list(partitions(50))
527 ms ± 8.86 ms per loop (mean ± std. dev. of 7 runs, 5 loops each)
In [24]: sorted(map(sorted, r2)) == sorted(map(sorted, r3))
Out[24]: True
You can find other versions on ActiveState: Generator For Integer Partitions (Python Recipe).
I use Python 3.6.1 and IPython 6.0.0.
While this answer is fine, I'd recommend skovorodkin's answer.
>>> def partition(number):
... answer = set()
... answer.add((number, ))
... for x in range(1, number):
... for y in partition(number - x):
... answer.add(tuple(sorted((x, ) + y)))
... return answer
...
>>> partition(4)
set([(1, 3), (2, 2), (1, 1, 2), (1, 1, 1, 1), (4,)])
If you want all permutations(ie (1, 3) and (3, 1)) change answer.add(tuple(sorted((x, ) + y)) to answer.add((x, ) + y)
I've compared the solution with perfplot (a little project of mine for such purposes) and found that Nolen's top-voted answer is also the slowest.
Both answers supplied by skovorodkin are much faster. (Note the log-scale.)
To to generate the plot:
import perfplot
import collections
def nolen(number):
answer = set()
answer.add((number,))
for x in range(1, number):
for y in nolen(number - x):
answer.add(tuple(sorted((x,) + y)))
return answer
def skovorodkin(n):
return set(skovorodkin_yield(n))
def skovorodkin_yield(n, I=1):
yield (n,)
for i in range(I, n // 2 + 1):
for p in skovorodkin_yield(n - i, i):
yield (i,) + p
def accel_asc(n):
return set(accel_asc_yield(n))
def accel_asc_yield(n):
a = [0 for i in range(n + 1)]
k = 1
y = n - 1
while k != 0:
x = a[k - 1] + 1
k -= 1
while 2 * x <= y:
a[k] = x
y -= x
k += 1
l = k + 1
while x <= y:
a[k] = x
a[l] = y
yield tuple(a[: k + 2])
x += 1
y -= 1
a[k] = x + y
y = x + y - 1
yield tuple(a[: k + 1])
def mct(n):
partitions_of = []
partitions_of.append([()])
partitions_of.append([(1,)])
for num in range(2, n + 1):
ptitions = set()
for i in range(num):
for partition in partitions_of[i]:
ptitions.add(tuple(sorted((num - i,) + partition)))
partitions_of.append(list(ptitions))
return partitions_of[n]
perfplot.show(
setup=lambda n: n,
kernels=[nolen, mct, skovorodkin, accel_asc],
n_range=range(1, 17),
logy=True,
# https://stackoverflow.com/a/7829388/353337
equality_check=lambda a, b: collections.Counter(set(a))
== collections.Counter(set(b)),
xlabel="n",
)
I needed to solve a similar problem, namely the partition of an integer n into d nonnegative parts, with permutations. For this, there's a simple recursive solution (see here):
def partition(n, d, depth=0):
if d == depth:
return [[]]
return [
item + [i]
for i in range(n+1)
for item in partition(n-i, d, depth=depth+1)
]
# extend with n-sum(entries)
n = 5
d = 3
lst = [[n-sum(p)] + p for p in partition(n, d-1)]
print(lst)
Output:
[
[5, 0, 0], [4, 1, 0], [3, 2, 0], [2, 3, 0], [1, 4, 0],
[0, 5, 0], [4, 0, 1], [3, 1, 1], [2, 2, 1], [1, 3, 1],
[0, 4, 1], [3, 0, 2], [2, 1, 2], [1, 2, 2], [0, 3, 2],
[2, 0, 3], [1, 1, 3], [0, 2, 3], [1, 0, 4], [0, 1, 4],
[0, 0, 5]
]
I'm a bit late to the game, but I can offer a contribution which might qualify as more elegant in a few senses:
def partitions(n, m = None):
"""Partition n with a maximum part size of m. Yield non-increasing
lists in decreasing lexicographic order. The default for m is
effectively n, so the second argument is not needed to create the
generator unless you do want to limit part sizes.
"""
if m is None or m >= n: yield [n]
for f in range(n-1 if (m is None or m >= n) else m, 0, -1):
for p in partitions(n-f, f): yield [f] + p
Only 3 lines of code. Yields them in lexicographic order. Optionally allows imposition of a maximum part size.
I also have a variation on the above for partitions with a given number of parts:
def sized_partitions(n, k, m = None):
"""Partition n into k parts with a max part of m.
Yield non-increasing lists. m not needed to create generator.
"""
if k == 1:
yield [n]
return
for f in range(n-k+1 if (m is None or m > n-k+1) else m, (n-1)//k, -1):
for p in sized_partitions(n-f, k-1, f): yield [f] + p
After composing the above, I ran across a solution I had created almost 5 years ago, but which I had forgotten about. Besides a maximum part size, this one offers the additional feature that you can impose a maximum length (as opposed to a specific length). FWIW:
def partitions(sum, max_val=100000, max_len=100000):
""" generator of partitions of sum with limits on values and length """
# Yields lists in decreasing lexicographical order.
# To get any length, omit 3rd arg.
# To get all partitions, omit 2nd and 3rd args.
if sum <= max_val: # Can start with a singleton.
yield [sum]
# Must have first*max_len >= sum; i.e. first >= sum/max_len.
for first in range(min(sum-1, max_val), max(0, (sum-1)//max_len), -1):
for p in partitions(sum-first, first, max_len-1):
yield [first]+p
Much quicker than the accepted response and not bad looking, either. The accepted response does lots of the same work multiple times because it calculates the partitions for lower integers multiple times. For example, when n=22 the difference is 12.7 seconds against 0.0467 seconds.
def partitions_dp(n):
partitions_of = []
partitions_of.append([()])
partitions_of.append([(1,)])
for num in range(2, n+1):
ptitions = set()
for i in range(num):
for partition in partitions_of[i]:
ptitions.add(tuple(sorted((num - i, ) + partition)))
partitions_of.append(list(ptitions))
return partitions_of[n]
The code is essentially the same except we save the partitions of smaller integers so we don't have to calculate them again and again.
Here is a recursive function, which uses a stack in which we store the numbers of the partitions in increasing order.
It is fast enough and very intuitive.
# get the partitions of an integer
Stack = []
def Partitions(remainder, start_number = 1):
if remainder == 0:
print(" + ".join(Stack))
else:
for nb_to_add in range(start_number, remainder+1):
Stack.append(str(nb_to_add))
Partitions(remainder - nb_to_add, nb_to_add)
Stack.pop()
When the stack is full (the sum of the elements of the stack then corresponds to the number we want the partitions), we print it,
remove its last value and test the next possible value to be stored in the stack. When all the next values have been tested, we pop the last value of the stack again and we go back to the last calling function.
Here is an example of the output (with 8):
Partitions(8)
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + 1 + 2
1 + 1 + 1 + 1 + 1 + 3
1 + 1 + 1 + 1 + 2 + 2
1 + 1 + 1 + 1 + 4
1 + 1 + 1 + 2 + 3
1 + 1 + 1 + 5
1 + 1 + 2 + 2 + 2
1 + 1 + 2 + 4
1 + 1 + 3 + 3
1 + 1 + 6
1 + 2 + 2 + 3
1 + 2 + 5
1 + 3 + 4
1 + 7
2 + 2 + 2 + 2
2 + 2 + 4
2 + 3 + 3
2 + 6
3 + 5
4 + 4
8
The structure of the recursive function is easy to understand and is illustrated below (for the integer 31):
remainder corresponds to the value of the remaining number we want a partition (31 and 21 in the example above).
start_number corresponds to the first number of the partition, its default value is one (1 and 5 in the example above).
If we wanted to return the result in a list and get the number of partitions, we could do this:
def Partitions2_main(nb):
global counter, PartitionList, Stack
counter, PartitionList, Stack = 0, [], []
Partitions2(nb)
return PartitionList, counter
def Partitions2(remainder, start_number = 1):
global counter, PartitionList, Stack
if remainder == 0:
PartitionList.append(list(Stack))
counter += 1
else:
for nb_to_add in range(start_number, remainder+1):
Stack.append(nb_to_add)
Partitions2(remainder - nb_to_add, nb_to_add)
Stack.pop()
Last, a big advantage of the function Partitions shown above is that it adapts very easily to find all the compositions of a natural number (two compositions can have the same set of numbers, but the order differs in this case):
we just have to drop the variable start_number and set it to 1 in the for loop.
# get the compositions of an integer
Stack = []
def Compositions(remainder):
if remainder == 0:
print(" + ".join(Stack))
else:
for nb_to_add in range(1, remainder+1):
Stack.append(str(nb_to_add))
Compositions(remainder - nb_to_add)
Stack.pop()
Example of output:
Compositions(4)
1 + 1 + 1 + 1
1 + 1 + 2
1 + 2 + 1
1 + 3
2 + 1 + 1
2 + 2
3 + 1
4
I think the recipe here may qualify as being elegant. It's lean (20 lines long), fast and based upon Kelleher and O'Sullivan's work which is referenced therein:
def aP(n):
"""Generate partitions of n as ordered lists in ascending
lexicographical order.
This highly efficient routine is based on the delightful
work of Kelleher and O'Sullivan.
Examples
========
>>> for i in aP(6): i
...
[1, 1, 1, 1, 1, 1]
[1, 1, 1, 1, 2]
[1, 1, 1, 3]
[1, 1, 2, 2]
[1, 1, 4]
[1, 2, 3]
[1, 5]
[2, 2, 2]
[2, 4]
[3, 3]
[6]
>>> for i in aP(0): i
...
[]
References
==========
.. [1] Generating Integer Partitions, [online],
Available: http://jeromekelleher.net/generating-integer-partitions.html
.. [2] Jerome Kelleher and Barry O'Sullivan, "Generating All
Partitions: A Comparison Of Two Encodings", [online],
Available: http://arxiv.org/pdf/0909.2331v2.pdf
"""
# The list `a`'s leading elements contain the partition in which
# y is the biggest element and x is either the same as y or the
# 2nd largest element; v and w are adjacent element indices
# to which x and y are being assigned, respectively.
a = [1]*n
y = -1
v = n
while v > 0:
v -= 1
x = a[v] + 1
while y >= 2 * x:
a[v] = x
y -= x
v += 1
w = v + 1
while x <= y:
a[v] = x
a[w] = y
yield a[:w + 1]
x += 1
y -= 1
a[v] = x + y
y = a[v] - 1
yield a[:w]
# -*- coding: utf-8 -*-
import timeit
ncache = 0
cache = {}
def partition(number):
global cache, ncache
answer = {(number,), }
if number in cache:
ncache += 1
return cache[number]
if number == 1:
cache[number] = answer
return answer
for x in range(1, number):
for y in partition(number - x):
answer.add(tuple(sorted((x, ) + y)))
cache[number] = answer
return answer
print('To 5:')
for r in sorted(partition(5))[::-1]:
print('\t' + ' + '.join(str(i) for i in r))
print(
'Time: {}\nCache used:{}'.format(
timeit.timeit(
"print('To 30: {} possibilities'.format(len(partition(30))))",
setup="from __main__ import partition",
number=1
), ncache
)
)
or https://gist.github.com/sxslex/dd15b13b28c40e695f1e227a200d1646
I don't know if my code is the most elegant, but I've had to solve this many times for research purposes. If you modify the
sub_nums
variable you can restrict what numbers are used in the partition.
def make_partitions(number):
out = []
tmp = []
sub_nums = range(1,number+1)
for num in sub_nums:
if num<=number:
tmp.append([num])
for elm in tmp:
sum_elm = sum(elm)
if sum_elm == number:
out.append(elm)
else:
for num in sub_nums:
if sum_elm + num <= number:
L = [i for i in elm]
L.append(num)
tmp.append(L)
return out
F(x,n) = \union_(i>=n) { {i}U g| g in F(x-i,i) }
Just implement this recursion. F(x,n) is the set of all sets that sum to x and their elements are greater than or equal to n.
Suppose we have a matrix of dimension N x M and we want to reduce its dimension preserving the values in each by summing the firs neighbors.
Suppose the matrix A is a 4x4 matrix:
A =
3 4 5 6
2 3 4 5
2 2 0 1
5 2 2 3
we want to reduce it to a 2x2 matrix as following:
A1 =
12 20
11 6
In particular my matrix represent the number of incident cases in an x-y plane. My matrix is A=103x159, if I plot it I get:
what I want to do is to aggregate those data to a bigger area, such as
Assuming you're using a numpy.matrix:
import numpy as np
A = np.matrix([
[3,4,5,6],
[2,3,4,5],
[2,2,0,1],
[5,2,2,3]
])
N, M = A.shape
assert N % 2 == 0
assert M % 2 == 0
A1 = np.empty((N//2, M//2))
for i in range(N//2):
for j in range(M//2):
A1[i,j] = A[2*i:2*i+2, 2*j:2*j+2].sum()
Though these loops can probably be optimized away by proper numpy functions.
I see that there is a solution using numpy.maxtrix, maybe you can test my solution too and return your feedbacks.
It works with a*b matrix if a and b are even. Otherwise, it may fail if a or b are odd.
Here is my solution:
v = [
[3,4,5,6],
[2,3,4,5],
[2,2,0,1],
[5,2,2,3]
]
def shape(v):
return len(v), len(v[0])
def chunks(v, step):
"""
Chunk list step per step and sum
Example: step = 2
[3,4,5,6] => [7,11]
[2,3,4,5] => [5,9]
[2,2,0,1] => [4,1]
[5,2,2,3] => [7,5]
"""
for i in v:
for k in range(0, len(i),step):
yield sum(j for j in i[k:k+step])
def sum_chunks(k, step):
"""
Sum near values with step
Example: step = 2
[
[7,11], [
[5,9], => [12, 11],
[4,1], [20, 6]
[7,5] ]
]
"""
a, c = [k[i::step] for i in range(step)], []
print(a)
for m in a:
# sum near values
c.append([sum(m[j:j+2]) for j in range(0, len(m), 2)])
return c
rows, columns = shape(v)
chunk_list = list(chunks(v, columns // 2))
final_sum = sum_chunks(chunk_list, rows // 2)
print(final_sum)
Output:
[[12, 11], [20, 6]]