Develop Reference

how to calculate the minimum unfairness sum of a list

I have tried to summarize the problem statement something like this::
Given n, k and an array(a list) arr where n = len(arr) and k is an integer in set (1, n) inclusive.
For an array (or list) myList, The Unfairness Sum is defined as the sum of the absolute differences between all possible pairs (combinations with 2 elements each) in myList.
To explain: if mylist = [1, 2, 5, 5, 6] then Minimum unfairness sum or MUS. Please note that elements are considered unique by their index in list not their values
MUS = |1-2| + |1-5| + |1-5| + |1-6| + |2-5| + |2-5| + |2-6| + |5-5| + |5-6| + |5-6|
If you actually need to look at the problem statement, It's HERE
My Objective
given n, k, arr(as described above), find the Minimum Unfairness Sum out of all of the unfairness sums of sub arrays possible with a constraint that each len(sub array) = k [which is a good thing to make our lives easy, I believe :) ]
what I have tried
well, there is a lot to be added in here, so I'll try to be as short as I can.
My First approach was this where i used itertools.combinations to get all the possible combinations and statistics.variance to check its spread of data (yeah, I know I'm a mess).
Before you see the code below, Do you think these variance and unfairness sum are perfectly related (i know they are strongly related) i.e. the sub array with minimum variance has to be the sub array with MUS??
You only have to check the LetMeDoIt(n, k, arr) function. If you need MCVE, check the second code snippet below.
from itertools import combinations as cmb
from statistics import variance as varn
def LetMeDoIt(n, k, arr):
v = []
s = []
subs = [list(x) for x in list(cmb(arr, k))] # getting all sub arrays from arr in a list
i = 0
for sub in subs:
if i != 0:
var = varn(sub) # the variance thingy
if float(var) < float(min(v)):
v.remove(v[0])
v.append(var)
s.remove(s[0])
s.append(sub)
else:
pass
elif i == 0:
var = varn(sub)
v.append(var)
s.append(sub)
i = 1
final = []
f = list(cmb(s[0], 2)) # getting list of all pairs (after determining sub array with least MUS)
for r in f:
final.append(abs(r[0]-r[1])) # calculating the MUS in my messy way
return sum(final)
The above code works fine for n<30 but raised a MemoryError beyond that.
In Python chat, Kevin suggested me to try generator which is memory efficient (it really is), but as generator also generates those combination on the fly as we iterate over them, it was supposed to take over 140 hours (:/) for n=50, k=8 as estimated.
I posted the same as a question on SO HERE (you might wanna have a look to understand me properly - it has discussions and an answer by fusion which takes me to my second approach - a better one(i should say fusion's approach xD)).
Second Approach
from itertools import combinations as cmb
def myvar(arr): # a function to calculate variance
l = len(arr)
m = sum(arr)/l
return sum((i-m)**2 for i in arr)/l
def LetMeDoIt(n, k, arr):
sorted_list = sorted(arr) # i think sorting the array makes it easy to get the sub array with MUS quickly
variance = None
min_variance_sub = None
for i in range(n - k + 1):
sub = sorted_list[i:i+k]
var = myvar(sub)
if variance is None or var<variance:
variance = var
min_variance_sub=sub
final = []
f = list(cmb(min_variance_sub, 2)) # again getting all possible pairs in my messy way
for r in f:
final.append(abs(r[0] - r[1]))
return sum(final)
def MainApp():
n = int(input())
k = int(input())
arr = list(int(input()) for _ in range(n))
result = LetMeDoIt(n, k, arr)
print(result)
if __name__ == '__main__':
MainApp()
This code works perfect for n up to 1000 (maybe more), but terminates due to time out (5 seconds is the limit on online judge :/ ) for n beyond 10000 (the biggest test case has n=100000).
=====
How would you approach this problem to take care of all the test cases in given time limits (5 sec) ? (problem was listed under algorithm & dynamic programming)
(for your references you can have a look on
successful submissions(py3, py2, C++, java) on this problem by other candidates - so that you can
explain that approach for me and future visitors)
an editorial by the problem setter explaining how to approach the question
a solution code by problem setter himself (py2, C++).
Input data (test cases) and expected output
Edit1 ::
For future visitors of this question, the conclusions I have till now are,
that variance and unfairness sum are not perfectly related (they are strongly related) which implies that among a lots of lists of integers, a list with minimum variance doesn't always have to be the list with minimum unfairness sum. If you want to know why, I actually asked that as a separate question on math stack exchange HERE where one of the mathematicians proved it for me xD (and it's worth taking a look, 'cause it was unexpected)
As far as the question is concerned overall, you can read answers by archer & Attersson below (still trying to figure out a naive approach to carry this out - it shouldn't be far by now though)
Thank you for any help or suggestions :)

You must work on your list SORTED and check only sublists with consecutive elements. This is because BY DEFAULT, any sublist that includes at least one element that is not consecutive, will have higher unfairness sum.
For example if the list is
[1,3,7,10,20,35,100,250,2000,5000] and you want to check for sublists with length 3, then solution must be one of [1,3,7] [3,7,10] [7,10,20] etc
Any other sublist eg [1,3,10] will have higher unfairness sum because 10>7 therefore all its differences with rest of elements will be larger than 7
The same for [1,7,10] (non consecutive on the left side) as 1<3
Given that, you only have to check for consecutive sublists of length k which reduces the execution time significantly
Regarding coding, something like this should work:
def myvar(array):
return sum([abs(i[0]-i[1]) for i in itertools.combinations(array,2)])
def minsum(n, k, arr):
res=1000000000000000000000 #alternatively make it equal with first subarray
for i in range(n-k):
res=min(res, myvar(l[i:i+k]))
return res

I see this question still has no complete answer. I will write a track of a correct algorithm which will pass the judge. I will not write the code in order to respect the purpose of the Hackerrank challenge. Since we have working solutions.
The original array must be sorted. This has a complexity of O(NlogN)
At this point you can check consecutive sub arrays as non-consecutive ones will result in a worse (or equal, but not better) "unfairness sum". This is also explained in archer's answer
The last check passage, to find the minimum "unfairness sum" can be done in O(N). You need to calculate the US for every consecutive k-long subarray. The mistake is recalculating this for every step, done in O(k), which brings the complexity of this passage to O(k*N). It can be done in O(1) as the editorial you posted shows, including mathematic formulae. It requires a previous initialization of a cumulative array after step 1 (done in O(N) with space complexity O(N) too).
It works but terminates due to time out for n<=10000.
(from comments on archer's question)
To explain step 3, think about k = 100. You are scrolling the N-long array and the first iteration, you must calculate the US for the sub array from element 0 to 99 as usual, requiring 100 passages. The next step needs you to calculate the same for a sub array that only differs from the previous by 1 element 1 to 100. Then 2 to 101, etc.
If it helps, think of it like a snake. One block is removed and one is added.
There is no need to perform the whole O(k) scrolling. Just figure the maths as explained in the editorial and you will do it in O(1).
So the final complexity will asymptotically be O(NlogN) due to the first sort.

How to calculate Delta F / F using python?

I've recently "taught" myself python in order to analyze data for my experiments. As such I'm pretty clueless on many aspects. I've managed to make my analysis work for certain files but in some cases it breaks down and I imagine it is a result of faulty programming.
Currently I export a file containing 3 numpy arrays. One of these arrays is my signal (float values from -10 to 10). What I wish to do is to normalize every datum in this array to a range of values that preceed it. (i.e. the 30001st value must have the average of the preceeding 3000 values subtracted from it and then the difference must then be divided by thisvery same average (the preceeding 3000 values). My data is collected at a rate of 100Hz thus to get a normalization of the alst 30s i must use the preceeding 3000values.
As it stand this is how I've managed to make it work:
this stores the signal into the variable photosignal
photosignal = np.array(seg.analogsignals[0], ndmin=1)
now this the part I use to get the delta F/F over a moving window of 30s
normalizedphotosignal = [(uu-(np.mean(photosignal[uu-3000:uu])))/abs(np.mean(photosignal[uu-3000:uu])) for uu in photosignal[3000:]]
The following adds 3000 values to the beginning to keep the array the same length since later on i must time lock it to another list that is the same length
holder =list(range(3000))
normalizedphotosignal = holder + normalizedphotosignal
What I have noticed is that in certain files this code gives me an error because it says that the"slice" is empty and therefore it cannot create a mean.
I think maybe there is a better way to program this that could avoid this problem altogether. Or this a correct way to approach this problem?
So i tried the solution but it is quite slow and it nevertheless still gives me the "empty slice error".
I went over the moving average post and found this method:
def running_mean(x, N):
cumsum = np.cumsum(np.insert(x, 0, 0))
return (cumsum[N:] - cumsum[:-N]) / N
however I'm having trouble accommodating it to my desired output. namely (x-running average)/running average

Allright so I finally figured it out thanks to your help and the posts you referred me to.
The calculation for my entire data (300 000 +) takes about a second!
I used the following code:
def runningmean(x,N):
cumsum =np.cumsum(np.insert(x,0,0))
return (cumsum[N:] -cumsum[:-N])/N
photosignal = np.array(seg.analogsignal[0], ndmin =1)
photosignalaverage = runningmean(photosignal, 3000)
holder = np.zeros(2999)
photosignalaverage = np.append(holder,photosignalaverage)
detalfsignal = (photosignal-photosignalaverage)/abs(photosignalaverage)
Photosignal stores my raw signal in a numpy array.
Photosignalaverage uses cumsum to calculate the running average of every datapoint in photosignal. I then add the first 2999 values as 0, to maintian the same list size as my photosignal.
I then use basic numpy calculations to get my delta F/F signal.
Thank you once more for the feedback, was truly helpful!

Your approach goes in the right direction. However, you made a mistake in your list comprehension: you are using uu as your index whereas uu are the elements of your input data photosignal.
You want something like this:
normalizedphotosignal2 = np.zeros((photosignal.shape[0]-3000))
for i, uu in enumerate(photosignal[3000:]):
normalizedphotosignal2 = (uu - (np.mean(photosignal[i-3000:i]))) / abs(np.mean(photosignal[i-3000:i]))
Keep in mind that for-loops are relatively slow in python. If performance is an issue here, you could try avoiding the for loop and use numpy methods instead (e.g. have a look at Moving average or running mean).
Hope this helps.

Avoid python loop in deleting error prone values

My timelines are stored in simple numpy Arrays, and they are long (>10 Million entrys)
I have to detect machine shutdowns, that show in jumps in the time vector . After that shutdown I want do delete the next 10 values (The sensors do give bad results for a while after being switched on) and continue.
I came up with the following code:
Keep_data=np.empty_like(Timestamp_new,dtype=np.bool)
Keep_data[0]=False
Keep_data[1:]=Timestamp_new[1:]>(Timestamp_new[:-1]+min_shutdown_length)
for item in np.nonzero(np.logical_not(Keep_data))[0]:
Keep_data[item:min(item+10,len(Keep_data)]=False
Timestampnew=Timestampnew[Keep_data]
Can anyone suggest a more effective code, without a pure python Loop?
Thank you.

Basically you are trying to spread/grow or in image-processing terms dilate the False regions. For the same, we have a built-in as scipy's binary_dilation. Now, you are trying to make it grow starting from each such False element in input array Keep_data towards higher indices. So, we need to use a different offset (or as scipy calls it : origin) than the default one as 0, which otherwise would have dilated across both ends for each element.
Thus, to sum up, an implementation with it to get rid of the loopy portion of the code, we would have an implementation like so -
N = 10 # Interval length
dilated_mask = binary_dilation(~Keep_data, structure=np.ones(N),origin=-int(N/2))
Keep_data[dilated_mask] = False
An alternative approach that would be closer to the one posted as the loopy code in the question, but vectorized with NumPy's broadcasting feature, would look something like this -
N = 10 # Interval length
idx = np.nonzero(np.logical_not(Keep_data[:-N]))[0]
Keep_datac[(idx + np.arange(N)[:,None]).ravel()] = False
rest = np.nonzero(np.logical_not(Keep_data[-N:]))[0]
if len(rest)>0:
Keep_datac[-N+rest[0]:] = False

Python - comparing elements of list with 'neighbour' elements

This may be more of an 'approach' or conceptual question.
Basically, I have a python a multi-dimensional list like so:
my_list = [[0,1,1,1,0,1], [1,1,1,0,0,1], [1,1,0,0,0,1], [1,1,1,1,1,1]]
What I have to do is iterate through the array and compare each element with those directly surrounding it as though the list was layed out as a matrix.
For instance, given the first element of the first row, my_list[0][0], I need to know know the value of my_list[0][1], my_list[1][0] and my_list[1][1]. The value of the 'surrounding' elements will determine how the current element should be operated on. Of course for an element in the heart of the array, 8 comparisons will be necessary.
Now I know I could simply iterate through the array and compare with the indexed values, as above. I was curious as to whether there was a more efficient way which limited the amount of iteration required? Should I iterate through the array as is, or iterate and compare only values to either side and then transpose the array and run it again. This, however would ignore those values to the diagonal. And should I store results of the element lookups, so I don't keep determining the value of the same element multiple times?
I suspect this may have a fundamental approach in Computer Science, and I am eager to get feedback on the best approach using Python as opposed to looking for a specific answer to my problem.

You may get faster, and possibly even simpler, code by using numpy, or other alternatives (see below for details). But from a theoretical point of view, in terms of algorithmic complexity, the best you can get is O(N*M), and you can do that with your design (if I understand it correctly). For example:
def neighbors(matrix, row, col):
for i in row-1, row, row+1:
if i < 0 or i == len(matrix): continue
for j in col-1, col, col+1:
if j < 0 or j == len(matrix[i]): continue
if i == row and j == col: continue
yield matrix[i][j]
matrix = [[0,1,1,1,0,1], [1,1,1,0,0,1], [1,1,0,0,0,1], [1,1,1,1,1,1]]
for i, row in enumerate(matrix):
for j, cell in enumerate(cell):
for neighbor in neighbors(matrix, i, j):
do_stuff(cell, neighbor)
This has takes N * M * 8 steps (actually, a bit less than that, because many cells will have fewer than 8 neighbors). And algorithmically, there's no way you can do better than O(N * M). So, you're done.
(In some cases, you can make things simpler—with no significant change either way in performance—by thinking in terms of iterator transformations. For example, you can easily create a grouper over adjacent triplets from a list a by properly zipping a, a[1:], and a[2:], and you can extend this to adjacent 2-dimensional nonets. But I think in this case, it would just make your code more complicated that writing an explicit neighbors iterator and explicit for loops over the matrix.)
However, practically, you can get a whole lot faster, in various ways. For example:
Using numpy, you may get an order of magnitude or so faster. When you're iterating a tight loop and doing simple arithmetic, that's one of the things that Python is particularly slow at, and numpy can do it in C (or Fortran) instead.
Using your favorite GPGPU library, you can explicitly vectorize your operations.
Using multiprocessing, you can break the matrix up into pieces and perform multiple pieces in parallel on separate cores (or even separate machines).
Of course for a single 4x6 matrix, none of these are worth doing… except possibly for numpy, which may make your code simpler as well as faster, as long as you can express your operations naturally in matrix/broadcast terms.
In fact, even if you can't easily express things that way, just using numpy to store the matrix may make things a little simpler (and save some memory, if that matters). For example, numpy can let you access a single column from a matrix naturally, while in pure Python, you need to write something like [row[col] for row in matrix].
So, how would you tackle this with numpy?
First, you should read over numpy.matrix and ufunc (or, better, some higher-level tutorial, but I don't have one to recommend) before going too much further.
Anyway, it depends on what you're doing with each set of neighbors, but there are three basic ideas.
First, if you can convert your operation into simple matrix math, that's always easiest.
If not, you can create 8 "neighbor matrices" just by shifting the matrix in each direction, then perform simple operations against each neighbor. For some cases, it may be easier to start with an N+2 x N+2 matrix with suitable "empty" values (usually 0 or nan) in the outer rim. Alternatively, you can shift the matrix over and fill in empty values. Or, for some operations, you don't need an identical-sized matrix, so you can just crop the matrix to create a neighbor. It really depends on what operations you want to do.
For example, taking your input as a fixed 6x4 board for the Game of Life:
def neighbors(matrix):
for i in -1, 0, 1:
for j in -1, 0, 1:
if i == 0 and j == 0: continue
yield np.roll(np.roll(matrix, i, 0), j, 1)
matrix = np.matrix([[0,0,0,0,0,0,0,0],
[0,0,1,1,1,0,1,0],
[0,1,1,1,0,0,1,0],
[0,1,1,0,0,0,1,0],
[0,1,1,1,1,1,1,0],
[0,0,0,0,0,0,0,0]])
while True:
livecount = sum(neighbors(matrix))
matrix = (matrix & (livecount==2)) | (livecount==3)
(Note that this isn't the best way to solve this problem, but I think it's relatively easy to understand, and likely to illuminate whatever your actual problem is.)

Python: sliding window of variable width

I'm writing a program in Python that's processing some data generated during experiments, and it needs to estimate the slope of the data. I've written a piece of code that does this quite nicely, but it's horribly slow (and I'm not very patient). Let me explain how this code works:
1) It grabs a small piece of data of size dx (starting with 3 datapoints)
2) It evaluates whether the difference (i.e. |y(x+dx)-y(x-dx)| ) is larger than a certain minimum value (40x std. dev. of noise)
3) If the difference is large enough, it will calculate the slope using OLS regression. If the difference is too small, it will increase dx and redo the loop with this new dx
4) This continues for all the datapoints
[See updated code further down]
For a datasize of about 100k measurements, this takes about 40 minutes, whereas the rest of the program (it does more processing than just this bit) takes about 10 seconds. I am certain there is a much more efficient way of doing these operations, could you guys please help me out?
Thanks
EDIT:
Ok, so I've got the problem solved by using only binary searches, limiting the number of allowed steps by 200. I thank everyone for their input and I selected the answer that helped me most.
FINAL UPDATED CODE:
def slope(self, data, time):
(wave1, wave2) = wt.dwt(data, "db3")
std = 2*np.std(wave2)
e = std/0.05
de = 5*std
N = len(data)
slopes = np.ones(shape=(N,))
data2 = np.concatenate((-data[::-1]+2*data[0], data, -data[::-1]+2*data[N-1]))
time2 = np.concatenate((-time[::-1]+2*time[0], time, -time[::-1]+2*time[N-1]))
for n in xrange(N+1, 2*N):
left = N+1
right = 2*N
for i in xrange(200):
mid = int(0.5*(left+right))
diff = np.abs(data2[n-mid+N]-data2[n+mid-N])
if diff >= e:
if diff < e + de:
break
right = mid - 1
continue
left = mid + 1
leftlim = n - mid + N
rightlim = n + mid - N
y = data2[leftlim:rightlim:int(0.05*(rightlim-leftlim)+1)]
x = time2[leftlim:rightlim:int(0.05*(rightlim-leftlim)+1)]
xavg = np.average(x)
yavg = np.average(y)
xlen = len(x)
slopes[n-N] = (np.dot(x,y)-xavg*yavg*xlen)/(np.dot(x,x)-xavg*xavg*xlen)
return np.array(slopes)

Your comments suggest that you need to find a better method to estimate ik+1 given ik. No knowledge of values in data would yield to the naive algorithm:
At each iteration for n, leave i at previous value, and see if the abs(data[start]-data[end]) value is less than e. If it is, leave i at its previous value, and find your new one by incrementing it by 1 as you do now. If it is greater, or equal, do a binary search on i to find the appropriate value. You can possibly do a binary search forwards, but finding a good candidate upper limit without knowledge of data can prove to be difficult. This algorithm won't perform worse than your current estimation method.
If you know that data is kind of smooth (no sudden jumps, and hence a smooth plot for all i values) and monotonically increasing, you can replace the binary search with a search backwards by decrementing its value by 1 instead.

How to optimize this will depend on some properties of your data, but here are some ideas:
Have you tried profiling the code? Using one of the Python profilers can give you some useful information about what's taking the most time. Often, a piece of code you've just written will have one biggest bottleneck, and it's not always obvious which piece it is; profiling lets you figure that out and attack the main bottleneck first.
Do you know what typical values of i are? If you have some idea, you can speed things up by starting with i greater than 0 (as #vhallac noted), or by increasing i by larger amounts — if you often see big values for i, increase i by 2 or 3 at a time; if the distribution of is has a long tail, try doubling it each time; etc.
Do you need all the data when doing the least squares regression? If that function call is the bottleneck, you may be able to speed it up by using only some of the data in the range. Suppose, for instance, that at a particular point, you need i to be 200 to see a large enough (above-noise) change in the data. But you may not need all 400 points to get a good estimate of the slope — just using 10 or 20 points, evenly spaced in the start:end range, may be sufficient, and might speed up the code a lot.

I work with Python for similar analyses, and have a few suggestions to make. I didn't look at the details of your code, just to your problem statement:
1) It grabs a small piece of data of size dx (starting with 3
datapoints)
2) It evaluates whether the difference (i.e. |y(x+dx)-y(x-dx)| ) is
larger than a certain minimum value (40x std. dev. of noise)
3) If the difference is large enough, it will calculate the slope
using OLS regression. If the difference is too small, it will increase
dx and redo the loop with this new dx
4) This continues for all the datapoints
I think the more obvious reason for slow execution is the LOOPING nature of your code, when perhaps you could use the VECTORIZED (array-based operations) nature of Numpy.
For step 1, instead of taking pairs of points, you can perform directly `data[3:] - data[-3:] and get all the differences in a single array operation;
For step 2, you can use the result from array-based tests like numpy.argwhere(data > threshold) instead of testing every element inside some loop;
Step 3 sounds conceptually wrong to me. You say that if the difference is too small, it will increase dx. But if the difference is small, the resulting slope would be small because it IS actually small. Then, getting a small value is the right result, and artificially increasing dx to get a "better" result might not be what you want. Well, it might actually be what you want, but you should consider this. I would suggest that you calculate the slope for a fixed dx across the whole data, and then take the resulting array of slopes to select your regions of interest (for example, using data_slope[numpy.argwhere(data_slope > minimum_slope)].
Hope this helps!