Trjtdtk

I have a matrix $b$ with elements:
$$b =
beginpmatrix
0.01 & 0.02 & cdots & 1 \
0.01 & 0.02 & cdots & 1 \
vdots& vdots & ddots & vdots \
0.01 & 0.02 & cdots & 1 \
endpmatrix
$$For which through a series of calculation which is vectorised, $b$ is used to calculate $a$ which is another matrix that has the same dimension/shape as $b$.
$$a =
beginpmatrix
3 & 5 & cdots & 17 \
2 & 6 & cdots & 23 \
vdots& vdots & ddots & vdots \
4 & 3 & cdots & 19 \
endpmatrix
$$
At this point it is important to note that the elements of $a$ and $b$ have a one to one correspondence. The different row values(let's call it $sigma$) $0.01, 0.02...$ are different parameters for a series of simulations that I'm running. Hence for a fixed value of say $sigma = 0.01$, the length of its column values correspond to the total number of "simulations" I'm running for that particular parameter. If you know python vectorisation then you'll start to understand what I'm doing.

It is known that higher the $sigma$, the more the number of simulations for that particular sigma will have a value higher than 5 i.e. more of the matrix element along a column will have value bigger than 5. Essentially what I'm doing is vectorising $N$(columns) different simulations for $M$(rows) different parameters. Now I wish to find out the value of $sigma$ for which the total number simulation that's bigger than 5, is bigger than 95% of the total simulation.

To put it more concisely, for a $sigma$ of 0.02, each simulation would have results of $$5, 6, ..., 3$$ with say a total simulation of $N$. So let $$kappa = sum (textall the simulations that have values bigger than 5),$$I wish to find out the FIRST $sigma$ for which
$$frackappaN > 0.95*N$$
i.e. the FIRST $sigma$ for which the proportion of total experiment for which its value $>5$ is bigger than 95% of the total number of experiment.

The code that I have written is:

# say 10000 simulations for a particular sigma 
SIMULATION = 10000

# say 100 different values of sigma ranging from 0.01 to 1
# this is equivalent to matrix b in mathjax above
SIGMA = np.ones((EXPERIMENTS,100))*np.linspace(0.01, 1, 100)

def return_sigma(matrix, simulation, sigma):
 """
 My idea here is I put in sigma and matrix and total number of simulation. 
 Each time using np.ndenumerate looping over i and j to compare if the 
 element values are greater than 5. If yes then I add 1 to counter, if no 
 then continue. If the number of experiments with result bigger than 5 is 
 bigger than 95% of total number of experiment then I return that particular 
 sigma.
 """
 counter = 0
 for (i, j), value in np.ndenumerate(matrix):
 if value[i, j] > 5:
 counter+=1
 if counter/experiments > 0.95*simulation:
 break
 return sigma[0, j] # sigma[:, j] should all be the same anyway
"""Now this can be ran by:"""
print(return_sigma(a, SIMULATION, SIGMA))

which doesn't seem to quite work as I'm not well-versed with 2D slicing comprehension so this is quite a challenging problem for me. Thanks in advance.

EDIT
I apologise on not giving away my calculation as it's sort of a coursework of mine. I have generated a for 15 different values of $sigma$ with 15 simulations each, and here they are:

array([[ 6, 2, 12, 12, 14, 14, 11, 11, 9, 23, 15, 3, 10, 12, 10],
 [ 7, 7, 6, 9, 13, 8, 11, 17, 13, 8, 10, 16, 11, 16, 8],
 [14, 6, 4, 8, 10, 9, 11, 14, 12, 14, 5, 8, 18, 29, 22],
 [ 4, 12, 12, 3, 7, 8, 5, 13, 13, 10, 14, 16, 22, 15, 22],
 [ 9, 8, 7, 12, 12, 6, 4, 13, 12, 12, 18, 20, 18, 14, 23],
 [ 8, 6, 8, 6, 12, 11, 11, 4, 9, 9, 13, 19, 13, 11, 20],
 [12, 8, 7, 17, 3, 9, 11, 5, 12, 24, 11, 12, 17, 9, 16],
 [ 4, 8, 7, 5, 6, 10, 9, 6, 4, 13, 13, 14, 18, 20, 23],
 [ 5, 10, 5, 6, 8, 4, 7, 7, 10, 11, 9, 22, 14, 30, 17],
 [ 6, 4, 5, 9, 8, 8, 4, 21, 14, 18, 21, 13, 14, 22, 10],
 [ 6, 2, 7, 7, 8, 3, 7, 19, 14, 7, 13, 12, 18, 8, 12],
 [ 5, 7, 6, 4, 13, 9, 4, 3, 20, 11, 11, 8, 12, 29, 14],
 [ 6, 3, 13, 6, 12, 10, 17, 6, 9, 15, 12, 12, 16, 12, 15],
 [ 2, 9, 8, 15, 5, 4, 5, 7, 16, 13, 20, 18, 14, 18, 14],
 [14, 10, 7, 11, 8, 13, 14, 13, 12, 19, 9, 10, 11, 17, 13]])

As you can see as $sigma$ gets higher the number of matrix elements in each column for which it is bigger than 5 is higher.

EDIT 2
So now condition is giving me the right thing, which is an array of booleans.

array([[False, False, False, False, False, False, False, False, True, True],
 ....................................................................,
 [False, False, False, False, False, False, False, True, True, True]])

So now the last row is the important thing here as it corresponds to the parameters, in this case,

array([[0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.35, 0.40, 0.45, 0.5],
 ...........................................................,
 [0.05, 0.1 , 0.15, 0.2 , 0.25, 0.3 , 0.35, 0.4 , 0.45, 0.5]])

Now the last row of condition is telling me that first True happens at $sigma$=0.4 i.e. for which all the > 95% of the total simulations for that $sigma$ have simulation result of > 5. So now I need to return the index of condition where the first True in the last row appeared i.e. [i, j]. Now doing b[i, j] should give me the parameter I want.(which I'm not sure if your next few line of codes are doing that.)

I think I have understood your problem (mostly from the comments added in your function).

I'll show step by step what the logic is, building upon each previous step to get the final solution.

First we want to find all position where the matrix is larger than 5:

a > 5 # returns a boolean array with true/false in each position

Now we want to check each row to count if the proportion of of matches (> 5) has reached a certain threshold; $N * 0.95$. We can divide by the number of simulations (number of columns) to essentially normalise by the number of simulations:

(a > 5) / SIMULATION # returns the value of one match

These values are required to sum to your threshold for an experiment to be valid.

Now we cumulatively sum across each row. As the True/False array is ones and zeros, we now have a running total of the numbers of matches for each experiment (each row).

np.cumsum((a > 5) / SIMULATION, axis=1) # still same shape as b

Now we just need to find out where (in each row) the sum of matches reaches your threshold. We can use np.where:

## EDIT: we only need to check the cumsum is greater than 0.95 and not (0.95 * SUMLATION)
## because we already "normalised" the values within the cumsum.
condition = np.cumsum((a > 5) / SIMULATION, axis=0) > 0.95
mask = np.where(condition)

I broke it down now as the expressions are getting long.

That gave us the i and j coordinates of places where the condition was True. We just want to find the place where we first breached the threshold, so we want to find the indices for the first time in each row:

valid_rows = np.unique(mask[0], return_index=True)[1] # [1] gets the indices themselves

Now we can simply use these indices to get the first index in each valid row, where the threshold was breached:

valid_cols = mask[1][valid_rows]

So now you can get the corresponding values from the parameter matrix using these valid rows/columns:

params = b[valid_rows, valid_cols]

If this is correct, it should be significantly faster than your solution because it avoids looping over the 2d array and instead utilises NumPy's vectorised method and ufuncs.

Is this helpful?

import numpy as np, numpy.random as npr
N_sims = 15 # sims per sigma
N_vals = 15 # num sigmas
# Parameters
SIGMA = np.ones( (N_sims, N_vals) ) * np.linspace(0.01, 1, N_vals)
# Generate "results" :3 (i.e., the matrix a)
RESULTS = npr.random_integers(low=1, high=10, size=SIGMA.shape)
for i in range(N_vals): 
 RESULTS[:, i] += npr.random_integers(low=0, high=1, size=(N_sims)) + i // 3
print("SIGMAn", SIGMA)
print("RESULTSn", RESULTS)
# Mark the positions > 5
more_than_five = RESULTS > 5
print("more_than_fiven", more_than_five)
# Count how many are greater than five, per column (i.e., per sigma)
counts = more_than_five.sum(axis=0)
print('COUNTSn', counts)
# Compute the proportions (so, 1 if all exps were > 5)
proportions = counts.astype(float) / N_sims
print('Proportionsn', proportions)
# Find the first time it is larger than 0.5
first_index = np.argmax( proportions > 0.95 )
print('---nFIRST INDEXn', first_index)