I am new to python and facing a problem in dealing with Matrices.
I have a matrix, let's say
A = [1 0 0 2; 3 3 3 2; 3 3 0 2; 3 4 4 4]
Now, I want to make all the elements in the matrix equal to zero, except elements which are repeated the maximum number of times in a matrix. (In this case, it is 3).
So the expected result is,
B = [0 0 0 0; 3 3 3 0; 3 3 0 0;3 0 0 0]
It would be really helpful if anyone can help me with the python code for this.
Get the most occurring number across the entire array using Scipy's mode with axis set to None. Compare that number against the input array to give us a mask, which could be used to set the rest to zeros by elementwise multiplication with the input array/most-occurring number or using np.where to do the choosing.
Thus, one approach would be -
from scipy.stats import mode
most_occ_num = mode(A, axis=None)[0][0]
out = most_occ_num*(A==most_occ_num)
With np.where for array output -
out = np.where(A==most_occ_num,A,0)
Sample run -
In [129]: A = np.matrix([[1, 0 ,0 ,2],[ 3, 3, 3, 2],[ 3 ,3 ,0 ,2],[ 3 ,4 ,4 ,4]])
In [140]: A
Out[140]:
matrix([[1, 0, 0, 2],
[3, 3, 3, 2],
[3, 3, 0, 2],
[3, 4, 4, 4]])
In [141]: most_occ_num = mode(A, axis=None)[0][0]
In [142]: most_occ_num*(A==most_occ_num)
Out[142]:
matrix([[0, 0, 0, 0],
[3, 3, 3, 0],
[3, 3, 0, 0],
[3, 0, 0, 0]])
In [143]: np.where(A==most_occ_num,A,0)
Out[143]:
array([[0, 0, 0, 0],
[3, 3, 3, 0],
[3, 3, 0, 0],
[3, 0, 0, 0]])
----------------------------------------------------
This is MATLAB syntax, not numpy:
A = [1 0 0 2; 3 3 3 2; 3 3 0 2; 3 4 4 4]
though np.matrix emulates it with:
In [172]: A = np.matrix('1 0 0 2; 3 3 3 2; 3 3 0 2; 3 4 4 4')
In [173]: A
Out[173]:
matrix([[1, 0, 0, 2],
[3, 3, 3, 2],
[3, 3, 0, 2],
[3, 4, 4, 4]])
Your task is 2 fold, finding that most frequent element, and then replacing all the others. Neither action depends on the matrix being 2d, or being matrix as opposed to array.
In [174]: A1=A.A1
In [175]: A1
Out[175]: array([1, 0, 0, 2, 3, 3, 3, 2, 3, 3, 0, 2, 3, 4, 4, 4])
np.unique can give us the frequency count, so we can fine the most frequent value with (unique requires the 1d):
In [179]: u,c = np.unique(A1, return_counts=True)
In [180]: u
Out[180]: array([0, 1, 2, 3, 4])
In [181]: c
Out[181]: array([3, 1, 3, 6, 3])
In [182]: np.argmax(c)
Out[182]: 3
In [183]: u[np.argmax(c)]
Out[183]: 3
I'm surprised that Divakar use the scipy mode instead of unique. He's something of an expert in using unique. :)
Divakar's use of np.where may be the simplest way of performing the replace.
Just for the fun of it, here's a masked array approach:
In [196]: np.ma.MaskedArray(A, A!=3)
Out[196]:
masked_matrix(data =
[[-- -- -- --]
[3 3 3 --]
[3 3 -- --]
[3 -- -- --]],
mask =
[[ True True True True]
[False False False True]
[False False True True]
[False True True True]],
fill_value = 999999)
In [197]: _.filled(0)
Out[197]:
matrix([[0, 0, 0, 0],
[3, 3, 3, 0],
[3, 3, 0, 0],
[3, 0, 0, 0]])
Or an inplace change:
In [199]: A[A!=3] = 0
In [200]: A
Out[200]:
matrix([[0, 0, 0, 0],
[3, 3, 3, 0],
[3, 3, 0, 0],
[3, 0, 0, 0]])
First, flatten it:
flat = A.flatten[0]
# flat = [1, 0, 0, 2, .. ]
Then, find the mode of the list:
replace none modes:
B = A.copy()
B[B != mode] = 0
