Python: Comparing all elements of two arrays and modifying 2nd array

Question

New to Python, and have been learning about arrays. I am stuck with a simple enough problem and need a solution. I have two arrays:

a = [2.0, 5.1, 6.2, 7.9, 23.0]     # always increasing
b = [5.1, 5.5, 5.7, 6.2, 00.0]     # also always increasing

and I want the resultant array to be:

c = [0.0, 5.1, 6.2, 0.0, 0.0]      # 5.5, 5.7, 00.0 from 'b' were dropped and rearranged such that position of equivalent elements as in 'a' are maintained

I have compared both 'a' & 'b' using Numpy as in:

y = np.isclose(a, b)
print y
# [False False False False False]

(Alternately,) I also tried something like this, which isn't the right way (I think):

c = np.zeros(len(a))
for i in range (len(a)):
    for j in range (len(a)):
        err = abs(a[i]-b[j])
        if err == 0.0 or err < abs(1):
            print (err, a[i], b[j], i, j)
        else:
            print (err, a[i], b[j], i, j)

How do I proceed from here towards obtaining 'c'?

Answer 1

These solutions work even when the arrays are of different size.

Simple version

c = []

for i in a:
    if any(np.isclose(i, b)):
        c.append(i)
    else:
        c.append(0.0)

Numpy version

aa = np.tile(a, (len(b), 1))
bb = np.tile(b, (len(a), 1))
cc = np.isclose(aa, bb.T)
np.any(cc, 0)
c = np.zeros(shape=a.shape)
result = np.where(np.any(cc, 0), a, c)

---

Explained:

I will be doing matrix comparison here. First you expand the arrays into matrices. Lengths are exchanged, which creates matrices having equal size of one dimension:

aa = np.tile(a, (len(b), 1))
bb = np.tile(b, (len(a), 1))

They look like this:

# aa
array([[  2. ,   5.1,   6.2,   7.9,  23. ],
       [  2. ,   5.1,   6.2,   7.9,  23. ],
       [  2. ,   5.1,   6.2,   7.9,  23. ],
       [  2. ,   5.1,   6.2,   7.9,  23. ],
       [  2. ,   5.1,   6.2,   7.9,  23. ]])

# bb
array([[ 5.1,  5.5,  5.7,  6.2,  0. ],
       [ 5.1,  5.5,  5.7,  6.2,  0. ],
       [ 5.1,  5.5,  5.7,  6.2,  0. ],
       [ 5.1,  5.5,  5.7,  6.2,  0. ],
       [ 5.1,  5.5,  5.7,  6.2,  0. ]])

Then compare them. Note that bb is transposed:

cc = np.isclose(aa, bb.T)

And you get:

array([[False,  True, False, False, False],
       [False, False, False, False, False],
       [False, False, False, False, False],
       [False, False,  True, False, False],
       [False, False, False, False, False]], dtype=bool)

You can aggregate this by axis 0:

np.any(cc, 0)

which returns

array([False,  True,  True, False, False], dtype=bool)

Now create array c:

c = np.zeros(shape=a.shape)

And select appropriate value, either from a or c:

np.where(np.any(cc, 0), a, c)

And the result:

array([ 0. ,  5.1,  6.2,  0. ,  0. ])

Answer 2

With np.isclose you already create an array where the "closest" elements are True. So you can use this result to set all other elements to zero.

import numpy as np
a = np.array([2.0, 5.1, 6.2, 7.9, 23.0])     # always increasing
b = np.array([5.1, 5.5, 5.7, 6.2, 00.0])     # also always increasing
a[~np.isclose(a,b, atol=0.5)] = 0
a

this returns array([ 0. , 5.1, 6.2, 0. , 0. ]).

But notice you want to set all elements that are not close, so you need to invert (~) the result.