Previous and next array should have two much common values pandas

Question

I have a dataframe like ,

Test              Test1
[1,1,1]          [1,2,2]
[1,2,2]          [1,0,1]
[1,0,1]          [1,1,0]
[2,2,0]          [0,2,2]
[1,2,0]          [1,0,2]

I am trying to compare the two arrays where 2 numbers should match and third one should be o.So like [1,0,1] [1,1,0] should be matched and return true as two numbers are matching and third one is 0 . same for [2,2,0] [0,2,2] but [1,2,0] [1,0,2] this should not match as it does not have the same numbers.will return false. So,Is there any way to do this ?

Each 3-tuple is judged individually? Or in pairs of 3-tuples? — Daniel F, Nov 07 '19 at 09:17
Can you explain what should be `True` and `False` in the comparison in more detail? Why is `[1,2,0]` and `[1,0,2]` evaluated to `False`. Contains two matching numbers and zeros. — AnsFourtyTwo, Nov 07 '19 at 09:19
I mean it does not have the same number it's 1 and 2. So because of that.. if it would have [1,0,1] and [1,1,0] then it would have been true — ganesh kaspate, Nov 07 '19 at 09:21
And you need the second 3-tuple to have the same numbers as well? So `[1,1,0]` and `[2,2,0]` would evaluate to false? — AnsFourtyTwo, Nov 07 '19 at 09:24

AnsFourtyTwo · Answer 1 · 2019-11-07T12:10:54.340

0

Using this function, you can compare two tuples and check if they fulfill your requirements:

def eval_tuples(tup1, tup2):
    # Check if zeros occures once in each tuple
    if tup1.count(0) != 1 or tup2.count(0) != 1:
        return False

    # Get non-zero values
    rem1 = [x for x in tup1 if x != 0]
    rem2 = [x for x in tup2 if x != 0]

    # Check if they are equal in both tuples
    if rem1[0] != rem1[1] or rem2[0] != rem2[1]:
        return False
    return True

Here some example runs of the function:

print(eval_tuples([1,1,0], [1,0,1]))
# True
print(eval_tuples([1,1,0], [2,0,1]))
# False
print(eval_tuples([2,2,0], [0,2,2]))
# True
print(eval_tuples([1,2,0], [1,0,2]))
# False

edited Nov 07 '19 at 12:10

answered Nov 07 '19 at 09:37

AnsFourtyTwo

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Hey, so what If I try to match [1,2,0] and [1,0,2] . How will it be ? – ganesh kaspate Nov 07 '19 at 10:28
Added some exmple runs in the answer – AnsFourtyTwo Nov 07 '19 at 12:11
No No, Here I was asking, If I want [1,2,0] and [1,0,2] this type of condition I want to be true that time – ganesh kaspate Nov 07 '19 at 12:19
In your question you write: "... `[1,2,0] [1,0,2]` this should not match as it does not have the same numbers.will return false. ...". – AnsFourtyTwo Nov 07 '19 at 12:30
If you want `[1,2,0]` and `[1,0,2]` to be evaluated to `True`, simply replace `if rem1[0] != rem1[1] or rem2[0] != rem2[1]:` with `set(rem1) == set(rem2):`! – AnsFourtyTwo Nov 07 '19 at 12:37

Daniel F · Accepted Answer · 2019-11-07T09:54:21.543

0

This is the testing function you need (I think, assuming all values are positive)

def test_func(x):
    A = x[0]
    B = x[1]
    f = lambda X: np.unique(X, return_counts = True)
    Au, Ac = f(A)
    Bu, Bc = f(B)
    return np.all(Au == Bu) and \ 
           Au.size == 2 and  \     
           Ac[0] == 1 and \        
           Bc[0] == 1

And to apply in in pandas (not a pandas expert, but I think this should work):

df['new_col'] = df[['Test', 'Test1']].apply(test_func, axis = 1)

in any case, this question should help you apply your function over two columns.

edited Nov 07 '19 at 09:54

answered Nov 07 '19 at 09:44

Daniel F

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Something unexpected char it is giving for \ – ganesh kaspate Nov 07 '19 at 09:52
get rid of the comments – Daniel F Nov 07 '19 at 09:53
could you please check https://stackoverflow.com/questions/58991053/create-a-sub-columns-in-the-dataframe-using-a-another-dataframe?noredirect=1#comment104233258_58991053 – ganesh kaspate Nov 22 '19 at 09:45

score 0 · Answer 3 · answered Nov 07 '19 at 09:54

Do you mean like this?

>>> def compare_tuples(x, y):
...     return (
...         # there is any series of size two which is common in both tuples
...         any(x[i:i+2] == y[j:j+2] for i in range(2) for j in range(2))
...         # there is at least one zero at third position
...         and 0 in (x[2], y[2])
...     )

>>> df = pd.DataFrame.from_records([
...     [(1, 1, 1), (1, 2, 2)],  # False
...     [(1, 2, 2), (1, 0, 1)],  # False
...     [(1, 0, 1), (1, 1, 0)],  # True - (1, 0) matches and 0 in third position
...     [(2, 2, 0), (0, 2, 2)],  # True - (2, 2) matches and 0 in third position
...     [(1, 2, 0), (1, 0, 2)],  # False
... ], columns=["Test1", "Test2"])

>>> df.apply(lambda x: compare_tuples(*x), axis=1)
0    False
1    False
2     True
3     True
4    False
dtype: bool

Previous and next array should have two much common values pandas

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