上一个和下一个数组应该具有两个很常见的值pandas

Question

我有一个数据框，如，

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]

我正在尝试比较两个数字应该匹配并且第三个数字应该为o的两个数组。因此，像[1,0,1] [1,1,0]应该被匹配并且当两个数字匹配并且第三个数字为0时返回true。对于[2,2,0] [0,2,2]相同，但[1,2,0] [1,0,2]不应匹配，因为它没有相同的数字。将返回false。那么，有什么办法可以做到这一点？

Answer 1

这是您需要的测试功能（我认为，假设所有值均为正）

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              

并在pandas中申请（不是熊猫专家，但我认为这应该可行）：

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

无论如何，this question应该可以帮助您将函数应用于两列。

Answer 2

你的意思是这样吗？

>>> 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

Answer 3

使用此功能，您可以比较两个元组并检查它们是否满足您的要求：

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

以下是该函数的一些示例运行：

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