How to delete particular array in 2 dimensional NumPy array by value?-CodePudding

Let the 2-dimensional array is as below:

In [1]: a = [[1, 2], [3, 4], [5, 6], [1, 2], [7, 8]]
        a = np.array(a)
        a, type(a)
Out [1]: (array([[1, 2],
                 [3, 4],
                 [5, 6],
                 [1, 2],
                 [7, 8]]),
         numpy.ndarray)

I have tried to do this procedure:

In [2]: a = a[a != [1, 2])
        a = np.reshape(a, (int(a.size/2), 2) # I have to do this since on the first line in In [2] change the dimension to 1 [3, 4, 5, 6, 7, 8] (the initial array is 2-dimensional array)
        a
Out[2]: array([[3, 4],
               [5, 6],
               [7, 8]])

My question is, is there any function in NumPy that can directly do that?

Updated Question

Here's the semi-full source code that I've been working on:

from sklearn import datasets
data = datasets.load_iris()
df = pd.DataFrame(data.data, columns=data.feature_names)
df['Target'] = pd.DataFrame(data.target)

bucket = df[df['Target'] == 0]
bucket = bucket.iloc[:,[0,1]].values
lp, rp = leftestRightest(bucket)
bucket = np.array([x for x in bucket if list(x) != lp])
bucket = np.array([x for x in bucket if list(x) != rp])

Notes:

leftestRightest(arg) is a function that returns 2 one-dimensional NumPy arrays of size 2 (which are lp and rp). For instances, lp = [1, 3], rp = [2, 4] and the parameter is 2-dimensional NumPy array

CodePudding user response：

You're approach is correct, but the mask needs to be single-dimensional:

a[(a != [1, 2]).all(-1)]

Output:

array([[3, 4],
       [5, 6],
       [7, 8]])

Alternatively, you can collect the elements and infer the dimension with -1:

a[a != [1, 2]].reshape(-1, 2)

CodePudding user response：

the boolean condition creates a 2D array of True/False. You have to apply and operation across the columns to make sure the match is not a partial match. Consider a row [5,2] in your above array, the script you wrote will add 5 and ignore 2 in the resultant 1D array. It can be done as follows:

a[np.all(a != [1, 2],axis=1)]

CodePudding user response：

There should be a more delicate approach, but here what I have come up with:

np.array([x for x in a if list(x) != [1,2]])

Output

[[3, 4], [5, 6], [7, 8]]

Note that I wouldn't recommend working with list comprehensions in the large array since it would be highly time-consuming.