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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

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1
  • What is list(bucket) != lp? If lp is an array, then this too is an array. You can't use that in the if clause. Why the list(bucket)? bucket is a values, an array. I assume lp is too. This may well be a case where applying all or any on an axis of bucket!-lp works. But you need to pay attention to array shape and dtype. Commented Feb 27, 2022 at 17:55

3 Answers 3

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1

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.

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11 Comments

This is not a good approach for numpy.
@mozway Thank you for the note. Would you mind explaining more? Referring to this link, this shouldn't be the worst.
You are reconstructing the full array instead of slicing. Try to run some timing on a large array, this should be slower.
I have tried this in my case (the explanation has been updated in the question). But it returns an error message. How to solve that?
@sempraEdic Not sure, but try editing your code from np.array([x for x in bucket if list(bucket) != lp]) to np.array([x for x in bucket if list(x) != lp])
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0

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)
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2 Comments

I have tried this too, but it delete an array that shouldn't be expected to be deleted. For instance, a = [[5.1, 3.5], [4.9, 3. ], [4.6, 3.1]]. And let variable lp = [4.3, 3.]. When I run a[(a != lp).all(-1)], it returns [[5.1, 3.5], [4.6, 3.1]] (with a[1] being deleted).
And for the larger case, it seems any array in that 2-dimensional array which has element 4.3 or 3. is deleted. How to handle that?
0

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)]

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