CODEWITHSUNDEEP
pythonarrays
caterina
caterina

Reputation: 1

Creating a new array from a subset of another one

I need to create an array, correct_data, that is a subset of another one, all_data. This is all_data:

all_data = np.array([[[ 0, 11, 12, 13, 14],
                      [ 0, 21, 22, 23, 24],
                      [ 1, 31, 32, 33, 34],
                      [ 0, 41, 42, 43, 44]]])

The array correct_data will have all the the rows in all_data where the first item is a 1. The way I was doing it is by creating a mask that identifies the rows starting with 1, such as:

correct_mask=all_data[:,0]==1

and then identifies the indexes of such rows (using "where").

Then I create a for loop that takes each row from the index and copy it onto correct_data (using vstack?). Is this a practical way of doing it? If so, I need to create at fist correct_data as an empty array: can this be done?

Alternatively, I have tried to do it in a different way, i.e., delete the rows starting with a 0 from all_data

This is what I was trying:

for row in range(len(all_data)):
    this_row=all_data[row]
    if (this_row[0])==0:
        np.delete(all_data,row,0)

This does delete the rows, but then again how do I store the remaining rows in a new (or the same) array?

Many thanks for any help and clarification (I have a very basic knowledge of Python).

Upvotes: 0

Views: 215

Answers (3)

Ajay Maity
Ajay Maity

Reputation: 760

correct_data = all_data[all_data[:, :, 0]==1]

The index here: all_data[:, :, 0]==1 returns a boolean array where the 0th element in 3rd dimension is 1. Just use this to index the original array. You will get array([[ 1, 31, 32, 33, 34]]) as the result.

Upvotes: 0

tzaman
tzaman

Reputation: 47790

You can just filter it with an index condition:

>>> all_data[all_data[:,:,0]==1]
array([[ 1, 31, 32, 33, 34]])

Upvotes: 1

aledel
aledel

Reputation: 23

To get all arrays starting with 1 you can use the following:

correct_data = all_data[all_data[:,:, 0] == 1]

It will reassign matching arrays to the desired variable.

Upvotes: 0

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