What is the fastest way to stack numpy arrays in a loop?

Question

I have a code that generates me within a for loop two numpy arrays (data_transform). In the first loop generates a numpy array of (40, 2) and in the second loop one of (175, 2). I want to concatenate these two arrays into one, to give me an array of (215, 2). I tried with np.concatenate and with np.append, but it gives me an error since the arrays must be the same size. Here is an example of how I am doing the code:

result_arr = np.array([])

for label in labels_set:
    data = [index for index, value in enumerate(labels_list) if value == label]
    for i in data:
        sub_corpus.append(corpus[i])
    data_sub_tfidf = vec.fit_transform(sub_corpus) 
    data_transform = pca.fit_transform(data_sub_tfidf) 
    #Append array
    sub_corpus = []

I have also used np.row_stack but nothing else gives me a value of (175, 2) which is the second array I want to concatenate.

Answer 1

What @hpaulj was trying to say with

Stick with list append when doing loops.

is

#use a normal list
result_arr = []

for label in labels_set:

    data_transform = pca.fit_transform(data_sub_tfidf) 

    # append the data_transform object to that list
    # Note: this is not np.append(), which is slow here
    result_arr.append(data_transform)

# and stack it after the loop
# This prevents slow memory allocation in the loop. 
# So only one large chunk of memory is allocated since
# the final size of the concatenated array is known.

result_arr = np.concatenate(result_arr)

# or 
result_arr = np.stack(result_arr, axis=0)

# or
result_arr = np.vstack(result_arr)

Your arrays don't really have different dimensions. They have one different dimension, the other one is identical. And in that case you can always stack along the "different" dimension.

Answer 2

Using concatenate, initializing "c":

a = np.array([[8,3,1],[2,5,1],[6,5,2]])
b = np.array([[2,5,1],[2,5,2]])
matrix = [a,b]

c = np.empty([0,matrix[0].shape[1]])

for v in matrix:
    c = np.append(c, v, axis=0)

Output:

[[8. 3. 1.]
 [2. 5. 1.]
 [6. 5. 2.]
 [2. 5. 1.]
 [2. 5. 2.]]