根据百分比水平拆分2D numpy数组

Question

I want to be able to split 2D numpy horizontally into two splits (80% and 20%). I have tried using np.vsplit() but it seems it is not made for such a case. For instance, suppose I have the following matrix of size (6,3). I want to split it horizontally into 80% and 20% [roughly (5,3), (1,3)], so I tried something like this:

M = [[1,2,3],[4,5,6],[7,8,9], [10,11,12], [77,54,11], [424,78,98]]
M = np.asarray(M)
arr1 = np.vsplit(M, int(M.shape[0]* 0.8))[0]  # 80% of data goes to arr1
arr2 = np.vsplit(M, int(M.shape[0]* 0.2))[1]  # 20% of data goes to arr2

I know this try is incorrect but I can't fix it (actually still learning python). Kindly if someone can help to modify this code. Thank you

Answer 1

You can do it like this using Indexing (or use train_test_split ):

M = [[1,2,3],[4,5,6],[7,8,9], [10,11,12], [77,54,11], [424,78,98]]
M = np.asarray(M)

split_horizontally_idx = int(M.shape[0]* 0.8) # integer for line selection (horizontal selection)

array1 = M[:split_horizontally_idx , :] # indexing/selection of the 80%
array2 = M[split_horizontally_idx: , :] # indexing/selection of the remaining 20% 

Answer 2

You can do it by slicing the list according to your wish.

M = [[1,2,3],[4,5,6],[7,8,9], [10,11,12], [77,54,11], [424,78,98]]

M = np.asarray(M)

a80=M[:(round(0.8*len(M[:,2]))),:]
a20=M[:(round(0.2*len(M[:,2]))),:]
print(a80,"\n\n",a20)