Numpy array manipulation

Question

I have an array like -

x =  array([0, 1, 2, 3,4,5])

And I want the output like this -

[]
[1]
[1 2]
[1 2 3]
[1 2 3 4]
[1 2 3 4 5]

I tried this code-

y = np.array([np.arange(1,i) for i in x+1])

But it makes a list with dtype object which I dont want. I want it ot be integer so that I can indexed it later.

You cannot have an ordinary numpy array with non-uniform shape. That is, each row must be the same size if you want a 2d numpy array with the handy numpy indexing and slicing. This is why you get a 1d array of 1d arrays, it's just like a list of lists, except each item is an array. — askewchan
– askewchan, Commented May 14, 2013 at 17:43

Bonlenfum · Accepted Answer · 2013-05-14 17:39:33Z

1

If I understand the question correctly, is

y =  [np.arange(1,i) for i in x+1]

suitable? You can access the lists that make up the rows with y[r], e.g.,

>>> y[2] 
array([1, 2])

or the whole lot with y:

>>> y
[array([], dtype=int64),
 array([1]),
 array([1, 2]),
 array([1, 2, 3]),
 array([1, 2, 3, 4]),
 array([1, 2, 3, 4, 5])]

Also note that you can control the data type of the arrays returned by arange here by setting dtype=int (or similar).

answered May 14, 2013 at 17:39

Bonlenfum

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Comments

Solkar · Accepted Answer · 2013-05-14 17:50:58Z

0

And I want the output like this

Just outputting it that way is simply an of slicing:

import numpy as np
x =  np.array([0, 1, 2, 3, 4, 5])

for i in range(1,len(x) + 1):
    print(x[1:i])

answered May 14, 2013 at 17:50

Solkar

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