Creating arrays with a loop (Python)

Question

I am trying to create several arrays from a big array that I have. What I mean is:

data = [[0, 1, 0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 1, 0, 0, 1, 0, 0, 0, 0], 
[0, 1, 1, 0, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 1, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 0, 0, 0, 0,1], [0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 1, 0, 0], [0, 1, 0, 0, 0, 1, 0, 0, 0, 0], 
[0, 0, 0, 0, 1, 0, 0, 0, 1, 0]]

I want to create 10 different arrays - using the 10 data's columns - with different names.

data1 = [0, 0, 0, 1, 0, 0, 1, 0, 0],
data2 = [1, 0, 1, 0, 0, 0, 0, 1, 0], and so on

I found a close solution here - Also I take the example data from there - However, when I tried the solution suggested:

for d in xrange(0,9):
exec 'x%s = data[:,%s]' %(d,d-1)

A error message appears:

exec(code_obj, self.user_global_ns, self.user_ns)

  File "", line 2, in 
    exec ('x%s = data[:,%s]') %(d,d-1)

  File "", line 1
    x%s = data[:,%s]
                 ^
SyntaxError: invalid syntax

Please, any comments will be highly appreciated. Regards

Don't. Just don't. Dynamically creating variables looks fancy but leads to unreadable and unmaintainable code. Use dictionaries or lists instead. — Serge Ballesta
– Serge Ballesta, Commented May 27, 2020 at 14:55
There really isn't a need to even use dictioary. Use numpy array indexing. — Scott Boston
– Scott Boston, Commented May 27, 2020 at 14:57

Scott Boston · Accepted Answer · 2020-05-27 18:09:52Z

1

Use numpy array index:

data = [[0, 1, 0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 1, 0, 0, 1, 0, 0, 0, 0], 
[0, 1, 1, 0, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 1, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 0, 0, 0, 0,1], [0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 1, 0, 0], [0, 1, 0, 0, 0, 1, 0, 0, 0, 0], 
[0, 0, 0, 0, 1, 0, 0, 0, 1, 0]]

d = np.array(data)

d[:, 0]
#array([0, 0, 0, 1, 0, 0, 1, 0, 0])

d[:, 1]
#array([1, 0, 1, 0, 0, 0, 0, 1, 0])

etc...

d[:, 9]
#array([0, 0, 1, 1, 1, 0, 0, 0, 0])

If you must, then dictionaries are the way to go:

val = {i:d[:,i] for i in range(d.shape[1])}

To access the arrays:

val[0]
#array([0, 0, 0, 1, 0, 0, 1, 0, 0])

...

val[9] 
#array([0, 0, 1, 1, 1, 0, 0, 0, 0])

edited May 27, 2020 at 18:09

answered May 27, 2020 at 14:59

Scott Boston

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3 Comments

Alí Meres Vargas Over a year ago

Thanks for the suggestion, however, I would like to automotive these outputs in several arrays with different names each. Regards

Scott Boston Over a year ago

@AlíMeresVargas Okay then, dictionaries is the way to go.

Alí Meres Vargas Over a year ago

I believe that there is a numpy way to do it. Hope to find it soon.

Code Pope · Accepted Answer · 2020-05-27 14:59:59Z

0

Use the following code (it is also more readable -- for python 3.x) if you really want to create dynamic variables:

for d in range(0,9):
  # exec 'x%s = data[:,%s]' %(d,d-1)
  exec(f"data{d} = {data[d]}" )

answered May 27, 2020 at 14:59

Code Pope

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1 Comment

Alí Meres Vargas Over a year ago

Hi! I just checked the answer and works fine in case I change the exec for print. When I run with exec this message appears: exec(f"data{d} = {data[:,d]}") File "<string>", line 1 data0 = [0 0 0 1 0 0 1 0 0] ^ SyntaxError: invalid syntax I would like to design this function in order to use data1 and data2, for example, later in other analysis. Is that possible? Regards

Anuj Sharma · Accepted Answer · 2020-05-27 17:36:52Z

0

Either use numpy array as shown by scott boston above or use dictionary like this:

a = {}

for i in range(0,9):
    a[i] = data[i][:]

Output:

{0: [0, 1, 0, 0, 0, 0, 0, 1, 0, 0],
 1: [0, 0, 1, 0, 0, 1, 0, 0, 0, 0],
 2: [0, 1, 1, 0, 0, 0, 0, 0, 0, 1],...

answered May 27, 2020 at 17:36

Anuj Sharma

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Comments

Antonios Marios Christonasis · Accepted Answer · 2020-05-28 10:28:02Z

0

I don't see the proper indentation in your for loop.
I suggest you don't use %s for the second argument (string) but rather %d (number) since you need a number to do the indexing of your array.

edited May 28, 2020 at 10:28

answered May 27, 2020 at 15:00

Antonios Marios Christonasis

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2 Comments

Anuj Sharma Over a year ago

Please add comment in comments only.

Antonios Marios Christonasis Over a year ago

I am providing my answer to the problem, that's why it's not a comment.

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