Initialize 2d array with variable size
Using the following method:
myArray = [0,1] * NUM_ITEMS
Desired result (2d array):
[[0,1],[0,1],[0,1]...]
Actual result (extended 1d array):
[0,1,0,1,0,1...]
How can I achieve the desired result preferably without using numpy?
python multidimensional-array
add a comment |
Using the following method:
myArray = [0,1] * NUM_ITEMS
Desired result (2d array):
[[0,1],[0,1],[0,1]...]
Actual result (extended 1d array):
[0,1,0,1,0,1...]
How can I achieve the desired result preferably without using numpy?
python multidimensional-array
6
Before someone suggests[[0, 1]]*NUM_ITEMS
, no, that doesn't work, even if it looks like it does.
– user2357112
Nov 20 '18 at 22:09
add a comment |
Using the following method:
myArray = [0,1] * NUM_ITEMS
Desired result (2d array):
[[0,1],[0,1],[0,1]...]
Actual result (extended 1d array):
[0,1,0,1,0,1...]
How can I achieve the desired result preferably without using numpy?
python multidimensional-array
Using the following method:
myArray = [0,1] * NUM_ITEMS
Desired result (2d array):
[[0,1],[0,1],[0,1]...]
Actual result (extended 1d array):
[0,1,0,1,0,1...]
How can I achieve the desired result preferably without using numpy?
python multidimensional-array
python multidimensional-array
asked Nov 20 '18 at 22:07
A__A__
4521518
4521518
6
Before someone suggests[[0, 1]]*NUM_ITEMS
, no, that doesn't work, even if it looks like it does.
– user2357112
Nov 20 '18 at 22:09
add a comment |
6
Before someone suggests[[0, 1]]*NUM_ITEMS
, no, that doesn't work, even if it looks like it does.
– user2357112
Nov 20 '18 at 22:09
6
6
Before someone suggests
[[0, 1]]*NUM_ITEMS
, no, that doesn't work, even if it looks like it does.– user2357112
Nov 20 '18 at 22:09
Before someone suggests
[[0, 1]]*NUM_ITEMS
, no, that doesn't work, even if it looks like it does.– user2357112
Nov 20 '18 at 22:09
add a comment |
2 Answers
2
active
oldest
votes
A list comprehension should do the trick:
>>> NUM_ITEMS = 5
>>> my_array = [[0, 1] for _ in range(NUM_ITEMS)]
>>> my_array
[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]
Huh, interesting thanks.
– A__
Nov 20 '18 at 22:10
add a comment |
Since you tagged arrays, here's an alternative numpy
solution using numpy.tile
.
>>> import numpy as np
>>> NUM_ITEMS = 10
>>> np.tile([0, 1], (NUM_ITEMS, 1))
array([[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1]])
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
A list comprehension should do the trick:
>>> NUM_ITEMS = 5
>>> my_array = [[0, 1] for _ in range(NUM_ITEMS)]
>>> my_array
[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]
Huh, interesting thanks.
– A__
Nov 20 '18 at 22:10
add a comment |
A list comprehension should do the trick:
>>> NUM_ITEMS = 5
>>> my_array = [[0, 1] for _ in range(NUM_ITEMS)]
>>> my_array
[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]
Huh, interesting thanks.
– A__
Nov 20 '18 at 22:10
add a comment |
A list comprehension should do the trick:
>>> NUM_ITEMS = 5
>>> my_array = [[0, 1] for _ in range(NUM_ITEMS)]
>>> my_array
[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]
A list comprehension should do the trick:
>>> NUM_ITEMS = 5
>>> my_array = [[0, 1] for _ in range(NUM_ITEMS)]
>>> my_array
[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]
answered Nov 20 '18 at 22:08
chrischris
898814
898814
Huh, interesting thanks.
– A__
Nov 20 '18 at 22:10
add a comment |
Huh, interesting thanks.
– A__
Nov 20 '18 at 22:10
Huh, interesting thanks.
– A__
Nov 20 '18 at 22:10
Huh, interesting thanks.
– A__
Nov 20 '18 at 22:10
add a comment |
Since you tagged arrays, here's an alternative numpy
solution using numpy.tile
.
>>> import numpy as np
>>> NUM_ITEMS = 10
>>> np.tile([0, 1], (NUM_ITEMS, 1))
array([[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1]])
add a comment |
Since you tagged arrays, here's an alternative numpy
solution using numpy.tile
.
>>> import numpy as np
>>> NUM_ITEMS = 10
>>> np.tile([0, 1], (NUM_ITEMS, 1))
array([[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1]])
add a comment |
Since you tagged arrays, here's an alternative numpy
solution using numpy.tile
.
>>> import numpy as np
>>> NUM_ITEMS = 10
>>> np.tile([0, 1], (NUM_ITEMS, 1))
array([[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1]])
Since you tagged arrays, here's an alternative numpy
solution using numpy.tile
.
>>> import numpy as np
>>> NUM_ITEMS = 10
>>> np.tile([0, 1], (NUM_ITEMS, 1))
array([[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1],
[0, 1]])
answered Nov 20 '18 at 22:19
timgebtimgeb
51k116693
51k116693
add a comment |
add a comment |
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6
Before someone suggests
[[0, 1]]*NUM_ITEMS
, no, that doesn't work, even if it looks like it does.– user2357112
Nov 20 '18 at 22:09