Mixing
Repeat values from a sequence up to n
Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:
seq = np.array([0,1])
a = np.tile(seq, math.ceil(n/2))[:n]
Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:
n = 6
seq = np.array([1,2,3,4])
np.tile(seq, math.ceil(n/2))[:n]
array([1, 2, 3, 4, 1, 2])
python list numpy
asked Jan 1 at 21:20
yatuyatu
13.4k31341
add a comment |
Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:
seq = np.array([0,1])
a = np.tile(seq, math.ceil(n/2))[:n]
Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:
n = 6
seq = np.array([1,2,3,4])
np.tile(seq, math.ceil(n/2))[:n]
array([1, 2, 3, 4, 1, 2])
python list numpy
asked Jan 1 at 21:20
yatuyatu
13.4k31341
add a comment |
Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:
seq = np.array([0,1])
a = np.tile(seq, math.ceil(n/2))[:n]
Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:
n = 6
seq = np.array([1,2,3,4])
np.tile(seq, math.ceil(n/2))[:n]
array([1, 2, 3, 4, 1, 2])
python list numpy
asked Jan 1 at 21:20
yatuyatu
13.4k31341
Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:
seq = np.array([0,1])
a = np.tile(seq, math.ceil(n/2))[:n]
Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:
n = 6
seq = np.array([1,2,3,4])
np.tile(seq, math.ceil(n/2))[:n]
array([1, 2, 3, 4, 1, 2])
python list numpy
python list numpy
asked Jan 1 at 21:20
yatuyatu
13.4k31341
asked Jan 1 at 21:20
yatuyatu
13.4k31341
asked Jan 1 at 21:20
yatuyatu
13.4k31341
asked Jan 1 at 21:20
yatuyatu
13.4k31341
asked Jan 1 at 21:20
yatuyatu
13.4k31341
13.4k31341
add a comment |
add a comment |
1 Answer
1
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oldest
votes
np.resize may work for you.
In [43]: seq = np.array([1,2,3,4])
In [44]: np.resize(seq, 6)
Out[44]: array([1, 2, 3, 4, 1, 2])
We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
Nice :) Didn't know it repeated it if the new size is greater
– yatu
Jan 1 at 21:28
The.resizemethod behaves differently.
– hpaulj
Jan 1 at 21:30
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
np.resize may work for you.
In [43]: seq = np.array([1,2,3,4])
In [44]: np.resize(seq, 6)
Out[44]: array([1, 2, 3, 4, 1, 2])
We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
Nice :) Didn't know it repeated it if the new size is greater
– yatu
Jan 1 at 21:28
The.resizemethod behaves differently.
– hpaulj
Jan 1 at 21:30
add a comment |
np.resize may work for you.
In [43]: seq = np.array([1,2,3,4])
In [44]: np.resize(seq, 6)
Out[44]: array([1, 2, 3, 4, 1, 2])
We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
Nice :) Didn't know it repeated it if the new size is greater
– yatu
Jan 1 at 21:28
The.resizemethod behaves differently.
– hpaulj
Jan 1 at 21:30
add a comment |
np.resize may work for you.
In [43]: seq = np.array([1,2,3,4])
In [44]: np.resize(seq, 6)
Out[44]: array([1, 2, 3, 4, 1, 2])
We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
np.resize may work for you.
In [43]: seq = np.array([1,2,3,4])
In [44]: np.resize(seq, 6)
Out[44]: array([1, 2, 3, 4, 1, 2])
We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
answered Jan 1 at 21:25
hpauljhpaulj
116k785156
116k785156
Nice :) Didn't know it repeated it if the new size is greater
– yatu
Jan 1 at 21:28
The.resizemethod behaves differently.
– hpaulj
Jan 1 at 21:30
add a comment |
Nice :) Didn't know it repeated it if the new size is greater
– yatu
Jan 1 at 21:28
The.resizemethod behaves differently.
– hpaulj
Jan 1 at 21:30
Nice :) Didn't know it repeated it if the new size is greater
– yatu
Jan 1 at 21:28
Nice :) Didn't know it repeated it if the new size is greater
– yatu
Jan 1 at 21:28
The
.resize method behaves differently.– hpaulj
Jan 1 at 21:30
The
.resize method behaves differently.– hpaulj
Jan 1 at 21:30
add a comment |
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