Getting a price array
In the current moment, I have a tensor ytrue with ytrue.shape equal to (4650, 30, 161).
I am taking information every 5 seconds on the NYSE for a specific stock between 9:00 AM and 4:30 PM. So I have 4650 layers in the tensor which represent each 5 seconds during that period. Each layer has dimension (30, 161). 30 represents the number of steps I look in the past and 161 represents how many features (160 features + the price as the target).
I tried to reshape it the get the current price with
X = ytrue[:, :, -1]
but I got
ipdb> X
array([[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ],
[262.605, 261.99 , 261.99 , ..., 263.75 , 263.75 , 263.71 ],
[261.99 , 261.99 , 262.015, ..., 263.75 , 263.71 , 263.63 ],
...,
[253.065, 253.03 , 252.93 , ..., 253.115, 253.18 , 253.27 ],
[253.03 , 252.93 , 253.1 , ..., 253.18 , 253.27 , 253.345],
[252.93 , 253.1 , 253.145, ..., 253.27 , 253.345, 253.35 ]])
I need to take the last element of each subarray to create an array of prices, e.g.
[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ] --> 263.75
How can I get an array of each of those elements from ytrue?
python numpy indexing
asked Nov 20 '18 at 17:02
user1050421user1050421
177
add a comment |
In the current moment, I have a tensor ytrue with ytrue.shape equal to (4650, 30, 161).
I am taking information every 5 seconds on the NYSE for a specific stock between 9:00 AM and 4:30 PM. So I have 4650 layers in the tensor which represent each 5 seconds during that period. Each layer has dimension (30, 161). 30 represents the number of steps I look in the past and 161 represents how many features (160 features + the price as the target).
I tried to reshape it the get the current price with
X = ytrue[:, :, -1]
but I got
ipdb> X
array([[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ],
[262.605, 261.99 , 261.99 , ..., 263.75 , 263.75 , 263.71 ],
[261.99 , 261.99 , 262.015, ..., 263.75 , 263.71 , 263.63 ],
...,
[253.065, 253.03 , 252.93 , ..., 253.115, 253.18 , 253.27 ],
[253.03 , 252.93 , 253.1 , ..., 253.18 , 253.27 , 253.345],
[252.93 , 253.1 , 253.145, ..., 253.27 , 253.345, 253.35 ]])
I need to take the last element of each subarray to create an array of prices, e.g.
[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ] --> 263.75
How can I get an array of each of those elements from ytrue?
python numpy indexing
asked Nov 20 '18 at 17:02
user1050421user1050421
177
2
ytrue[:, -1, -1]?
– Divakar
Nov 20 '18 at 17:16
add a comment |
In the current moment, I have a tensor ytrue with ytrue.shape equal to (4650, 30, 161).
I am taking information every 5 seconds on the NYSE for a specific stock between 9:00 AM and 4:30 PM. So I have 4650 layers in the tensor which represent each 5 seconds during that period. Each layer has dimension (30, 161). 30 represents the number of steps I look in the past and 161 represents how many features (160 features + the price as the target).
I tried to reshape it the get the current price with
X = ytrue[:, :, -1]
but I got
ipdb> X
array([[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ],
[262.605, 261.99 , 261.99 , ..., 263.75 , 263.75 , 263.71 ],
[261.99 , 261.99 , 262.015, ..., 263.75 , 263.71 , 263.63 ],
...,
[253.065, 253.03 , 252.93 , ..., 253.115, 253.18 , 253.27 ],
[253.03 , 252.93 , 253.1 , ..., 253.18 , 253.27 , 253.345],
[252.93 , 253.1 , 253.145, ..., 253.27 , 253.345, 253.35 ]])
I need to take the last element of each subarray to create an array of prices, e.g.
[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ] --> 263.75
How can I get an array of each of those elements from ytrue?
python numpy indexing
asked Nov 20 '18 at 17:02
user1050421user1050421
177
In the current moment, I have a tensor ytrue with ytrue.shape equal to (4650, 30, 161).
I am taking information every 5 seconds on the NYSE for a specific stock between 9:00 AM and 4:30 PM. So I have 4650 layers in the tensor which represent each 5 seconds during that period. Each layer has dimension (30, 161). 30 represents the number of steps I look in the past and 161 represents how many features (160 features + the price as the target).
I tried to reshape it the get the current price with
X = ytrue[:, :, -1]
but I got
ipdb> X
array([[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ],
[262.605, 261.99 , 261.99 , ..., 263.75 , 263.75 , 263.71 ],
[261.99 , 261.99 , 262.015, ..., 263.75 , 263.71 , 263.63 ],
...,
[253.065, 253.03 , 252.93 , ..., 253.115, 253.18 , 253.27 ],
[253.03 , 252.93 , 253.1 , ..., 253.18 , 253.27 , 253.345],
[252.93 , 253.1 , 253.145, ..., 253.27 , 253.345, 253.35 ]])
I need to take the last element of each subarray to create an array of prices, e.g.
[262.655, 262.605, 261.99 , ..., 263.615, 263.75 , 263.75 ] --> 263.75
How can I get an array of each of those elements from ytrue?
python numpy indexing
python numpy indexing
asked Nov 20 '18 at 17:02
user1050421user1050421
177
asked Nov 20 '18 at 17:02
user1050421user1050421
177
edited Nov 20 '18 at 17:12
user1050421
asked Nov 20 '18 at 17:02
user1050421user1050421
177
asked Nov 20 '18 at 17:02
user1050421user1050421
177
asked Nov 20 '18 at 17:02
user1050421user1050421
177
177
2
ytrue[:, -1, -1]?
– Divakar
Nov 20 '18 at 17:16
add a comment |
2
ytrue[:, -1, -1]?
– Divakar
Nov 20 '18 at 17:16
2
2
ytrue[:, -1, -1]?– Divakar
Nov 20 '18 at 17:16
ytrue[:, -1, -1]?– Divakar
Nov 20 '18 at 17:16
add a comment |
1 Answer
1
active
oldest
votes
Use fancy indexing.
X[:, -1] should do the trick.
A smaller example you can easily eyeball:
In [4]: x = np.arange(25).reshape(5, 5)
In [5]: x
Out[5]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
In [6]: x[:, -1]
Out[6]: array([ 4, 9, 14, 19, 24])
HTH.
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
I thinkX[:, -1]should still work. I've got to run, but I'll check back later...
– Matt Messersmith
Nov 20 '18 at 17:14
Sorry, I made a mistake in my question.Xis already equal toytrue[:, :, -1]. So I guess X has to be equal toytrue[:, :, -1] [:, -1]?
– user1050421
Nov 20 '18 at 17:16
Yes, that works. You could also just do it in two steps withXas a temp. Alternatively I believeytrue[:, -1,-1]will work if you want it all on one slice.
– Matt Messersmith
Nov 20 '18 at 17:59
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Use fancy indexing.
X[:, -1] should do the trick.
A smaller example you can easily eyeball:
In [4]: x = np.arange(25).reshape(5, 5)
In [5]: x
Out[5]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
In [6]: x[:, -1]
Out[6]: array([ 4, 9, 14, 19, 24])
HTH.
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
I thinkX[:, -1]should still work. I've got to run, but I'll check back later...
– Matt Messersmith
Nov 20 '18 at 17:14
Sorry, I made a mistake in my question.Xis already equal toytrue[:, :, -1]. So I guess X has to be equal toytrue[:, :, -1] [:, -1]?
– user1050421
Nov 20 '18 at 17:16
Yes, that works. You could also just do it in two steps withXas a temp. Alternatively I believeytrue[:, -1,-1]will work if you want it all on one slice.
– Matt Messersmith
Nov 20 '18 at 17:59
add a comment |
Use fancy indexing.
X[:, -1] should do the trick.
A smaller example you can easily eyeball:
In [4]: x = np.arange(25).reshape(5, 5)
In [5]: x
Out[5]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
In [6]: x[:, -1]
Out[6]: array([ 4, 9, 14, 19, 24])
HTH.
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
I thinkX[:, -1]should still work. I've got to run, but I'll check back later...
– Matt Messersmith
Nov 20 '18 at 17:14
Sorry, I made a mistake in my question.Xis already equal toytrue[:, :, -1]. So I guess X has to be equal toytrue[:, :, -1] [:, -1]?
– user1050421
Nov 20 '18 at 17:16
Yes, that works. You could also just do it in two steps withXas a temp. Alternatively I believeytrue[:, -1,-1]will work if you want it all on one slice.
– Matt Messersmith
Nov 20 '18 at 17:59
add a comment |
Use fancy indexing.
X[:, -1] should do the trick.
A smaller example you can easily eyeball:
In [4]: x = np.arange(25).reshape(5, 5)
In [5]: x
Out[5]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
In [6]: x[:, -1]
Out[6]: array([ 4, 9, 14, 19, 24])
HTH.
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
Use fancy indexing.
X[:, -1] should do the trick.
A smaller example you can easily eyeball:
In [4]: x = np.arange(25).reshape(5, 5)
In [5]: x
Out[5]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
In [6]: x[:, -1]
Out[6]: array([ 4, 9, 14, 19, 24])
HTH.
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
edited Nov 20 '18 at 17:12
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
answered Nov 20 '18 at 17:08
Matt MessersmithMatt Messersmith
5,98921730
5,98921730
I thinkX[:, -1]should still work. I've got to run, but I'll check back later...
– Matt Messersmith
Nov 20 '18 at 17:14
Sorry, I made a mistake in my question.Xis already equal toytrue[:, :, -1]. So I guess X has to be equal toytrue[:, :, -1] [:, -1]?
– user1050421
Nov 20 '18 at 17:16
Yes, that works. You could also just do it in two steps withXas a temp. Alternatively I believeytrue[:, -1,-1]will work if you want it all on one slice.
– Matt Messersmith
Nov 20 '18 at 17:59
add a comment |
I thinkX[:, -1]should still work. I've got to run, but I'll check back later...
– Matt Messersmith
Nov 20 '18 at 17:14
Sorry, I made a mistake in my question.Xis already equal toytrue[:, :, -1]. So I guess X has to be equal toytrue[:, :, -1] [:, -1]?
– user1050421
Nov 20 '18 at 17:16
Yes, that works. You could also just do it in two steps withXas a temp. Alternatively I believeytrue[:, -1,-1]will work if you want it all on one slice.
– Matt Messersmith
Nov 20 '18 at 17:59
I think
X[:, -1] should still work. I've got to run, but I'll check back later...– Matt Messersmith
Nov 20 '18 at 17:14
I think
X[:, -1] should still work. I've got to run, but I'll check back later...– Matt Messersmith
Nov 20 '18 at 17:14
Sorry, I made a mistake in my question.
X is already equal to ytrue[:, :, -1]. So I guess X has to be equal to ytrue[:, :, -1] [:, -1]?– user1050421
Nov 20 '18 at 17:16
Sorry, I made a mistake in my question.
X is already equal to ytrue[:, :, -1]. So I guess X has to be equal to ytrue[:, :, -1] [:, -1]?– user1050421
Nov 20 '18 at 17:16
Yes, that works. You could also just do it in two steps with
X as a temp. Alternatively I believe ytrue[:, -1,-1] will work if you want it all on one slice.– Matt Messersmith
Nov 20 '18 at 17:59
Yes, that works. You could also just do it in two steps with
X as a temp. Alternatively I believe ytrue[:, -1,-1] will work if you want it all on one slice.– Matt Messersmith
Nov 20 '18 at 17:59
add a comment |
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2
ytrue[:, -1, -1]?– Divakar
Nov 20 '18 at 17:16