How does concat works using ipython?
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Can somebody explain me the code below?
Asume n_a and n_x are known.
Questions I have for the below code are:
for the second concat, why is there no , between : and n_a
and idem for the third contact (no , between n_a and :)?
What is the difference between : n_a and n_a : ?
What is the difference in : before and after the , (in both concats ?)
concat = np.zeros((n_a + n_x, m))
concat[: n_a, :] = a_prev
concat[n_a :, :] = xt
ipython
asked Nov 22 '18 at 10:56
IlseIlse
218
add a comment |
0
Can somebody explain me the code below?
Asume n_a and n_x are known.
Questions I have for the below code are:
for the second concat, why is there no , between : and n_a
and idem for the third contact (no , between n_a and :)?
What is the difference between : n_a and n_a : ?
What is the difference in : before and after the , (in both concats ?)
concat = np.zeros((n_a + n_x, m))
concat[: n_a, :] = a_prev
concat[n_a :, :] = xt
ipython
asked Nov 22 '18 at 10:56
IlseIlse
218
:nandn:are both Python slices. The general slice pattern isstart:stop:step. You should be familiar with this from working with Python lists.numpygeneralizes this by allowing slices for several dimensions, separated by the comma. Usually we writex[n:, :]but the extra space in 'n :' doesn't matter.
– hpaulj
Nov 22 '18 at 16:50
@hpaulj can you clarify with an example. Eg: with the code I write in my question?
– Ilse
Nov 23 '18 at 7:46
@hpaulj can you give fictive value to concat and then tell me what the value will be for a_prev and xt ? It should really help me further.
– Ilse
Nov 26 '18 at 11:21
Have you experimented with lists and arrays? What do those different slicing expressions produce? You've tagged this withipython, which means you can and should be doing these calculations interactively. You can see for yourself whatconcat[:3,:]produces. An for that testing, start with something likenp.arange(12).reshape(3,4)so you can see right off which rows you select.
– hpaulj
Nov 26 '18 at 17:14
add a comment |
0
0
0
Can somebody explain me the code below?
Asume n_a and n_x are known.
Questions I have for the below code are:
for the second concat, why is there no , between : and n_a
and idem for the third contact (no , between n_a and :)?
What is the difference between : n_a and n_a : ?
What is the difference in : before and after the , (in both concats ?)
concat = np.zeros((n_a + n_x, m))
concat[: n_a, :] = a_prev
concat[n_a :, :] = xt
ipython
asked Nov 22 '18 at 10:56
IlseIlse
218
Can somebody explain me the code below?
Asume n_a and n_x are known.
Questions I have for the below code are:
for the second concat, why is there no , between : and n_a
and idem for the third contact (no , between n_a and :)?
What is the difference between : n_a and n_a : ?
What is the difference in : before and after the , (in both concats ?)
concat = np.zeros((n_a + n_x, m))
concat[: n_a, :] = a_prev
concat[n_a :, :] = xt
ipython
ipython
asked Nov 22 '18 at 10:56
IlseIlse
218
asked Nov 22 '18 at 10:56
IlseIlse
218
asked Nov 22 '18 at 10:56
IlseIlse
218
asked Nov 22 '18 at 10:56
IlseIlse
218
asked Nov 22 '18 at 10:56
IlseIlse
218
218
:nandn:are both Python slices. The general slice pattern isstart:stop:step. You should be familiar with this from working with Python lists.numpygeneralizes this by allowing slices for several dimensions, separated by the comma. Usually we writex[n:, :]but the extra space in 'n :' doesn't matter.
– hpaulj
Nov 22 '18 at 16:50
@hpaulj can you clarify with an example. Eg: with the code I write in my question?
– Ilse
Nov 23 '18 at 7:46
@hpaulj can you give fictive value to concat and then tell me what the value will be for a_prev and xt ? It should really help me further.
– Ilse
Nov 26 '18 at 11:21
Have you experimented with lists and arrays? What do those different slicing expressions produce? You've tagged this withipython, which means you can and should be doing these calculations interactively. You can see for yourself whatconcat[:3,:]produces. An for that testing, start with something likenp.arange(12).reshape(3,4)so you can see right off which rows you select.
– hpaulj
Nov 26 '18 at 17:14
add a comment |
:nandn:are both Python slices. The general slice pattern isstart:stop:step. You should be familiar with this from working with Python lists.numpygeneralizes this by allowing slices for several dimensions, separated by the comma. Usually we writex[n:, :]but the extra space in 'n :' doesn't matter.
– hpaulj
Nov 22 '18 at 16:50
@hpaulj can you clarify with an example. Eg: with the code I write in my question?
– Ilse
Nov 23 '18 at 7:46
@hpaulj can you give fictive value to concat and then tell me what the value will be for a_prev and xt ? It should really help me further.
– Ilse
Nov 26 '18 at 11:21
Have you experimented with lists and arrays? What do those different slicing expressions produce? You've tagged this withipython, which means you can and should be doing these calculations interactively. You can see for yourself whatconcat[:3,:]produces. An for that testing, start with something likenp.arange(12).reshape(3,4)so you can see right off which rows you select.
– hpaulj
Nov 26 '18 at 17:14
:n and n: are both Python slices. The general slice pattern is start:stop:step. You should be familiar with this from working with Python lists. numpy generalizes this by allowing slices for several dimensions, separated by the comma. Usually we write x[n:, :] but the extra space in 'n :' doesn't matter.– hpaulj
Nov 22 '18 at 16:50
:n and n: are both Python slices. The general slice pattern is start:stop:step. You should be familiar with this from working with Python lists. numpy generalizes this by allowing slices for several dimensions, separated by the comma. Usually we write x[n:, :] but the extra space in 'n :' doesn't matter.– hpaulj
Nov 22 '18 at 16:50
@hpaulj can you clarify with an example. Eg: with the code I write in my question?
– Ilse
Nov 23 '18 at 7:46
@hpaulj can you clarify with an example. Eg: with the code I write in my question?
– Ilse
Nov 23 '18 at 7:46
@hpaulj can you give fictive value to concat and then tell me what the value will be for a_prev and xt ? It should really help me further.
– Ilse
Nov 26 '18 at 11:21
@hpaulj can you give fictive value to concat and then tell me what the value will be for a_prev and xt ? It should really help me further.
– Ilse
Nov 26 '18 at 11:21
Have you experimented with lists and arrays? What do those different slicing expressions produce? You've tagged this with
ipython, which means you can and should be doing these calculations interactively. You can see for yourself what concat[:3,:] produces. An for that testing, start with something like np.arange(12).reshape(3,4) so you can see right off which rows you select.– hpaulj
Nov 26 '18 at 17:14
Have you experimented with lists and arrays? What do those different slicing expressions produce? You've tagged this with
ipython, which means you can and should be doing these calculations interactively. You can see for yourself what concat[:3,:] produces. An for that testing, start with something like np.arange(12).reshape(3,4) so you can see right off which rows you select.– hpaulj
Nov 26 '18 at 17:14
add a comment |
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:nandn:are both Python slices. The general slice pattern isstart:stop:step. You should be familiar with this from working with Python lists.numpygeneralizes this by allowing slices for several dimensions, separated by the comma. Usually we writex[n:, :]but the extra space in 'n :' doesn't matter.– hpaulj
Nov 22 '18 at 16:50
@hpaulj can you clarify with an example. Eg: with the code I write in my question?
– Ilse
Nov 23 '18 at 7:46
@hpaulj can you give fictive value to concat and then tell me what the value will be for a_prev and xt ? It should really help me further.
– Ilse
Nov 26 '18 at 11:21
Have you experimented with lists and arrays? What do those different slicing expressions produce? You've tagged this with
ipython, which means you can and should be doing these calculations interactively. You can see for yourself whatconcat[:3,:]produces. An for that testing, start with something likenp.arange(12).reshape(3,4)so you can see right off which rows you select.– hpaulj
Nov 26 '18 at 17:14