how can I reshape a numpy array of (100,) to (250,100)
Imagine that you have created an array with 100 dimensions and then you calculate something and fill this array. for whatever reason, you have not created 2d array, what is wrong with this question that you want to assign another dimension to this data, with this justification that for example 250 samples should have this calculated data?!!
I have searched this but I could not find any solution. Maybe I am not searching with correct keyword!
Actually I want to reshape a numpy array of (100,) to (250,100).
I have read this link and a couple of other links but did not help me.
I have also tried this way:
numpyarray = (100,)
transformed_numpyarray = np.reshape(numpyarray,(100,-1)).T
which gives me this output:
(1, 100)
but I really do not want 1 as the first item of 2d array.
what Im trying to do is to either convert to (,100) or at least something like this (250,100). "250" is a constant number I know already so I want to say for example for 250 samples with 100 dimension.
Thanks.
python numpy
asked Nov 21 at 1:27
sariii
4161316
|
show 6 more comments
Imagine that you have created an array with 100 dimensions and then you calculate something and fill this array. for whatever reason, you have not created 2d array, what is wrong with this question that you want to assign another dimension to this data, with this justification that for example 250 samples should have this calculated data?!!
I have searched this but I could not find any solution. Maybe I am not searching with correct keyword!
Actually I want to reshape a numpy array of (100,) to (250,100).
I have read this link and a couple of other links but did not help me.
I have also tried this way:
numpyarray = (100,)
transformed_numpyarray = np.reshape(numpyarray,(100,-1)).T
which gives me this output:
(1, 100)
but I really do not want 1 as the first item of 2d array.
what Im trying to do is to either convert to (,100) or at least something like this (250,100). "250" is a constant number I know already so I want to say for example for 250 samples with 100 dimension.
Thanks.
python numpy
asked Nov 21 at 1:27
sariii
4161316
3
(100,)is standard python notation for a 1 element tuple.(,100)is not valid python syntax.
– hpaulj
Nov 21 at 1:54
np.reshape(numpyarray,(100,-1))gives an error,cannot reshape array of size 1 into shape (100,newaxis). Your description confuses arrays and their shapes.
– hpaulj
Nov 21 at 2:42
Your link is to akerasquestion. there(None, 100)is a valid shape. There isn't anumpyequivalent.
– hpaulj
Nov 21 at 2:43
@hpaulj thank you so much for your explanations. though (,100) may not be a valid but I need it to be in this way. Or Im gonna to update my question to another way. I will be ok to update (100,) to (250,100). 250 is something constant. is in this way is doable?
– sariii
Nov 21 at 3:30
1
Question has nothing to do withmachine-learningorscikit-learn- kindly do not spam the tags (removed).
– desertnaut
Nov 21 at 12:14
|
show 6 more comments
Imagine that you have created an array with 100 dimensions and then you calculate something and fill this array. for whatever reason, you have not created 2d array, what is wrong with this question that you want to assign another dimension to this data, with this justification that for example 250 samples should have this calculated data?!!
I have searched this but I could not find any solution. Maybe I am not searching with correct keyword!
Actually I want to reshape a numpy array of (100,) to (250,100).
I have read this link and a couple of other links but did not help me.
I have also tried this way:
numpyarray = (100,)
transformed_numpyarray = np.reshape(numpyarray,(100,-1)).T
which gives me this output:
(1, 100)
but I really do not want 1 as the first item of 2d array.
what Im trying to do is to either convert to (,100) or at least something like this (250,100). "250" is a constant number I know already so I want to say for example for 250 samples with 100 dimension.
Thanks.
python numpy
asked Nov 21 at 1:27
sariii
4161316
Imagine that you have created an array with 100 dimensions and then you calculate something and fill this array. for whatever reason, you have not created 2d array, what is wrong with this question that you want to assign another dimension to this data, with this justification that for example 250 samples should have this calculated data?!!
I have searched this but I could not find any solution. Maybe I am not searching with correct keyword!
Actually I want to reshape a numpy array of (100,) to (250,100).
I have read this link and a couple of other links but did not help me.
I have also tried this way:
numpyarray = (100,)
transformed_numpyarray = np.reshape(numpyarray,(100,-1)).T
which gives me this output:
(1, 100)
but I really do not want 1 as the first item of 2d array.
what Im trying to do is to either convert to (,100) or at least something like this (250,100). "250" is a constant number I know already so I want to say for example for 250 samples with 100 dimension.
Thanks.
python numpy
python numpy
asked Nov 21 at 1:27
sariii
4161316
asked Nov 21 at 1:27
sariii
4161316
edited Nov 22 at 17:11
asked Nov 21 at 1:27
sariii
4161316
asked Nov 21 at 1:27
sariii
4161316
asked Nov 21 at 1:27
sariii
4161316
4161316
3
(100,)is standard python notation for a 1 element tuple.(,100)is not valid python syntax.
– hpaulj
Nov 21 at 1:54
np.reshape(numpyarray,(100,-1))gives an error,cannot reshape array of size 1 into shape (100,newaxis). Your description confuses arrays and their shapes.
– hpaulj
Nov 21 at 2:42
Your link is to akerasquestion. there(None, 100)is a valid shape. There isn't anumpyequivalent.
– hpaulj
Nov 21 at 2:43
@hpaulj thank you so much for your explanations. though (,100) may not be a valid but I need it to be in this way. Or Im gonna to update my question to another way. I will be ok to update (100,) to (250,100). 250 is something constant. is in this way is doable?
– sariii
Nov 21 at 3:30
1
Question has nothing to do withmachine-learningorscikit-learn- kindly do not spam the tags (removed).
– desertnaut
Nov 21 at 12:14
|
show 6 more comments
3
(100,)is standard python notation for a 1 element tuple.(,100)is not valid python syntax.
– hpaulj
Nov 21 at 1:54
np.reshape(numpyarray,(100,-1))gives an error,cannot reshape array of size 1 into shape (100,newaxis). Your description confuses arrays and their shapes.
– hpaulj
Nov 21 at 2:42
Your link is to akerasquestion. there(None, 100)is a valid shape. There isn't anumpyequivalent.
– hpaulj
Nov 21 at 2:43
@hpaulj thank you so much for your explanations. though (,100) may not be a valid but I need it to be in this way. Or Im gonna to update my question to another way. I will be ok to update (100,) to (250,100). 250 is something constant. is in this way is doable?
– sariii
Nov 21 at 3:30
1
Question has nothing to do withmachine-learningorscikit-learn- kindly do not spam the tags (removed).
– desertnaut
Nov 21 at 12:14
3
3
(100,) is standard python notation for a 1 element tuple. (,100) is not valid python syntax.– hpaulj
Nov 21 at 1:54
(100,) is standard python notation for a 1 element tuple. (,100) is not valid python syntax.– hpaulj
Nov 21 at 1:54
np.reshape(numpyarray,(100,-1)) gives an error, cannot reshape array of size 1 into shape (100,newaxis). Your description confuses arrays and their shapes.– hpaulj
Nov 21 at 2:42
np.reshape(numpyarray,(100,-1)) gives an error, cannot reshape array of size 1 into shape (100,newaxis). Your description confuses arrays and their shapes.– hpaulj
Nov 21 at 2:42
Your link is to a
keras question. there (None, 100) is a valid shape. There isn't a numpy equivalent.– hpaulj
Nov 21 at 2:43
Your link is to a
keras question. there (None, 100) is a valid shape. There isn't a numpy equivalent.– hpaulj
Nov 21 at 2:43
@hpaulj thank you so much for your explanations. though (,100) may not be a valid but I need it to be in this way. Or Im gonna to update my question to another way. I will be ok to update (100,) to (250,100). 250 is something constant. is in this way is doable?
– sariii
Nov 21 at 3:30
@hpaulj thank you so much for your explanations. though (,100) may not be a valid but I need it to be in this way. Or Im gonna to update my question to another way. I will be ok to update (100,) to (250,100). 250 is something constant. is in this way is doable?
– sariii
Nov 21 at 3:30
1
1
Question has nothing to do with
machine-learning or scikit-learn - kindly do not spam the tags (removed).– desertnaut
Nov 21 at 12:14
Question has nothing to do with
machine-learning or scikit-learn - kindly do not spam the tags (removed).– desertnaut
Nov 21 at 12:14
|
show 6 more comments
2 Answers
2
active
oldest
votes
I'm still confused about what you are trying to do. So far I can picture two alternatives - reshape and repeat. To illustrate:
In [148]: x = np.arange(16)
In [149]: x
Out[149]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
In [150]: x.reshape(4,4)
Out[150]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
In [151]: np.repeat(x[None,:], 4, axis=0)
Out[151]:
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]])
answered Nov 22 at 16:59
hpaulj
110k774140
exactly this answer meets my requirement. @vlizana also answered correcty but I did not want to apply for loop and make it expensive. Thank you :)
– sariii
Nov 22 at 17:10
Do you mind explaining why do we need this part x.reshape(4,4)?
– sariii
Nov 22 at 17:13
No sorry, it seems you have not used it ;)
– sariii
Nov 22 at 17:14
add a comment |
numpy's arrays are static sized, you can't have an array with a variable shape. If you don't know beforehand how many samples you will have you can gradually add them with vstack:
In [4]: numpyarray.shape
Out[4]: (3, 4)
In [5]: new_sample.shape
Out[5]: (4,)
In [6]: numpyarray = np.vstack([numpyarray, new_sample])
In [7]: numpyarray.shape
Out[7]: (4, 4)
you can also first define the size by creating an array full of zeros and then progressively fill it with samples.
numpyarray = np.zeros((250,100))
...
numpyarray[i] = new_sample
answered Nov 21 at 2:22
vlizana
638215
thanks for the answer, but I got this error after running your example ValueError: all the input array dimensions except for the concatenation axis must match exactly. do you have any idea of this?
– sariii
Nov 21 at 3:25
I also updated my question. in case it will be easier to convert to a constant number like 250
– sariii
Nov 21 at 3:34
1
The first error means that the length of the samples must be the same as thenumpyarrayconcatenation axis, in this case, all of your samples must have shape(100,)or(1,100)andnumpyarraymust be(n,100). About the update on the question, when you reshape an array the resulting array must have the same number of cells, in this case the resulting array would have a lot more. I edited my answer about how this can be done.
– vlizana
Nov 21 at 3:45
1
thank you so much. for the explanations of the error, I followed exactly like your example but got that error. your second answer seems doable but I think I need to for loop through the data?. I'm not at work right now. I will try on that and then get back to mark as answer. again thanks for taking the time :)
– sariii
Nov 21 at 4:08
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
I'm still confused about what you are trying to do. So far I can picture two alternatives - reshape and repeat. To illustrate:
In [148]: x = np.arange(16)
In [149]: x
Out[149]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
In [150]: x.reshape(4,4)
Out[150]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
In [151]: np.repeat(x[None,:], 4, axis=0)
Out[151]:
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]])
answered Nov 22 at 16:59
hpaulj
110k774140
exactly this answer meets my requirement. @vlizana also answered correcty but I did not want to apply for loop and make it expensive. Thank you :)
– sariii
Nov 22 at 17:10
Do you mind explaining why do we need this part x.reshape(4,4)?
– sariii
Nov 22 at 17:13
No sorry, it seems you have not used it ;)
– sariii
Nov 22 at 17:14
add a comment |
I'm still confused about what you are trying to do. So far I can picture two alternatives - reshape and repeat. To illustrate:
In [148]: x = np.arange(16)
In [149]: x
Out[149]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
In [150]: x.reshape(4,4)
Out[150]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
In [151]: np.repeat(x[None,:], 4, axis=0)
Out[151]:
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]])
answered Nov 22 at 16:59
hpaulj
110k774140
exactly this answer meets my requirement. @vlizana also answered correcty but I did not want to apply for loop and make it expensive. Thank you :)
– sariii
Nov 22 at 17:10
Do you mind explaining why do we need this part x.reshape(4,4)?
– sariii
Nov 22 at 17:13
No sorry, it seems you have not used it ;)
– sariii
Nov 22 at 17:14
add a comment |
I'm still confused about what you are trying to do. So far I can picture two alternatives - reshape and repeat. To illustrate:
In [148]: x = np.arange(16)
In [149]: x
Out[149]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
In [150]: x.reshape(4,4)
Out[150]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
In [151]: np.repeat(x[None,:], 4, axis=0)
Out[151]:
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]])
answered Nov 22 at 16:59
hpaulj
110k774140
I'm still confused about what you are trying to do. So far I can picture two alternatives - reshape and repeat. To illustrate:
In [148]: x = np.arange(16)
In [149]: x
Out[149]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
In [150]: x.reshape(4,4)
Out[150]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
In [151]: np.repeat(x[None,:], 4, axis=0)
Out[151]:
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]])
answered Nov 22 at 16:59
hpaulj
110k774140
answered Nov 22 at 16:59
hpaulj
110k774140
answered Nov 22 at 16:59
hpaulj
110k774140
answered Nov 22 at 16:59
hpaulj
110k774140
110k774140
exactly this answer meets my requirement. @vlizana also answered correcty but I did not want to apply for loop and make it expensive. Thank you :)
– sariii
Nov 22 at 17:10
Do you mind explaining why do we need this part x.reshape(4,4)?
– sariii
Nov 22 at 17:13
No sorry, it seems you have not used it ;)
– sariii
Nov 22 at 17:14
add a comment |
exactly this answer meets my requirement. @vlizana also answered correcty but I did not want to apply for loop and make it expensive. Thank you :)
– sariii
Nov 22 at 17:10
Do you mind explaining why do we need this part x.reshape(4,4)?
– sariii
Nov 22 at 17:13
No sorry, it seems you have not used it ;)
– sariii
Nov 22 at 17:14
exactly this answer meets my requirement. @vlizana also answered correcty but I did not want to apply for loop and make it expensive. Thank you :)
– sariii
Nov 22 at 17:10
exactly this answer meets my requirement. @vlizana also answered correcty but I did not want to apply for loop and make it expensive. Thank you :)
– sariii
Nov 22 at 17:10
Do you mind explaining why do we need this part x.reshape(4,4)?
– sariii
Nov 22 at 17:13
Do you mind explaining why do we need this part x.reshape(4,4)?
– sariii
Nov 22 at 17:13
No sorry, it seems you have not used it ;)
– sariii
Nov 22 at 17:14
No sorry, it seems you have not used it ;)
– sariii
Nov 22 at 17:14
add a comment |
numpy's arrays are static sized, you can't have an array with a variable shape. If you don't know beforehand how many samples you will have you can gradually add them with vstack:
In [4]: numpyarray.shape
Out[4]: (3, 4)
In [5]: new_sample.shape
Out[5]: (4,)
In [6]: numpyarray = np.vstack([numpyarray, new_sample])
In [7]: numpyarray.shape
Out[7]: (4, 4)
you can also first define the size by creating an array full of zeros and then progressively fill it with samples.
numpyarray = np.zeros((250,100))
...
numpyarray[i] = new_sample
answered Nov 21 at 2:22
vlizana
638215
thanks for the answer, but I got this error after running your example ValueError: all the input array dimensions except for the concatenation axis must match exactly. do you have any idea of this?
– sariii
Nov 21 at 3:25
I also updated my question. in case it will be easier to convert to a constant number like 250
– sariii
Nov 21 at 3:34
1
The first error means that the length of the samples must be the same as thenumpyarrayconcatenation axis, in this case, all of your samples must have shape(100,)or(1,100)andnumpyarraymust be(n,100). About the update on the question, when you reshape an array the resulting array must have the same number of cells, in this case the resulting array would have a lot more. I edited my answer about how this can be done.
– vlizana
Nov 21 at 3:45
1
thank you so much. for the explanations of the error, I followed exactly like your example but got that error. your second answer seems doable but I think I need to for loop through the data?. I'm not at work right now. I will try on that and then get back to mark as answer. again thanks for taking the time :)
– sariii
Nov 21 at 4:08
add a comment |
numpy's arrays are static sized, you can't have an array with a variable shape. If you don't know beforehand how many samples you will have you can gradually add them with vstack:
In [4]: numpyarray.shape
Out[4]: (3, 4)
In [5]: new_sample.shape
Out[5]: (4,)
In [6]: numpyarray = np.vstack([numpyarray, new_sample])
In [7]: numpyarray.shape
Out[7]: (4, 4)
you can also first define the size by creating an array full of zeros and then progressively fill it with samples.
numpyarray = np.zeros((250,100))
...
numpyarray[i] = new_sample
answered Nov 21 at 2:22
vlizana
638215
thanks for the answer, but I got this error after running your example ValueError: all the input array dimensions except for the concatenation axis must match exactly. do you have any idea of this?
– sariii
Nov 21 at 3:25
I also updated my question. in case it will be easier to convert to a constant number like 250
– sariii
Nov 21 at 3:34
1
The first error means that the length of the samples must be the same as thenumpyarrayconcatenation axis, in this case, all of your samples must have shape(100,)or(1,100)andnumpyarraymust be(n,100). About the update on the question, when you reshape an array the resulting array must have the same number of cells, in this case the resulting array would have a lot more. I edited my answer about how this can be done.
– vlizana
Nov 21 at 3:45
1
thank you so much. for the explanations of the error, I followed exactly like your example but got that error. your second answer seems doable but I think I need to for loop through the data?. I'm not at work right now. I will try on that and then get back to mark as answer. again thanks for taking the time :)
– sariii
Nov 21 at 4:08
add a comment |
numpy's arrays are static sized, you can't have an array with a variable shape. If you don't know beforehand how many samples you will have you can gradually add them with vstack:
In [4]: numpyarray.shape
Out[4]: (3, 4)
In [5]: new_sample.shape
Out[5]: (4,)
In [6]: numpyarray = np.vstack([numpyarray, new_sample])
In [7]: numpyarray.shape
Out[7]: (4, 4)
you can also first define the size by creating an array full of zeros and then progressively fill it with samples.
numpyarray = np.zeros((250,100))
...
numpyarray[i] = new_sample
answered Nov 21 at 2:22
vlizana
638215
numpy's arrays are static sized, you can't have an array with a variable shape. If you don't know beforehand how many samples you will have you can gradually add them with vstack:
In [4]: numpyarray.shape
Out[4]: (3, 4)
In [5]: new_sample.shape
Out[5]: (4,)
In [6]: numpyarray = np.vstack([numpyarray, new_sample])
In [7]: numpyarray.shape
Out[7]: (4, 4)
you can also first define the size by creating an array full of zeros and then progressively fill it with samples.
numpyarray = np.zeros((250,100))
...
numpyarray[i] = new_sample
answered Nov 21 at 2:22
vlizana
638215
edited Nov 21 at 3:38
answered Nov 21 at 2:22
vlizana
638215
answered Nov 21 at 2:22
vlizana
638215
answered Nov 21 at 2:22
vlizana
638215
638215
thanks for the answer, but I got this error after running your example ValueError: all the input array dimensions except for the concatenation axis must match exactly. do you have any idea of this?
– sariii
Nov 21 at 3:25
I also updated my question. in case it will be easier to convert to a constant number like 250
– sariii
Nov 21 at 3:34
1
The first error means that the length of the samples must be the same as thenumpyarrayconcatenation axis, in this case, all of your samples must have shape(100,)or(1,100)andnumpyarraymust be(n,100). About the update on the question, when you reshape an array the resulting array must have the same number of cells, in this case the resulting array would have a lot more. I edited my answer about how this can be done.
– vlizana
Nov 21 at 3:45
1
thank you so much. for the explanations of the error, I followed exactly like your example but got that error. your second answer seems doable but I think I need to for loop through the data?. I'm not at work right now. I will try on that and then get back to mark as answer. again thanks for taking the time :)
– sariii
Nov 21 at 4:08
add a comment |
thanks for the answer, but I got this error after running your example ValueError: all the input array dimensions except for the concatenation axis must match exactly. do you have any idea of this?
– sariii
Nov 21 at 3:25
I also updated my question. in case it will be easier to convert to a constant number like 250
– sariii
Nov 21 at 3:34
1
The first error means that the length of the samples must be the same as thenumpyarrayconcatenation axis, in this case, all of your samples must have shape(100,)or(1,100)andnumpyarraymust be(n,100). About the update on the question, when you reshape an array the resulting array must have the same number of cells, in this case the resulting array would have a lot more. I edited my answer about how this can be done.
– vlizana
Nov 21 at 3:45
1
thank you so much. for the explanations of the error, I followed exactly like your example but got that error. your second answer seems doable but I think I need to for loop through the data?. I'm not at work right now. I will try on that and then get back to mark as answer. again thanks for taking the time :)
– sariii
Nov 21 at 4:08
thanks for the answer, but I got this error after running your example ValueError: all the input array dimensions except for the concatenation axis must match exactly. do you have any idea of this?
– sariii
Nov 21 at 3:25
thanks for the answer, but I got this error after running your example ValueError: all the input array dimensions except for the concatenation axis must match exactly. do you have any idea of this?
– sariii
Nov 21 at 3:25
I also updated my question. in case it will be easier to convert to a constant number like 250
– sariii
Nov 21 at 3:34
I also updated my question. in case it will be easier to convert to a constant number like 250
– sariii
Nov 21 at 3:34
1
1
The first error means that the length of the samples must be the same as the
numpyarray concatenation axis, in this case, all of your samples must have shape (100,) or (1,100) and numpyarray must be (n,100). About the update on the question, when you reshape an array the resulting array must have the same number of cells, in this case the resulting array would have a lot more. I edited my answer about how this can be done.– vlizana
Nov 21 at 3:45
The first error means that the length of the samples must be the same as the
numpyarray concatenation axis, in this case, all of your samples must have shape (100,) or (1,100) and numpyarray must be (n,100). About the update on the question, when you reshape an array the resulting array must have the same number of cells, in this case the resulting array would have a lot more. I edited my answer about how this can be done.– vlizana
Nov 21 at 3:45
1
1
thank you so much. for the explanations of the error, I followed exactly like your example but got that error. your second answer seems doable but I think I need to for loop through the data?. I'm not at work right now. I will try on that and then get back to mark as answer. again thanks for taking the time :)
– sariii
Nov 21 at 4:08
thank you so much. for the explanations of the error, I followed exactly like your example but got that error. your second answer seems doable but I think I need to for loop through the data?. I'm not at work right now. I will try on that and then get back to mark as answer. again thanks for taking the time :)
– sariii
Nov 21 at 4:08
add a comment |
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3
(100,)is standard python notation for a 1 element tuple.(,100)is not valid python syntax.– hpaulj
Nov 21 at 1:54
np.reshape(numpyarray,(100,-1))gives an error,cannot reshape array of size 1 into shape (100,newaxis). Your description confuses arrays and their shapes.– hpaulj
Nov 21 at 2:42
Your link is to a
kerasquestion. there(None, 100)is a valid shape. There isn't anumpyequivalent.– hpaulj
Nov 21 at 2:43
@hpaulj thank you so much for your explanations. though (,100) may not be a valid but I need it to be in this way. Or Im gonna to update my question to another way. I will be ok to update (100,) to (250,100). 250 is something constant. is in this way is doable?
– sariii
Nov 21 at 3:30
1
Question has nothing to do with
machine-learningorscikit-learn- kindly do not spam the tags (removed).– desertnaut
Nov 21 at 12:14