Creating an array by adding to shape of existing array
0
I have a numpy array:
X = np.array([[1,0,1],
[1,1,1],
[0,1,0],
[1,0,1]])
which has a shape of (4,3)
I would like to change this shape into (4,4) by adding 1 to the second dimension of the array, via:
X_b = np.ones((X.shape+(0,1)))
but what I get is:
ValueError: could not broadcast input array from shape (4,3) into shape (4,2,0,1)
What is the right way to do it?
Basically I want X_b to have a shape of (4,4) if X.shape = (4,3)
python arrays numpy shapes
edited Nov 21 '18 at 20:35
tel
7,47621431
asked Nov 21 '18 at 20:11
oakcaoakca
392114
add a comment |
0
I have a numpy array:
X = np.array([[1,0,1],
[1,1,1],
[0,1,0],
[1,0,1]])
which has a shape of (4,3)
I would like to change this shape into (4,4) by adding 1 to the second dimension of the array, via:
X_b = np.ones((X.shape+(0,1)))
but what I get is:
ValueError: could not broadcast input array from shape (4,3) into shape (4,2,0,1)
What is the right way to do it?
Basically I want X_b to have a shape of (4,4) if X.shape = (4,3)
python arrays numpy shapes
edited Nov 21 '18 at 20:35
tel
7,47621431
asked Nov 21 '18 at 20:11
oakcaoakca
392114
add a comment |
0
0
0
I have a numpy array:
X = np.array([[1,0,1],
[1,1,1],
[0,1,0],
[1,0,1]])
which has a shape of (4,3)
I would like to change this shape into (4,4) by adding 1 to the second dimension of the array, via:
X_b = np.ones((X.shape+(0,1)))
but what I get is:
ValueError: could not broadcast input array from shape (4,3) into shape (4,2,0,1)
What is the right way to do it?
Basically I want X_b to have a shape of (4,4) if X.shape = (4,3)
python arrays numpy shapes
edited Nov 21 '18 at 20:35
tel
7,47621431
asked Nov 21 '18 at 20:11
oakcaoakca
392114
I have a numpy array:
X = np.array([[1,0,1],
[1,1,1],
[0,1,0],
[1,0,1]])
which has a shape of (4,3)
I would like to change this shape into (4,4) by adding 1 to the second dimension of the array, via:
X_b = np.ones((X.shape+(0,1)))
but what I get is:
ValueError: could not broadcast input array from shape (4,3) into shape (4,2,0,1)
What is the right way to do it?
Basically I want X_b to have a shape of (4,4) if X.shape = (4,3)
python arrays numpy shapes
python arrays numpy shapes
edited Nov 21 '18 at 20:35
tel
7,47621431
asked Nov 21 '18 at 20:11
oakcaoakca
392114
edited Nov 21 '18 at 20:35
tel
7,47621431
asked Nov 21 '18 at 20:11
oakcaoakca
392114
edited Nov 21 '18 at 20:35
tel
7,47621431
edited Nov 21 '18 at 20:35
tel
7,47621431
edited Nov 21 '18 at 20:35
tel
7,47621431
7,47621431
asked Nov 21 '18 at 20:11
oakcaoakca
392114
asked Nov 21 '18 at 20:11
oakcaoakca
392114
asked Nov 21 '18 at 20:11
oakcaoakca
392114
392114
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
To fix your code, do this instead:
X_b = np.ones(X.shape + np.array((0,1)))
The catch here is that X.shape returns a plain Python tuple. By adding (0,1) you were actually performing tuple concatenation, instead of pairwise addition like you intended.
Of course, you could also just stick an extra column on to your existing array with append:
X_b = np.append(X, [[1]]*X.shape[0], axis=1)
answered Nov 21 '18 at 20:13
teltel
7,47621431
X_b[:,:-1] = Xand then useX_bas input... probably doing the same thing. your answer looks good. Just waiting the minutes
– oakca
Nov 21 '18 at 20:17
@oakca assumingX_bwas created withnp.ones, then yep. The one "gotcha" is thatX_bwill be of dtypefloat64, which is the default fornp.ones. Thus thedtypes ofXandX_bmay not match (eg ifX.dtype==int)
– tel
Nov 21 '18 at 20:41
Ornp.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.
– xnx
Nov 21 '18 at 20:42
@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, bothappendandhstackare thin (5-10 lines) wrappers aroundconcatenate.
– tel
Nov 21 '18 at 20:48
Here you can find more about the thema stackoverflow.com/questions/8486294/…
– oakca
Nov 21 '18 at 21:03
|
show 3 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
To fix your code, do this instead:
X_b = np.ones(X.shape + np.array((0,1)))
The catch here is that X.shape returns a plain Python tuple. By adding (0,1) you were actually performing tuple concatenation, instead of pairwise addition like you intended.
Of course, you could also just stick an extra column on to your existing array with append:
X_b = np.append(X, [[1]]*X.shape[0], axis=1)
answered Nov 21 '18 at 20:13
teltel
7,47621431
X_b[:,:-1] = Xand then useX_bas input... probably doing the same thing. your answer looks good. Just waiting the minutes
– oakca
Nov 21 '18 at 20:17
@oakca assumingX_bwas created withnp.ones, then yep. The one "gotcha" is thatX_bwill be of dtypefloat64, which is the default fornp.ones. Thus thedtypes ofXandX_bmay not match (eg ifX.dtype==int)
– tel
Nov 21 '18 at 20:41
Ornp.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.
– xnx
Nov 21 '18 at 20:42
@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, bothappendandhstackare thin (5-10 lines) wrappers aroundconcatenate.
– tel
Nov 21 '18 at 20:48
Here you can find more about the thema stackoverflow.com/questions/8486294/…
– oakca
Nov 21 '18 at 21:03
|
show 3 more comments
1
To fix your code, do this instead:
X_b = np.ones(X.shape + np.array((0,1)))
The catch here is that X.shape returns a plain Python tuple. By adding (0,1) you were actually performing tuple concatenation, instead of pairwise addition like you intended.
Of course, you could also just stick an extra column on to your existing array with append:
X_b = np.append(X, [[1]]*X.shape[0], axis=1)
answered Nov 21 '18 at 20:13
teltel
7,47621431
X_b[:,:-1] = Xand then useX_bas input... probably doing the same thing. your answer looks good. Just waiting the minutes
– oakca
Nov 21 '18 at 20:17
@oakca assumingX_bwas created withnp.ones, then yep. The one "gotcha" is thatX_bwill be of dtypefloat64, which is the default fornp.ones. Thus thedtypes ofXandX_bmay not match (eg ifX.dtype==int)
– tel
Nov 21 '18 at 20:41
Ornp.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.
– xnx
Nov 21 '18 at 20:42
@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, bothappendandhstackare thin (5-10 lines) wrappers aroundconcatenate.
– tel
Nov 21 '18 at 20:48
Here you can find more about the thema stackoverflow.com/questions/8486294/…
– oakca
Nov 21 '18 at 21:03
|
show 3 more comments
1
1
1
To fix your code, do this instead:
X_b = np.ones(X.shape + np.array((0,1)))
The catch here is that X.shape returns a plain Python tuple. By adding (0,1) you were actually performing tuple concatenation, instead of pairwise addition like you intended.
Of course, you could also just stick an extra column on to your existing array with append:
X_b = np.append(X, [[1]]*X.shape[0], axis=1)
answered Nov 21 '18 at 20:13
teltel
7,47621431
To fix your code, do this instead:
X_b = np.ones(X.shape + np.array((0,1)))
The catch here is that X.shape returns a plain Python tuple. By adding (0,1) you were actually performing tuple concatenation, instead of pairwise addition like you intended.
Of course, you could also just stick an extra column on to your existing array with append:
X_b = np.append(X, [[1]]*X.shape[0], axis=1)
answered Nov 21 '18 at 20:13
teltel
7,47621431
edited Nov 21 '18 at 21:55
answered Nov 21 '18 at 20:13
teltel
7,47621431
answered Nov 21 '18 at 20:13
teltel
7,47621431
answered Nov 21 '18 at 20:13
teltel
7,47621431
7,47621431
X_b[:,:-1] = Xand then useX_bas input... probably doing the same thing. your answer looks good. Just waiting the minutes
– oakca
Nov 21 '18 at 20:17
@oakca assumingX_bwas created withnp.ones, then yep. The one "gotcha" is thatX_bwill be of dtypefloat64, which is the default fornp.ones. Thus thedtypes ofXandX_bmay not match (eg ifX.dtype==int)
– tel
Nov 21 '18 at 20:41
Ornp.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.
– xnx
Nov 21 '18 at 20:42
@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, bothappendandhstackare thin (5-10 lines) wrappers aroundconcatenate.
– tel
Nov 21 '18 at 20:48
Here you can find more about the thema stackoverflow.com/questions/8486294/…
– oakca
Nov 21 '18 at 21:03
|
show 3 more comments
X_b[:,:-1] = Xand then useX_bas input... probably doing the same thing. your answer looks good. Just waiting the minutes
– oakca
Nov 21 '18 at 20:17
@oakca assumingX_bwas created withnp.ones, then yep. The one "gotcha" is thatX_bwill be of dtypefloat64, which is the default fornp.ones. Thus thedtypes ofXandX_bmay not match (eg ifX.dtype==int)
– tel
Nov 21 '18 at 20:41
Ornp.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.
– xnx
Nov 21 '18 at 20:42
@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, bothappendandhstackare thin (5-10 lines) wrappers aroundconcatenate.
– tel
Nov 21 '18 at 20:48
Here you can find more about the thema stackoverflow.com/questions/8486294/…
– oakca
Nov 21 '18 at 21:03
X_b[:,:-1] = X and then use X_b as input... probably doing the same thing. your answer looks good. Just waiting the minutes– oakca
Nov 21 '18 at 20:17
X_b[:,:-1] = X and then use X_b as input... probably doing the same thing. your answer looks good. Just waiting the minutes– oakca
Nov 21 '18 at 20:17
@oakca assuming
X_b was created with np.ones, then yep. The one "gotcha" is that X_b will be of dtype float64, which is the default for np.ones. Thus the dtypes of X and X_b may not match (eg if X.dtype==int)– tel
Nov 21 '18 at 20:41
@oakca assuming
X_b was created with np.ones, then yep. The one "gotcha" is that X_b will be of dtype float64, which is the default for np.ones. Thus the dtypes of X and X_b may not match (eg if X.dtype==int)– tel
Nov 21 '18 at 20:41
Or
np.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.– xnx
Nov 21 '18 at 20:42
Or
np.hstack((X, np.ones((4,1), dtype=np.int))), which I find easier to visualize.– xnx
Nov 21 '18 at 20:42
@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, both
append and hstack are thin (5-10 lines) wrappers around concatenate.– tel
Nov 21 '18 at 20:48
@xnx There's a whole bunch of these array-growing functions, and they're all roughly equivalent. Ideally you use none of them and instead just preallocate all your arrays (whose size you magically know in advance). Internally, both
append and hstack are thin (5-10 lines) wrappers around concatenate.– tel
Nov 21 '18 at 20:48
Here you can find more about the thema stackoverflow.com/questions/8486294/…
– oakca
Nov 21 '18 at 21:03
Here you can find more about the thema stackoverflow.com/questions/8486294/…
– oakca
Nov 21 '18 at 21:03
|
show 3 more comments
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