Fill a column of a numpy array with another array

Question

I have:

x = np.zeros((96,11,11,2,10),dtype=np.float64)
y = np.array([0,10,20,30,40,50,60,70,80,90,100],dtype=np.float64)
x[:,:,:,0,0] = y
print x[0,:,:,0,0]

i get:

[[   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]
 [   0.   10.   20.   30.   40.   50.   60.   70.   80.   90.  100.]]

but I want the output (for any i in first dimension) to be the transpose of this. ie fill columns instead of rows

Any suggestions?

Is this your actual example? Because I don't see how you can assign a 1D array to a 3D slice, and when I try it, as expected, x[:,:,:,0,0] = y raises ValueError: operands could not be broadcast together with shapes (20,11,11) (10). — abarnert
– abarnert, Commented Sep 12, 2013 at 9:06
I asked the question again in a better way here: stackoverflow.com/questions/18763717/… — Sammy
– Sammy, Commented Sep 12, 2013 at 12:42
@Sammy It is recommend that you edit your question then to ask a new one if the question is, in effect, the same. — Daniel
– Daniel, Commented Sep 12, 2013 at 12:54

Lee · Accepted Answer · 2013-09-12 16:28:55Z

3

You need to change y from 1D to 2D (with one column):

x[:,:,:,0,0] = y[:, np.newaxis]

or,

x[:,:,:,0,0] = y.reshape(11,1)

edited Sep 12, 2013 at 16:28

answered Sep 12, 2013 at 16:23

Lee

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jabaldonedo · Accepted Answer · 2013-09-12 09:42:08Z

2

If you want the output to be the transpose, just do:

>>> import numpy as np
>>> x = np.zeros((96,11,11,2,10),dtype=np.float64)
>>> y = np.array([0,10,20,30,40,50,60,70,80,90,100],dtype=np.float64)
>>> for i in range(x.shape[0]):
>>>    x[i,:,:,0,0] = x[i,:,:,0,0].T
>>> print x[0,:,:,0,0]
 [[   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
  [  10.   10.   10.   10.   10.   10.   10.   10.   10.   10.   10.]
  [  20.   20.   20.   20.   20.   20.   20.   20.   20.   20.   20.]
  [  30.   30.   30.   30.   30.   30.   30.   30.   30.   30.   30.]
  [  40.   40.   40.   40.   40.   40.   40.   40.   40.   40.   40.]
  [  50.   50.   50.   50.   50.   50.   50.   50.   50.   50.   50.]
  [  60.   60.   60.   60.   60.   60.   60.   60.   60.   60.   60.]
  [  70.   70.   70.   70.   70.   70.   70.   70.   70.   70.   70.]
  [  80.   80.   80.   80.   80.   80.   80.   80.   80.   80.   80.]
  [  90.   90.   90.   90.   90.   90.   90.   90.   90.   90.   90.]
  [ 100.  100.  100.  100.  100.  100.  100.  100.  100.  100.  100.]]

It updates the first dimension, this is the output for 34th index:

>>> print x[34,:,:,0,0]
 [[   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
  [  10.   10.   10.   10.   10.   10.   10.   10.   10.   10.   10.]
  [  20.   20.   20.   20.   20.   20.   20.   20.   20.   20.   20.]
  [  30.   30.   30.   30.   30.   30.   30.   30.   30.   30.   30.]
  [  40.   40.   40.   40.   40.   40.   40.   40.   40.   40.   40.]
  [  50.   50.   50.   50.   50.   50.   50.   50.   50.   50.   50.]
  [  60.   60.   60.   60.   60.   60.   60.   60.   60.   60.   60.]
  [  70.   70.   70.   70.   70.   70.   70.   70.   70.   70.   70.]
  [  80.   80.   80.   80.   80.   80.   80.   80.   80.   80.   80.]
  [  90.   90.   90.   90.   90.   90.   90.   90.   90.   90.   90.]
  [ 100.  100.  100.  100.  100.  100.  100.  100.  100.  100.  100.]]

edited Sep 12, 2013 at 9:42

answered Sep 12, 2013 at 9:14

jabaldonedo

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4 Comments

Sammy Over a year ago

Sorry for my confusing wording. I actually want the output to look like the transpose of this (don't want the actual transpose)

jabaldonedo Over a year ago

ok, take a look now, you only need to reassign x[0,:,:,0,0].T to x[0,:,:,0,0]

Sammy Over a year ago

My point is I want x[1,:,:,0,0] , x[2,:,:,0,0], x[3,:,:,0,0] .....x[95,:,:,0,0] to look like this too. so Im looking for something in the way I initially write to x

jabaldonedo Over a year ago

You don't mention that on your question, you didn't say anything about all dimension but only the first one. In any case, I have updated the answer so now all x[i,:,:,0,0] dimensions are updated

abarnert · Accepted Answer · 2013-09-12 18:47:09Z

The problem is simple: you're using a row vector for y instead of a column vector, so it's filling by row instead of by column.

More technically, you've got an array of shape (11,), instead of an array of (11, 1), so it broadcasts to (1, 11) when filling a 2D array.

Compare:

>>> x = np.zeros((96,11,11,2,10),dtype=np.float64)
>>> y = np.array([[0],[10],[20],[30],[40],[50],[60],[70],[80],[90],[100]],dtype=np.float64)
>>> x[:,:,:,0,0]=y
>>> print x[0,:,:,0,0]
[[   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
 [  10.   10.   10.   10.   10.   10.   10.   10.   10.   10.   10.]
 [  20.   20.   20.   20.   20.   20.   20.   20.   20.   20.   20.]
 [  30.   30.   30.   30.   30.   30.   30.   30.   30.   30.   30.]
 [  40.   40.   40.   40.   40.   40.   40.   40.   40.   40.   40.]
 [  50.   50.   50.   50.   50.   50.   50.   50.   50.   50.   50.]
 [  60.   60.   60.   60.   60.   60.   60.   60.   60.   60.   60.]
 [  70.   70.   70.   70.   70.   70.   70.   70.   70.   70.   70.]
 [  80.   80.   80.   80.   80.   80.   80.   80.   80.   80.   80.]
 [  90.   90.   90.   90.   90.   90.   90.   90.   90.   90.   90.]
 [ 100.  100.  100.  100.  100.  100.  100.  100.  100.  100.  100.]]

Of course in your real code, y probably isn't a literal, but a result of some earlier computation. (And even if it is a literal, you don't want to type all those extra brackets.) So, assume y is inherently a row vector, as we have to deal with it.

So, just reshape it on the fly:

>>> x = np.zeros((96,11,11,2,10),dtype=np.float64)
>>> y = np.array([0,10,20,30,40,50,60,70,80,90,100],dtype=np.float64)
>>> x[:,:,:,0,0] = y.reshape((11, 1))

Same result.

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