2

I want to add a vector as the first column of my 2D array which looks like : 

[[  1.     0.     0.      nan]
 [  4.     4.     9.97   1.  ]
 [  4.     4.    27.94   1.  ]
 [  2.     1.     4.17   1.  ]
 [  3.     2.    38.22   1.  ]
 [  4.     4.    31.83   1.  ]
 [  3.     4.    41.87   1.  ]
 [  2.     1.    18.33   1.  ]
 [  4.     4.    33.96   1.  ]
 [  2.     1.     5.65   1.  ]
 [  3.     3.    40.74   1.  ]
 [  2.     1.    10.04   1.  ]
 [  2.     2.    53.15   1.  ]]

I want to add an aray [] of 13 elements as the first column of the matrix. I tried with np.stack_column, np.append but it is for 1D vector or doesn't work because I can't chose axis=1 and only do np.append(peak_values, results)

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2
  • This is an old post, but for anyone still looking, make numpy.c_ your friend. Essentially it handles any reshape and stacking operations necessary to combine arrays along their last axis. For the example above, if arr is the 2D array and v is the vector, all you need is np.c_[v, arr]. Note the square brackets. Commented Aug 27, 2021 at 18:47
  • Note that is equivalent to np.hstack((v.reshape(-1,1), arr)) but more powerful because it works in many other cases too. Commented Aug 27, 2021 at 18:53

4 Answers 4

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1

I have a very simple option for you using numpy -

x = np.array( [[ 3.9427767, -4.297677 ],
 [ 3.9427767, -4.297677 ],
 [ 3.9427767, -4.297677 ],
 [ 3.9427767, -4.297677 ],
 [ 3.942777 , -4.297677 ],
 [ 3.9427767, -4.297677 ],
 [ 3.9427767, -4.297677 ],
 [ 3.9427767 ,-4.297677 ],
 [ 3.9427767, -4.297677 ],
 [ 3.9427772 ,-4.297677 ]])

b = np.arange(10).reshape(-1,1)
np.concatenate((b.T, x), axis=1)

Output-

array([[ 0.       ,  3.9427767, -4.297677 ],
       [ 1.       ,  3.9427767, -4.297677 ],
       [ 2.       ,  3.9427767, -4.297677 ],
       [ 3.       ,  3.9427767, -4.297677 ],
       [ 4.       ,  3.942777 , -4.297677 ],
       [ 5.       ,  3.9427767, -4.297677 ],
       [ 6.       ,  3.9427767, -4.297677 ],
       [ 7.       ,  3.9427767, -4.297677 ],
       [ 8.       ,  3.9427767, -4.297677 ],
       [ 9.       ,  3.9427772, -4.297677 ]])
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1 Comment

This raises ValueError: all the input array dimensions for the concatenation axis must match exactly, but along dimension 0, the array at index 0 has size 1 and the array at index 1 has size 10. I think you meant (b, x) instead of (b.T, x).
1

Improving on this answer by removing the unnecessary transposition, you can indeed use reshape(-1, 1) to transform the 1d array you'd like to prepend along axis 1 to the 2d array to a 2d array with a single column. At this point, the arrays only differ in shape along the second axis and np.concatenate accepts the arguments:

>>> import numpy as np
>>> a = np.arange(12).reshape(3, 4)
>>> b = np.arange(3)
>>> a
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])
>>> b
array([0, 1, 2])
>>> b.reshape(-1, 1) # preview the reshaping...
array([[0],
       [1],
       [2]])
>>> np.concatenate((b.reshape(-1, 1), a), axis=1)
array([[ 0,  0,  1,  2,  3],
       [ 1,  4,  5,  6,  7],
       [ 2,  8,  9, 10, 11]])
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Comments

0

For the simplest answer, you probably don't even need numpy.

Try the following:

new_array = []
new_array.append(your_array)

That's it.

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3 Comments

Afterward, you could simply transform this new_array into numpy format as you wish, by doing np.array(new_array).
I don't get where you add the vector. My 2D is array and 1D is vector, I have to do array.append(vector) ?
Nah don't worry about that. From the start, let's say your array name is my_2d_array. We do my_2d_list = [ ] and then my_2d_list.append(my_2d_array). Then you got a 2D list with your array as the first column. Finally, please do my_final_array = np.array(my_2d_list), which will transform the python list into numpy array format. Hope that helps!
-1

I would suggest using Numpy. It will allow you to easily do what you want.

Here is an example of squaring the entire set. you can use something like nums[0].

nums = [0, 1, 2, 3, 4]
even_squares = [x ** 2 for x in nums if x % 2 == 0] 
print even_squares  # Prints "[0, 4, 16]"
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1 Comment

I don't understand how squaring a matrix is relevant to the question. The code shown isn't numpy either.

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