Append Numpy array to correct shape

Question

I'm having a very hard time appending my x_ array to x while maintaining the correct shape. I tried vstack, but it gave me an error. The axis=0 doesn't seem to be doing anything like it's supposed to. I want an array with dimension (:,len(x_)).

Edit:

The code at the end of the post gives arrays x_ of the following shape:

array([3, 0, 2, 1, 0], dtype=int32)

I tried:

x_ = np.append(x_,np.array([5,4,6,7,8]), axis = 0)

But gives:

array([3, 0, 2, 1, 0, 5, 4, 6, 7, 8])

However, I want:

array([[3, 0, 2, 1, 0],
       [5, 4, 6, 7, 8]])

I tried vstack(x,x_), but got:

   x = np.vstack(x, x_)

TypeError: vstack() takes 1 positional argument but 2 were given

.

for k in range(2,9):
    temp_ = (2*k)+1
    x = np.zeros(shape=(1,temp_))
    y = []
    for i in range(k, len(number_list)-k-1):
        newk = k
        x_ = []
        while newk >= -k:
            x_.append(name[i-newk])
            newk-=1
        le = preprocessing.LabelEncoder()
        le.fit(x_)
        x_ = le.transform(x_)
        x = np.append(x, x_ , axis=0)
        y.append(residue_area[i])

Can you update your question with the full traceback of the error? — Kasravnd
– Kasravnd, Commented Apr 18, 2017 at 6:45
Appending to a list is faster and less error prone. But if you must join arrays, use the basic np.concatenate. It requires correct array shapes, and doesn't try to compensate or second guess your intentions. — hpaulj
– hpaulj, Commented Apr 18, 2017 at 7:09

Guillermo Ventura · Accepted Answer · 2017-10-11 13:54:43Z

4

np.append must be abolished; it distorts the thinking of too many beginners.

np.concatenate and all the np.?stack take a list as the first argument. The only other argument is an axis keyword. np.append changes that convention by replacing that list with 2 arguments - which it then tweaks, puts in a list and calls concatenate. And unless you read the source code you don't know for sure how it modifies the dimensions first.

Pull up an interactive session, and play with np.concatenate until you understand how the dimensions must match. Then revisit this problem.

edit: spelling error concatenate instead of concatenante

edited Oct 11, 2017 at 13:54

Guillermo Ventura

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answered Apr 18, 2017 at 7:54

hpaulj

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2 Comments

Lora Sinha Over a year ago

Thanks, but one question remains. By creating a zeros array with certain shape, I have to create a zeros first entry, which I don't want. How can I create an empty array of certain shape?

hpaulj Over a year ago

If you MUST go the iterative concatenate route, you could start with np.zeros(shape=(0,temp_)). Getting the starting array right is part of why I dislike np.append. Starting a list append with [] is a well known Python idiom. The numpy equivalent is awkward. Much better to collect ALL subarrays in a list, and do concatenate just once.

MSeifert · Accepted Answer · 2017-04-18 07:53:20Z

2

But you can use np.vstack:

>>> import numpy as np
>>> x_ = np.array([3, 0, 2, 1, 0]) 
>>> x_ = np.vstack([x_, np.array([5, 4, 6, 7, 8])])
>>> x_
array([[3, 0, 2, 1, 0],
       [5, 4, 6, 7, 8]])

But generally it's a bad idea (because it's really inefficient) to append or stack arrays regularly. Often it's better to create an array of the final shape and insert into it:

>>> x_ = np.empty((2, 5), dtype=int)
>>> x_[0] = [3, 0, 2, 1, 0]
>>> x_[1] = [5, 4, 6, 7, 8]
>>> x_
array([[3, 0, 2, 1, 0],
       [5, 4, 6, 7, 8]])

answered Apr 18, 2017 at 7:53

MSeifert

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2 Comments

Lora Sinha Over a year ago

Thanks, vstack worked for me, but one question remains. By creating a zeros array with certain shape, I have to create a zeros first entry, which I don't want. How can I create an empty array of certain shape?

MSeifert Over a year ago

No, don't create an empty array and append to it. Like I said that's a very inefficient solution (and I mean terrible inefficient). You can either use a normal list [] and append to it and convert it to an array after you finished appending or you create an empty array of the final shape and then insert into it.

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