Develop Reference

Appending a new row to a numpy array

I am trying to append a new row to an existing numpy array in a loop. I have tried the methods involving append, concatenate and also vstack none of them end up giving me the result I want.
I have tried the following:
for _ in col_change:
if (item + 2 < len(col_change)):
arr=[col_change[item], col_change[item + 1], col_change[item + 2]]
array=np.concatenate((array,arr),axis=0)
item+=1
I have also tried it in the most basic format and it still gives me an empty array.
array=np.array([])
newrow = [1, 2, 3]
newrow1 = [4, 5, 6]
np.concatenate((array,newrow), axis=0)
np.concatenate((array,newrow1), axis=0)
print(array)
I want the output to be [[1,2,3][4,5,6]...]

The correct way to build an array incrementally is to not start with an array:
alist = []
alist.append([1, 2, 3])
alist.append([4, 5, 6])
arr = np.array(alist)
This is essentially the same as
arr = np.array([ [1,2,3], [4,5,6] ])
the most common way of making a small (or large) sample array.
Even if you have good reason to use some version of concatenate (hstack, vstack, etc), it is better to collect the components in a list, and perform the concatante once.

If you want [[1,2,3],[4,5,6]] I could present you an alternative without append: np.arange and then reshape it:
>>> import numpy as np
>>> np.arange(1,7).reshape(2, 3)
array([[1, 2, 3],
[4, 5, 6]])
Or create a big array and fill it manually (or in a loop):
>>> array = np.empty((2, 3), int)
>>> array[0] = [1,2,3]
>>> array[1] = [4,5,6]
>>> array
array([[1, 2, 3],
[4, 5, 6]])
A note on your examples:
In the second one you forgot to save the result, make it array = np.concatenate((array,newrow1), axis=0) and it works (not exactly like you want it but the array is not empty anymore). The first example seems badly indented and without know the variables and/or the problem there it's hard to debug.

best way to create a numpy array from a list and additional individual values

I want to create an array from list entries and some additional individual values.
I am using the following approach which seems clumsy:
x=[1,2,3]
y=some_variable1
z=some_variable2
x.append(y)
x.append(z)
arr = np.array(x)
#print arr --> [1 2 3 some_variable1 some_variable2]
is there a better solution to the problem?

You can use list addition to add the variables all placed in a list to the larger one, like so:
arr = np.array(x + [y, z])

Appending or concatenating lists is fine, and probably fastest.
Concatenating at the array level works as well
In [456]: np.hstack([x,y,z])
Out[456]: array([1, 2, 3, 4, 5])
This is compact, but under the covers it does
np.concatenate([np.array(x),np.array([y]),np.array([z])])

Delete line from 2D array in Python

I'm have a 585L, 2L numpy array in Python.
The data is organized like in this example.
0,0
1,0
2,1
3,0
4,0
5,1
...
I would like to create an array deleting all the lines where 0 is present on the seconds column. Thus, the final result pretended is:
2,1
5,1
I have read some related examples but I'm still struggling with this.

Since you mention your structure being a numpy array, rather than a list, I would use numpy logical indexing to select only the values you care for.
>>> import numpy as np
>>> x = [[0,0], [1,0], [2,1], [3,0], [4,0], [5,1]] # Create dummy list
>>> x = np.asarray(x) # Convert list to numpy array
>>> x[x[:, 1] != 0] # Select out items whose second column don't contain zeroes
array([[2, 1],
[5, 1]])

here is my answer
if your list like this [[0,0],[2,1],[4,3],[2,0]]
if your list structure isnt like this please tell me
my answer prints the list whose second column num dont equal to 0
print [x for x in your_list if x[1] != 0] #your_list is the variable of the list

You could use an list comprehension. These are described on the Python Data Structures page (Python 2, Python 3).
If your array is:
x = [[0,0],
[1,0],
[2,1],
[3,0],
[4,0],
[5,1]]
Then the following command
[y for y in x if y[1] != 0]
Would return the desired result of:
[[2, 1], [5, 1]]
Edit: I overlooked that it was a numpy array. Taking that into account, JoErNanO's answer is better.

How to simplify a for loop in python

I have this piece of Python code that fills up a 2d matrix in a for loop
img=zeros((len(bins_x),len(bins_y)))
for i in arange(0,len(ix)):
img[ix[i]][iy[i]]=dummy[i]
Is it possible to use a vectorial operation for the last two lines of code? Is there also something that might speed up the calculation?

If ix, iy are index sequences:
img[ix, iy] = dummy

It might be useful to use numpy. In particular, the reshape method might be useful. Here is an example (adapted from the second link):
>>> import numpy as np
>>> a = np.array([1,2,3,4,5,6])
>>> np.reshape(a, (3,2))
array([[1, 2],
[3, 4],
[5, 6]])

Syntax in Python (.T)

In the help resource for the multivariate normal sampling function in SciPy, they give the following example:
x,y = np.random.multivariate_normal(mean,cov,5000).T
My question is rather basic: what does the final .T actually do?
Thanks a lot, I know it is fairly simple, but it is hard to look in Google for ".T".

The .T accesses the attribute T of the object, which happens to be a NumPy array. The T attribute is the transpose of the array, see the documentation.
Apparently you are creating random coordinates in the plane. The output of multivariate_normal() might look like this:
>>> np.random.multivariate_normal([0, 0], [[1, 0], [0, 1]], 5)
array([[ 0.59589335, 0.97741328],
[-0.58597307, 0.56733234],
[-0.69164572, 0.17840394],
[-0.24992978, -2.57494471],
[ 0.38896689, 0.82221377]])
The transpose of this matrix is:
array([[ 0.59589335, -0.58597307, -0.69164572, -0.24992978, 0.38896689],
[ 0.97741328, 0.56733234, 0.17840394, -2.57494471, 0.82221377]])
which can be conveniently separated in x and y parts by sequence unpacking.

.T is just np.transpose().
Best of luck

Example
import numpy as np
a = [[1, 2, 3]]
b = np.array(a).T # ndarray.T The transposed array. [[1,2,3]] -> [[1][2][3]]
print("a=", a, "\nb=", b)
for i in range(3):
print(" a=", a[0][i]) # prints 1 2 3
for i in range(3):
print(" b=", b[i][0]) # prints 1 2 3