I have an array like this
import numpy as np
a = np.zeros((2,2), dtype=np.int)
I want to replace the first column by the value 1. I did the following:
a[:][0] = [1, 1] # not working
a[:][0] = [[1], [1]] # not working
Contrariwise, when I replace the rows it worked!
a[0][:] = [1, 1] # working
I have a big array, so I cannot replace value by value.
You can replace the first column as follows:
>>> a = np.zeros((2,2), dtype=np.int)
>>> a[:, 0] = 1
>>> a
array([[1, 0],
[1, 0]])
Here a[:, 0] means "select all rows from column 0". The value 1 is broadcast across this selected column, producing the desired array (it's not necessary to use a list [1, 1], although you can).
Your syntax a[:][0] means "select all the rows from the array a and then select the first row". Similarly, a[0][:] means "select the first row of a and then select this entire row again". This is why you could replace the rows successfully, but not the columns - it's necessary to make a selection for axis 1, not just axis 0.
You can do something like this:
import numpy as np
a = np.zeros((2,2), dtype=np.int)
a[:,0] = np.ones((1,2), dtype=np.int)
Please refer to Accessing np matrix columns
Select the intended column using a proper indexing and just assign the value to it using =. Numpy will take care of the rest for you.
>>> a[::,0] = 1
>>> a
array([[1, 0],
[1, 0]])
Read more about numpy indexing.
