Operation on 2d array columns
Without numpy it can be done like this:
map(lambda x: x[:2] + [1] + x[3:], array)
map(lambda x: x[:2] + my_func(x[2]) + x[3:], array)
It would be very simple in numpy and you can do it with a simple assignment :
>>> numpy.array[:,column_number]=value
But if you are looking for a python approach you can use zip
function and itertools.repeat()
:
>>> from itertools import repeat
>>> def replacer(l,index,value):
... z=zip(*l)
... z[index]=list(repeat(value,len(l)))
... return zip(*z)
Demo:
>>> l=[range(4) for _ in range(3)]
>>> replacer(l,2,'*')
[(0, 1, '*', 3), (0, 1, '*', 3), (0, 1, '*', 3)]
Note that since in python 3.X zip
returns an iterator you can use list
function to return a list also since iterators doesn't support indexing inside the function you need to call the list too.
>>> def replacer(l,index,value):
... z=list(zip(*l))
... z[index]=list(repeat(value,len(l)))
... return zip(*z)
>>> list(replacer(l,2,'*'))
[(0, 1, '*', 3), (0, 1, '*', 3), (0, 1, '*', 3)]
You can do this easily with numpy
arrays. Example -
In [2]: import numpy as np
In [3]: na = np.array([[1,2,3],[3,4,5]])
In [4]: na
Out[4]:
array([[1, 2, 3],
[3, 4, 5]])
In [5]: na[:,2] = 10
In [6]: na
Out[6]:
array([[ 1, 2, 10],
[ 3, 4, 10]])
In [7]: na[:,2]
Out[7]: array([10, 10])
In [8]: def func(a):
...: for i,x in enumerate(a):
...: a[i] = x + 1
...:
In [9]: na
Out[9]:
array([[ 1, 2, 10],
[ 3, 4, 10]])
In [10]: func(na[:,1])
In [11]: na
Out[11]:
array([[ 1, 3, 10],
[ 3, 5, 10]])
You can find more details about this here . Please do be careful , for numpy arrays, as stated in documentation -
All arrays generated by basic slicing are always views of the original array.
This is why when changing the sliced array inside the function, the actual array got changed.