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)

You can do this easily with numpy arrays. Example -

In [2]: import numpy as npIn [3]: na = np.array([[1,2,3],[3,4,5]])In [4]: naOut[4]:array([[1, 2, 3],       [3, 4, 5]])In [5]: na[:,2] = 10In [6]: naOut[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]: naOut[9]:array([[ 1,  2, 10],       [ 3,  4, 10]])In [10]: func(na[:,1])In [11]: naOut[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.

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)]