Replacing a vertical sublist in a list of lists

Question

This question is an extension to this question.

I'm representing a two-dimensional array using list of lists, L , say:

[ [1, 2, 3, 4],
  [1, 2, 3, 4],
  [1, 2, 3, 4],
  [1, 2, 3, 4] ]

For a given sub-list, say [9, 99] , I want to replace a specific sub-list in the "2-D" list by this sublist using something intuitive like:

L[1][0:2] = sublist

# which updates `L` to:

[ [1, 2, 3, 4],
  [1, 9, 99, 4],
  [1, 2, 3, 4],
  [1, 2, 3, 4] ] # not in this format, but written like this for clarity

This works for horizontal replacements, but not for vertical replacements since, as we can't slice to separate lists like this: L[0:2][0] . If I had to use this slicing system, I could transpose L ( Transpose list of lists ), then use this slicing method, then transpose it back. But that's not efficient, even for the sake of simplicity.

What would be an efficient way to replicate L[0:2][0] and get this output?

[ [1, 2, 3, 4],
  [1, 9, 3, 4],
  [1, 99, 3, 4],
  [1, 2, 3, 4] ]

Note: Assume len(sublist) <= len(L) , for vertical replacements (which is the focus of this question).

Answer 1

Looping approach:

def replaceVert(al : list, repl:list, oIdx:int, iIdx:int):
    for pos in range(len(repl)):
        al[oIdx+pos][iIdx] =  repl[pos]

a = [ [1, 2, 3, 4],
      [5, 6, 7, 8],
      [9, 10, 11, 12],
      [13, 14, 15, 16] ]

print(a)   # [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

replaceVert(a,['ä','ü'],2,2)  # this is a one liner ;)

print(a)   # [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 'ä', 12], [13, 14, 'ü', 16]]

---

Transpose/slice/transpose approach:

I overread the mentioning of "no transposing". This is using transpose, change, transpose method with slicing which is not wanted by the Q . It is a answer for the title of this question, so I decided to leave it in for future people search SO and stumble over this Q:

a = [ [1, 2, 3, 4],
      [5, 6, 7, 8],
      [9, 10, 11, 12],
      [13, 14, 15, 16] ] 

b = list(map(list,zip(*a)))  # will make [ [1,5,9,13], ... ,[4,8,12,16]]
b[1][0:2]=['a','b']          # replaces what you want here (using a and b for clarity)
c = list(map(list,zip(*b)))  # inverts b back to a's form

print(a)
print(b)
print(c) 

Output:

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]     # a
[[1, 5, 9, 13], ['a', 'b', 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]] # b replaced 
[[1, 'a', 3, 4], [5, 'b', 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]] # c

---

Timing 4x4 list, 2 replaces :

setuptxt = """
def replaceVert(al : list, repl:list, oIdx:int, iIdx:int):
    for pos in range(len(repl)):
        al[oIdx+pos][iIdx] =  repl[pos]

a = [ [1, 2, 3, 4],
      [5, 6, 7, 8],
      [9, 10, 11, 12],
      [13, 14, 15, 16] ]
"""
zipp = """b = list(map(list,zip(*a)))  
b[1][0:2]=['a','b']           
c = list(map(list,zip(*b)))
"""

import timeit

print(timeit.timeit("replaceVert(a,['ä','ü'],2,2)",setup = setuptxt))
print(timeit.timeit(stmt=zipp, setup=setuptxt))

Output:

looping: 12.450226907037592
zipping: 7.50479947070815

The method wit ZIPPing (transpose/slice/transpose) needs roughly 60% of the time for 4x4 lists.

---

Bigger list 1000x1000 and ~70 elements replaced:

setuptxt = """
def replaceVert(al : list, repl:list, oIdx:int, iIdx:int):
    for pos in range(len(repl)):
        al[oIdx+pos][iIdx] =  repl[pos]

a = [ [kk for kk in range(1+pp,1000+pp)] for pp in range(1,1000)] 
repl = [chr(mm) for mm in range(32,100)]
"""

import timeit


print(timeit.timeit("replaceVert(a,repl,20,5)",number=500, setup = setuptxt))

zipp = """b = list(map(list,zip(*a)))  
b[20][5:5+len(repl)]=repl           
c = list(map(list,zip(*b)))
"""

print(timeit.timeit(stmt=zipp, setup=setuptxt,number=500))

Output:

looping: 0.07702917579216137
zipping: 69.4807168493871 

Looping wins. Thanks @Sphinx for his comment