Python: Bug in writing a transpose function using lists

Question

I have an assignment to do, which is:

Write a function transpose which takes in a matrix and transposes it. Basically, this converts a m x n matrix into a n x m matrix.

I wrote a code which seems sensible, but it doesnt get me the result I want. Can anyone point out what is wrong with my code?

def transpose(matrix):
    new_matrix=[[]]*len(matrix[0])
    for row in matrix:
        i=0
        for j in row:
            new_matrix[i]+=[j]
            i+=1
    return new_matrix

Test case:

print(transpose([[ 1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]))

Duplicate the original matrix, loop over i,j and keep doing new[j][i] = original[i][j] — Ubdus Samad
– Ubdus Samad, Commented Mar 5, 2018 at 12:23
[list(i) for i in zip(*a)] try this. this will transpose your matrix. — Rohit Chopra
– Rohit Chopra, Commented Mar 5, 2018 at 12:31
Possible duplicate of Initialize a list of objects in Python — Patrick Artner
– Patrick Artner, Commented Mar 5, 2018 at 12:39
Is there other methods other than zip, dont think its expected of me to use zip? @UbdusSamad How would u predetermine the dimensions of the new matrix? I tried it with 2 ranges and it gives me Index Error: index out of range — Prashin Jeevaganth
– Prashin Jeevaganth, Commented Mar 5, 2018 at 12:56

Patrick Artner · Accepted Answer · 2018-03-05 13:12:29Z

1

If you use the * to multiply some values in a list-initialisation, be careful. You might end up with references that point multiple times to the same value:

l = [ [] ]*3
print(l)

l[1].append(34)  # change only "the first" list by appending smth
print(l)

Output:

[[], [], []]
[[34], [34], [34]]  # they are all "the same data" reference

There is an built-in zip() that does exactly your transposing:

l = [[ 1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]] 


l_t = [ list(x) for x in zip(*l)]  # one-line solutions for transposing ;)

print(l)
print(l_t) # transposed

Zip has the limitation that it only works to the length of the smallest sublists - yours are all equal so all is fine.

Output:

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

If you ever need a zip that takes the longest list, itertools.zip_longest(..) can be used, it takes a default param that is substituted for any shorter list-items that are not there.

Btw. just list(zip(l)) looks like this: [(1,5,9),(2,6,10),(3,7,11),(4,8,12)] - it create tuples over the same indexes of the parts of the iterable you put into it.

By hand:

l = [[ 1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]] 


l2 = []
for colIdx in range(len(l[0])):  # 0-3 iterate over the inner indexes first
    newRow = []
    for rowIdx in range(len(l)): # 0-2 then over the outer ones 
        newRow.append(l[rowIdx][colIdx])
    l2.append(newRow)

print(l2)  # [[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

edited Mar 5, 2018 at 13:12

answered Mar 5, 2018 at 12:26

Patrick Artner

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11 Comments

Prashin Jeevaganth Over a year ago

Is there a way to do it without built in functions? I think I'm not expected to do it with them. I dont quite understand Ubdus Samad's solution too

Prashin Jeevaganth Over a year ago

actually reading the comments made me think of simply applying a list comprehension instead of multiplication to get the size of list I want, then my code works. Could mutability of lists be a cause for repeated referencing? ( haven't rlly tried making a transpose function in tuple form, so food for thought) My code:

def transpose(matrix):         new_matrix=[[] for i in range(len(matrix[0]))]         for row in matrix:         i=0         for j in row:             new_matrix[i]+=[j]             i+=1         return new_matrix

Patrick Artner Over a year ago

@PrashinJeevaganth Learn the built-ins and list comprehensions, they make python fun. Edited answer for a "by hand" solution. Use a debugger to step through it to understand it. Read: how-to-debug-small-programs

Patrick Artner Over a year ago

@PrashinJeevaganth The multiplication is what messes your code up this time, list comp is "a" solution - but still far longer then the zip() one

Prashin Jeevaganth Over a year ago

is there a way to learn the built in function? Dont understand what I'm reading by looking at help(zip) for eg

|

Ubdus Samad · Accepted Answer · 2018-03-05 13:27:52Z

1

What i meant was something like this:

def t(array): #The original array has size mxn
    duplicate = [[0 for x in range(len(array))] for y in range(len(array[1]))] #You create an array of size nxm, which is filled with zeros
    for i in range(len(array)): #Loop over the rows
        for j in range(len(array[i])): #Then loop over the columns
            duplicate[j][i] = array[i][j] #Replace j,i or duplicate with i,j th element of original
    return duplicate

Now,

>>> t([[ 1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

edited Mar 5, 2018 at 13:27

answered Mar 5, 2018 at 13:22

Ubdus Samad

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