How to get the transpose of this matrix..Any easier ,algorithmic way to do this...
1st question:
Input a=[[1,2,3],[4,5,6],[7,8,9]]
Expected output a=[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
2nd Question:
Zip gives me the following output said below,how can i zip when i dont know how many elements are there in the array,in this case i know 3 elements a[0],a[1],a[2] but how can i zip a[n] elements
>>> zip(a[0],a[1],a[2])
[(1, 4, 7), (2, 5, 8), (3, 6, 9)]
Use zip(*a):
>>> zip(*a)
[(1, 4, 7), (2, 5, 8), (3, 6, 9)]
How it works: zip(*a) is equal to zip(a[0], a[1], a[2]).
map(list, zip(*a)) since that would have answered the question without using a non-standard module.question answers:
>>> import numpy as np
>>> first_answer = np.transpose(a)
>>> second_answer = [list(i) for i in zip(*a)]
thanks to afg for helping out
zip(*a) is matrix transposition and so is its own inverse.map(list, zip(*a)) or [list(row) for row in zip(*a)] if you really need the rows to be lists. Often you don't.
[list(i) for i in zip(*a)] as zip(*a) already produces tuples. edit: I agree with agf, not sure why you are getting numpy involved here when you don't need to. zip(*a) with a cast to list is a much more elegant(and correct!) solution.You can use numpy.transpose
>>> import numpy
>>> a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> numpy.transpose(a)
array([[1, 4, 7],
[2, 5, 8],
[3, 6, 9]])
Solution is to use tuple() function.
Here is an example how to do that in your case :
a = [[1,2,3],[4,5,6],[7,8,9]]
output = tuple(zip(*a))
print(output)
You can use list(zip(*a)).
By using *a, your list of lists can have any number of entries.
Without zip
def transpose(A):
res = []
for col in range(len(A[0])):
tmp = []
for row in range(len(A)):
tmp.append(A[row][col])
res.append(tmp)
return res
Using zip
def transpose(A):
return map(list, zip(*A))
Try this replacing appropriate variable
import numpy as np
data = ((0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11))
data_transpose = np.transpose(data) # replace with your code
print(data_transpose)
