Python implementation of matrix transpose
Transposing a matrix means that we change its columns into rows. Let's understand it through an example of what it looks like if transposed.
Assuming you have the original matrix, for example -x = [[1,2][3,4][5,6]]
In the matrix " x" above, we have two columns, 1, 3, 5 and 2, 4, 6.
Thus, when we transpose above the matrix "x", the columns become rows. Thus, the transposed version of the matrix above looks like -x1 = [[1, 3, 5][2, 4, 6]]
Thus, we have another matrix " x1 ", organized differently with different values at different locations.
Method 1, using numpy transpose
import numpy as np A = ([[1,2,3],[4,5,6],[7,8,9]]) print() print((0, 1))
# Both outputs
# [[1 4 7]
# [2 5 8]
# [3 6 9]]
import numpy as np A = [[1,2,3],[4,5,6],[7,8,9]] print((A))
# Output
# [[1 4 7]
# [2 5 8]
# [3 6 9]]
Method two, using the zip() function
- The zip() function is used to take an iterable object as an argument, pack the corresponding elements of the object into a tuple, and then return the object composed of those tuples, which has the advantage of saving a lot of memory.
- You can use the list() conversion to output a list. Differences between the zip method in Python 2 and Python 3: In Python, zip() returns an object to minimize memory. To display a list, you need to convert list() manually.]
- If the number of elements is not the same across iterators, the list is returned with the same length as the shortest object, utilizing the
*No. operator that decompresses a tuple into a list.
zip(A) Equivalent to packing, packing isList of tuples：
>>> a = [1,2,3] >>> b = [4,5,6] >>> c = [4,5,6,7,8] >>> A = zip(a,b) # Packed as a list of tuples [(1, 4), (2, 5), (3, 6)] >>> zip(a,c) # of elements to match the shortest list [(1, 4), (2, 5), (3, 6)] >>> zip(*A) # In contrast to zip, *A can be interpreted as decompression and returns a two-dimensional matrix equation [(1, 2, 3), (4, 5, 6)]
A = [[1,2,3],[4,5,6],[7,8,9]] print(*A) #[1, 2, 3] [4, 5, 6] [7, 8, 9] #zip() returns an object. To display a list, you need to convert list() manually. #print(zip(*A)) #<zip object at 0x000001CD7733A2C8> print(list(zip(*A)))
# Output
# [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
Here the asterisk (*) in python serves to disassemble the elements of an iterable object in a variable.
Method 3: Using python list expressions
Doesn't take up extra space, "modifies in place"
A = [[1,2,3],[4,5,6],[7,8,9]] #print(len(A)) #matrix rows #print(len(A[0])) #matrix columns B = [[A[j][i] for j in range(len(A))] for i in range(len(A[0]))] print(B)
# Output
# [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
B = [[A[j][i] for j in range(len(A))] for i in range(len(A[0]))]
A clearer way to write this sentence is:
A = [[1,2,3],[4,5,6],[7,8,9]]
#print(len(A)) #matrix rows
#print(len(A[0])) #matrix columns
for i in range(len(A[0])):#len(A[0]) number of matrix columns
for j in range(i,len(A)):#len(A) number of matrix rows
#Transpose is A[i][j] and A[j][i] interchanged
A[j][i], A[i][j] = A[i][j], A[j][i]
print(A)# Output
# [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
Because of the symmetry of the transposed matrixfor j in range(i,len(A)) The restriction to traverse only the upper triangles of the matrix must be restricted, as failure to do so will result in repeated exchanges.
Method 4: Create a new list B and add elements using a double loop
A = [[1,2,3],[4,5,6],[7,8,9]]
B=[]
for i in range(len(A[0])):#len(A[0]) number of matrix columns
temp = []
for j in range(len(A)):#len(A) number of matrix rows
(A[j][i])
(temp)
print(B)# Output
# [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
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