在Python中将3x4矩阵转置为4x3

Question

我需要从命令行读取一个文件，并设置矩阵的行和列。当我运行我的程序（如下所示）：

python3 transpose.py               matrix                                                 3 4
        (current code below) (file with numbers 1 2 3 4 5 6 7 8 9 10 11 12 13)  (matrix row,cols)

打印出来的内容应该返回：

The empty matrix is:[[0,0,0,0],[0,0,0,0],[0,0,0,0]]
The original matrix is:
1 2 3 4
5 6 7 8
9 10 11 12

The transposed matrix is:
1 5 9
2 6 10
3 7 11
4 8 12

这是我的transpose.py，我花了大约10个小时在上面，但就是无法让它工作，我实在想不出还有什么办法了！

 import sys
from scanner import *


def main():
    readinput(sys.argv[1],[2])
    size = 3
    rows = size
    cols = rows
    makeList(size)
    matrix(rows,cols)
    manipulateMatrix(matrix(rows,cols))
    print(manipulateMatrix(matrix(rows,cols)))
    print("The original matrix is: ")
    displayMatrix(manipulateMatrix(matrix(rows,cols)))
    transposeSquare(manipulateMatrix(matrix(rows,cols)),size)
    print (transposeSquare(manipulateMatrix(matrix(rows,cols)),size))
    print("The transposed matrix: ")
    displayMatrix(transposeSquare(manipulateMatrix(matrix(rows,cols)),size))    
def readinput(filename,matrix):
    s = Scanner(filename)
    r = s.readtoken()
    while r != "":
        r = int(r)
        c = s.readint()
        v = s.readint()
        matrix[r][c]=v
        r = s.readtoken()
    s.close()

def makeList(size):
    lyst = []
    for i in range(size):
        lyst = lyst + [None]
    return lyst

def matrix(rows,cols):
    matrix = makeList(rows)
    for i in range(rows):
        matrix[i] = makeList(cols)
        return matrix

def manipulateMatrix(m):
    rows = len(m)
    cols = len(m[0])
    count = 1
    for r in range(0,rows,1):
        for c in range(0,cols,1):
            m[r][c] = count
            count += 1
    return m

def transposeSquare(m,size):
    for r in range(0,size):
         for c in range(0,size):
            m[r][c],m[c][r] = m[c][r],m[r][c]
    return m

def displayMatrix(m):
    m == rows == cols
    for r in range(0,size):
        for c in range(0,cols,1):
            print(m[r][c],end = "")
        print()
    return          
main()

现在我卡在了

 Traceback (most recent call last):
  File "transpose.py", line 66, in <module>
    main()
  File "transpose.py", line 6, in main
    readinput(sys.argv[1],[2])
  File "transpose.py", line 27, in readinput
    matrix[r][c]=v
IndexError: list index out of range

Answer 1

使用numpy库。这个库就是为这个目的设计的，它在C/C++代码中运行数值计算，比在Python中做这些事情要快得多，而且不需要调试。

import numpy as np
m = np.array(<your list matrix here>)
print m.shape # to see the shape of your matrix
mt = m.T #m.T transpose the matrix
print m.shape
print m

Answer 2

把一个二维列表转置其实很简单。你可以使用 zip() 函数，再加上 * 这个解包操作符：

>>> matrix = [[1,2,3,4], [5,6,7,8], [9,10,11,12]]
>>> new = list(zip(*matrix))  # call to list not necessary in Python 2
>>> new
[(1, 5, 9), (2, 6, 10), (3, 7, 11), (4, 8, 12)]

或者，如果你不想要元组的话：

>>> new = [list(tup) for tup in zip(*matrix)]
>>> new
[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]