Python 作业求教,矩阵转置 - V2EX
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HHH01
V2EX    Python

Python 作业求教,矩阵转置

  •  
  •   HHH01 2019-11-08 16:40:02 +08:00 4672 次点击
    这是一个创建于 2221 天前的主题,其中的信息可能已经有所发展或是发生改变。

    又有一题不是特别会做,求教大家, 需要用 python 写一个 function 去得出某个二维矩阵的转置矩阵。 好像看起来挺简单,转置矩阵就是 Nij=Nji,但是还没想到思路。因此求教大神。

    如题,

    Exercise 4. Within the module matrix_utils.py write a function transpose() . The function should accept one parameter matrix which is a two-dimensional matrix (in Python it is a list of lists of the same size). The function should return a transposed matrix. Example usages of this functions: matrix1 = [[0, 1, 2, 3, 4], [0, 1, 2, 3, 4] ] print("Transposed:", transpose(matrix1)) Transposed: [[0, 0], [1, 1], [2, 2], [3, 3], [4, 4]] matrix2 = [[ ' a ' , ' b ' ], [ ' c ' , ' d ' ] ] print("Transposed:", transpose(matrix2)) Transposed: [[ 'a ' , ' c '] , [ 'b ' , ' d ']]

    20 条回复    2019-11-11 16:58:11 +08:00
    KylinJiang
        1
    KylinJiang  
       2019-11-08 16:46:56 +08:00
    emmm 不太在意性能的话:
    KylinJiang
        2
    KylinJiang  
       2019-11-08 16:47:12 +08:00
    map(list,zip(*N))
    N 为你的多维数组
    HHH01
        3
    HHH01  
    OP
       2019-11-08 16:51:42 +08:00
    @KylinJiang 不能用 zip,还没学到,用 lists 的 loop。。。
    HHH01
        4
    HHH01  
    OP
       2019-11-08 16:57:26 +08:00
    我想到的思路,但是有问题
    def transpose(matrix):
    matrix == list
    n = len(matrix)
    tran_m = []
    m=0
    for m in range(0,n):
    m += len(matrix[m])

    temp=[]
    for i in range(n):
    tran_m=tran_m+[temp]

    for j in range(m):
    temp=temp+[matrix[j][i]]

    return tran_m
    HHH01
        5
    HHH01  
    OP
       2019-11-08 17:09:34 +08:00
    想到了!
    def transpose(matrix):
    matrix == list

    tran_m = []

    for i in range(len(matrix[0])):
    temp=[]

    for j in range(len(matrix)):
    temp=temp+[matrix[j][i]]
    tran_m=tran_m+[temp]

    return tran_m
    cherbim
        6
    cherbim  
       2019-11-08 17:09:49 +08:00   1
    ```
    def transpose(m):
    result = []
    row = len(m)
    col = len(m[0])
    for i in range(col):
    item = []
    for index in range(row):
    item.append(m[index][i])
    result.append(item)
    return result


    matrix1 = [[0, 1, 2, 3, 4], [0, 1, 2, 3, 4]]
    print("Transposed:", transpose(matrix1))
    ```
    cherbim
        7
    cherbim  
       2019-11-08 17:11:14 +08:00
    获得行数和列数,然后调一下行和列存入另一个矩阵
    TimePPT
        8
    TimePPT  
    PRO
       2019-11-08 17:37:31 +08:00
    进来学习。

    以及,每次看到这种题就感慨 numpy 真香
    Trim21
        9
    Trim21  
       2019-11-08 17:45:51 +08:00 via Android
    @TimePPT import numpy 交作业可还行 233
    009694
        10
    009694  
       2019-11-08 17:47:02 +08:00
    t = [[1,2,3],[4,5,6]]
    t_t = [[row[i] for row in t] for i in range(max(len(t) for t in t))]
    009694
        11
    009694  
       2019-11-08 17:49:03 +08:00
    transpose = lambda t:[[row[i] for row in t] for i in range(max(len(t) for t in t))]
    TimePPT
        12
    TimePPT  
    PRO
       2019-11-08 18:11:24 +08:00
    @Trim21 老师让你用锤子锯子钳子做把椅子,你转身去库房拉了个数控机床
    ddzzhen
        13
    ddzzhen  
       2019-11-08 18:15:10 +08:00 via Android
    pandas 直接 T 就行了
    xiri
        14
    xiri  
       2019-11-08 18:24:39 +08:00
    emmmm,如果不是作业的话 numpy 分分钟就 over 了
    coordinate
        15
    coordinate  
       2019-11-08 19:03:37 +08:00
    for i in range(len(mat)):
    for j in range(i+1, len(mat[0])):
    mat[i][j], mat[j][i] = mat[j][i], mat[i][j]
    necomancer
        16
    necomancer  
       2019-11-08 20:59:31 +08:00
    traspose = lambda x : [ [x[i][j] for i in range(len(x))] for j in range(len(x[0]))]

    转置只能对 (mxn) 形状的列表有效。
    HHH01
        17
    HHH01  
    OP
       2019-11-08 22:48:24 +08:00 via Android
    @TimePPT 还没学到,不能用啊
    HHH01
        18
    HHH01  
    OP
       2019-11-08 22:49:59 +08:00 via Android
    还是众人拾柴火焰高
    cshlxm
        19
    cshlxm  
       2019-11-09 14:03:25 +08:00
    import numpy 哈哈哈
    list 的话,只能读一行,再循环每一行 append 一个对应数
    yuuiasuka
        20
    yuuiasuka  
       2019-11-11 16:58:11 +08:00
    可以参考下 numpy 里面的 transpose 方法是怎么实现的。
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