如何在二维数组中切换行和列?

Day*_*man 5 python arrays algorithm multidimensional-array pascals-triangle

我正在编写一个代码,它将创建一个可视化的谢尔宾斯基三角形以进行 3D 打印,为了使其工作,我必须使用将创建数组的 Pascal 三角形算法,以便我可以用来告诉我的算法将创建我的三角形不放三角形的地方。

无论如何,问题是,我排列三角形的代码按列而不是像 Pascal 算法那样按行创建三角形,所以我只是想通过重新排列 Pascal 数组的子例程来快速修复。我只是被难住了如何去做,因为我不知道如何避免index out of range错误。

这是为帕斯卡三角形创建数组的代码。

TL:DR我正在尝试重新排列一个数组,其中行是列,列是行

def pascal(n):
    """Prints out n rows of Pascal's triangle."""
    row = [1]
    global array
    array = [[0 for x in range(int(n))] for y in range(int(n))]
    array[0]=row
    k = [0]
    for x in range(int(max(n,0)-1)):
       row=[l+r for l,r in zip(row+k,k+row)]
       array[x+1]=row
    return 1
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这是打印数组的输出。我只希望行是列,列是行

def pascal(n):
    """Prints out n rows of Pascal's triangle."""
    row = [1]
    global array
    array = [[0 for x in range(int(n))] for y in range(int(n))]
    array[0]=row
    k = [0]
    for x in range(int(max(n,0)-1)):
       row=[l+r for l,r in zip(row+k,k+row)]
       array[x+1]=row
    return 1
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如果您对该项目感到好奇,这里是完整代码,但它需要rhinoscriptsyntax制作模型。

import rhinoscriptsyntax as rhino
import math

obj = rhino.GetObject("Select object to transform", preselect=True)
scale = 3
n=math.pow(3,scale)
def pascal(n):
   """Prints out n rows of Pascal's triangle."""
   row = [1]
   global array
   array = [[0 for x in range(int(n))] for y in range(int(n))]
   array[0]=row
   k = [0]
   for x in range(int(max(n,0)-1)):
      row=[l+r for l,r in zip(row+k,k+row)]
      array[x+1]=row
   return 1
pascal(math.pow(2,scale))
print array
def remakePascal():
    pass
my_horizontalVector = [[1,0,0,6],
           [0,1,0,0],
           [0,0,1,0],
           [0,0,0,1]]

my_tsfm = [[1,0,0,0],  #identity
           [0,1,0,0],
           [0,0,1,0],
           [0,0,0,1]]


def makeTriangle(scale,obj):
    w=1/scale
    h=1/scale
    tsfm= [[w,0,0,0],   #scale about origin
            [0,h,0,0],
            [0,0,1,0],
            [0,0,0,1]]
    output= rhino.XformMultiply(my_tsfm,tsfm)
    new_obj=rhino.TransformObject(obj,output,copy=True)
    return new_obj
def placeObj(i):
    my_moveUpVector = [[1,0,0,(3/scale)*i],
           [0,1,0,(4/scale)*i],
           [0,0,1,0],
           [0,0,0,1]]
    vector = rhino.XformMultiply(my_tsfm,my_moveUpVector)
    return vector
n=0
for i in range(int(math.pow(2,scale))):
    if(i>0):
       hPlace=rhino.XformMultiply(my_tsfm,my_horizontalVector)
       obj = rhino.TransformObject(obj,hPlace)
    factor = int(math.pow(2,scale))-n
    for j in range(factor):
        if():
            pass
        else:
            Vertobj=makeTriangle(scale,obj)
            tsfm = rhino.TransformObject(Vertobj,placeObj(j),copy=True)
    n=n+1
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ram*_*gal 6

对于转置方形阵列,简单的解决方案是

transposed_array = zip(*array)
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这种方法不适合三角形数据,因为zip当行长度不等时不会插入填充。itertools.izip_longest确实如此,但是:

import itertools
transposed_array = itertools.izip_longest(*array)
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默认情况下,使用izip_longest填充None,因此您会得到如下结果:

[(1, 1, 1, 1, 1),
 (None, 1, 2, 3, 4),
 (None, None, 1, 3, 6),
 (None, None, None, 1, 4),
 (None, None, None, None, 1)]
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如果您愿意,可以None使用列表理解删除条目:

no_nones = [[item for item in row if item is not None] for row in transposed_array]
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这让你有:

[[1, 1, 1, 1, 1],
 [1, 2, 3, 4],
 [1, 3, 6],
 [1, 4],
 [1]]
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如果你喜欢用0(或其他任何东西)垫,指定为fillvalue关键字参数itertools.izip_longest; 例如

list(itertools.izip_longest(*array, fillvalue=0))
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返回

[(1, 1, 1, 1, 1),
 (0, 1, 2, 3, 4),
 (0, 0, 1, 3, 6),
 (0, 0, 0, 1, 4),
 (0, 0, 0, 0, 1)]
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我在这里假设元组列表适用于您的目的。如果您需要内部可变,您可以通过列表理解来获得它,例如:

list_of_lists = [list(row) for row in transposed_array]
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  • 是的你可以。如果你想获取列表列表(不是元组列表),你可以在一行中写:`list_of_lists = list(map(list, itertools.izip_longest(*array, fillvalue=0)))` (2认同)

mar*_*eau 1

你可以如图所示进行。它的工作原理是首先将数组变成正方形,以便所有行都具有相同数量的元素。然后使用内置zip()函数转置行和列,最后删除最初添加的元素。

另请注意,我删除了全局变量的使用array。最好避免全局变量。

from pprint import pprint

def pascal(n):
    """Creates n rows of Pascal's triangle."""
    array = [None for y in range(n)]

    row = [1]
    array[0] = row
    k = [0]
    for x in range(max(n, 0)-1):
       row = [l+r for l,r in zip(row+k, k+row)]
       array[x+1] = row

    return array

def transpose(array):
    array = array[:]  # make copy to avoid changing original
    n = len(array)
    for i, row in enumerate(array):
        array[i] = row + [None for _ in range(n - len(row))]

    array = zip(*array)

    for i, row in enumerate(array):
        array[i] = [elem for elem in row if elem is not None]

    return array

array = pascal(8)
array = transpose(array)
pprint(array)
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输出:

from pprint import pprint

def pascal(n):
    """Creates n rows of Pascal's triangle."""
    array = [None for y in range(n)]

    row = [1]
    array[0] = row
    k = [0]
    for x in range(max(n, 0)-1):
       row = [l+r for l,r in zip(row+k, k+row)]
       array[x+1] = row

    return array

def transpose(array):
    array = array[:]  # make copy to avoid changing original
    n = len(array)
    for i, row in enumerate(array):
        array[i] = row + [None for _ in range(n - len(row))]

    array = zip(*array)

    for i, row in enumerate(array):
        array[i] = [elem for elem in row if elem is not None]

    return array

array = pascal(8)
array = transpose(array)
pprint(array)
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