Pascal的Python三角形

时间：2014-06-07 03:35:53

标签： python pascals-triangle

作为Python的学习经验，我正在尝试编写自己版本的Pascal三角形。我花了几个小时（因为我刚刚开始），但我出来了这段代码：

pascals_triangle = []

def blank_list_gen(x):
    while len(pascals_triangle) < x:
        pascals_triangle.append([0])

def pascals_tri_gen(rows):
    blank_list_gen(rows)
    for element in range(rows):
        count = 1
        while count < rows - element:
            pascals_triangle[count + element].append(0)
            count += 1
    for row in pascals_triangle:
        row.insert(0, 1)
        row.append(1)
    pascals_triangle.insert(0, [1, 1])
    pascals_triangle.insert(0, [1])

pascals_tri_gen(6)

for row in pascals_triangle:
    print(row)

返回

[1]
[1, 1]
[1, 0, 1]
[1, 0, 0, 1]
[1, 0, 0, 0, 1]
[1, 0, 0, 0, 0, 1]
[1, 0, 0, 0, 0, 0, 1]
[1, 0, 0, 0, 0, 0, 0, 1]

但是，我不知道从哪里开始。我一直在墙上撞了好几个小时。我想强调一点，我不希望你为我做这件事;只是把我推向正确的方向。作为列表，我的代码返回

[[1], [1, 1], [1, 0, 1], [1, 0, 0, 1], [1, 0, 0, 0, 1], [1, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1]]

感谢。

编辑：我提出了一些很好的建议，我完全重写了我的代码，但我现在遇到了另一个问题。这是我的代码。

import math

pascals_tri_formula = []

def combination(n, r):
    return int((math.factorial(n)) / ((math.factorial(r)) * math.factorial(n - r)))

def for_test(x, y):
    for y in range(x):
        return combination(x, y)

def pascals_triangle(rows):
    count = 0
    while count <= rows:
        for element in range(count + 1):
            [pascals_tri_formula.append(combination(count, element))]
        count += 1

pascals_triangle(3)

print(pascals_tri_formula)

但是，我发现输出有点不受欢迎：

[1, 1, 1, 1, 2, 1, 1, 3, 3, 1]

我该如何解决这个问题？

12 个答案:

答案 0 :(得分：14)

确定代码审核：

import math

# pascals_tri_formula = [] # don't collect in a global variable.

def combination(n, r): # correct calculation of combinations, n choose k
    return int((math.factorial(n)) / ((math.factorial(r)) * math.factorial(n - r)))

def for_test(x, y): # don't see where this is being used...
    for y in range(x):
        return combination(x, y)

def pascals_triangle(rows):
    result = [] # need something to collect our results in
    # count = 0 # avoidable! better to use a for loop, 
    # while count <= rows: # can avoid initializing and incrementing 
    for count in range(rows): # start at 0, up to but not including rows number.
        # this is really where you went wrong:
        row = [] # need a row element to collect the row in
        for element in range(count + 1): 
            # putting this in a list doesn't do anything.
            # [pascals_tri_formula.append(combination(count, element))]
            row.append(combination(count, element))
        result.append(row)
        # count += 1 # avoidable
    return result

# now we can print a result:
for row in pascals_triangle(3):
    print(row)

打印：

[1]
[1, 1]
[1, 2, 1]

Pascal三角形的说明：

这是"n choose k"的公式（即有多少种不同的方式（无视顺序），从n项的有序列表中，我们可以选择k项）：

from math import factorial

def combination(n, k): 
    """n choose k, returns int"""
    return int((factorial(n)) / ((factorial(k)) * factorial(n - k)))

一位评论者询问这是否与itertools.combinations有关 - 事实确实如此。 ＆＃34; n选择k＆＃34;可以通过从组合中获取元素列表的长度来计算：

from itertools import combinations

def pascals_triangle_cell(n, k):
    """n choose k, returns int"""
    result = len(list(combinations(range(n), k)))
    # our result is equal to that returned by the other combination calculation:
    assert result == combination(n, k)
    return result

让我们看看这个证明：

from pprint import pprint

ptc = pascals_triangle_cell

>>> pprint([[ptc(0, 0),], 
            [ptc(1, 0), ptc(1, 1)], 
            [ptc(2, 0), ptc(2, 1), ptc(2, 2)],
            [ptc(3, 0), ptc(3, 1), ptc(3, 2), ptc(3, 3)],
            [ptc(4, 0), ptc(4, 1), ptc(4, 2), ptc(4, 3), ptc(4, 4)]],
           width = 20)
[[1],
 [1, 1],
 [1, 2, 1],
 [1, 3, 3, 1],
 [1, 4, 6, 4, 1]]

我们可以避免重复使用嵌套列表理解：

def pascals_triangle(rows):
    return [[ptc(row, k) for k in range(row + 1)] for row in range(rows)]

>>> pprint(pascals_triangle(15))
[[1],
 [1, 1],
 [1, 2, 1],
 [1, 3, 3, 1],
 [1, 4, 6, 4, 1],
 [1, 5, 10, 10, 5, 1],
 [1, 6, 15, 20, 15, 6, 1],
 [1, 7, 21, 35, 35, 21, 7, 1],
 [1, 8, 28, 56, 70, 56, 28, 8, 1],
 [1, 9, 36, 84, 126, 126, 84, 36, 9, 1],
 [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1],
 [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1],
 [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1],
 [1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1],
 [1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]]

递归定义：

我们可以使用三角形所示的关系递归地定义（效率较低，但可能更具数学优雅的定义）：

 def choose(n, k): # note no dependencies on any of the prior code
     if k in (0, n):
         return 1
     return choose(n-1, k-1) + choose(n-1, k)

为了好玩，您可以看到每一行的执行时间越来越长，因为每行必须每次重新计算前一行中的每个元素两次：

for row in range(40):
    for k in range(row + 1):
        # flush is a Python 3 only argument, you can leave it out,
        # but it lets us see each element print as it finishes calculating
        print(choose(row, k), end=' ', flush=True) 
    print()


1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 ...

当你厌倦了观看它时退出Ctrl-C，它会变得非常慢......

答案 1 :(得分：10)

我知道你想要自己实现，但我解释的最好方法是完成一个实现。这是我如何做到的，这个实现依赖于我对Python函数如何工作的相当完整的知识，所以你可能不想自己想使用这个代码，但它可能会让你指向正确的方向。

def pascals_triangle(n_rows):
    results = [] # a container to collect the rows
    for _ in range(n_rows): 
        row = [1] # a starter 1 in the row
        if results: # then we're in the second row or beyond
            last_row = results[-1] # reference the previous row
            # this is the complicated part, it relies on the fact that zip
            # stops at the shortest iterable, so for the second row, we have
            # nothing in this list comprension, but the third row sums 1 and 1
            # and the fourth row sums in pairs. It's a sliding window.
            row.extend([sum(pair) for pair in zip(last_row, last_row[1:])])
            # finally append the final 1 to the outside
            row.append(1)
        results.append(row) # add the row to the results.
    return results

用法：

>>> for i in pascals_triangle(6):
...     print(i)
... 
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]

答案 2 :(得分：5)

不使用zip，但使用生成器： 

def gen(n,r=[]):
    for x in range(n):
        l = len(r)
        r = [1 if i == 0 or i == l else r[i-1]+r[i] for i in range(l+1)]
        yield r

示例： 

print(list(gen(15)))

输出 

[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1], [1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56, 28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], [1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1], [1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1]]

以三角形显示

以漂亮的三角形绘制（仅适用于n <7，超出此范围时会被分散。参考draw_beautiful为n＆gt; 7）

对于n＆lt; 7 

def draw(n):
    for p in gen(n):
        print(' '.join(map(str,p)).center(n*2)+'\n')

<强>例如

draw(10）

输出 

      1       

     1 1      

    1 2 1     

   1 3 3 1    

  1 4 6 4 1   

1 5 10 10 5 1

适用于任何尺寸

因为我们需要知道最大宽度，所以我们无法使用发生器 

def draw_beautiful(n):
    ps = list(gen(n))
    max = len(' '.join(map(str,ps[-1])))
    for p in ps:
        print(' '.join(map(str,p)).center(max)+'\n')

例子（2）： 适用于任何数字：

draw_beautiful(100)

example of n = 100

答案 3 :(得分：2)

这是我的尝试：

def generate_pascal_triangle(rows):
    if rows == 1: return [[1]]

    triangle = [[1], [1, 1]] # pre-populate with the first two rows

    row = [1, 1] # Starts with the second row and calculate the next

    for i in range(2, rows):
        row = [1] + [sum(column) for column in zip(row[1:], row)] + [1]
        triangle.append(row)

    return triangle

for row in generate_pascal_triangle(6):
    print row

讨论

  • 三角形的前两行是硬编码的
  • zip()来电基本上将两个相邻的数字组合在一起
  • 我们仍然需要在开头添加1，在结尾添加1，因为zip()调用仅生成下一行的中间

答案 4 :(得分：1)

def pascal(n):
    if n==0:
        return [1]
    else:
        N = pascal(n-1)
        return [1] + [N[i] + N[i+1] for i in range(n-1)] + [1]


def pascal_triangle(n):
    for i in range(n):
        print pascal(i)

答案 5 :(得分：1)

这是一个优雅而有效的递归解决方案。我正在使用非常方便的toolz库。

from toolz import memoize, sliding_window

@memoize
def pascals_triangle(n):
    """Returns the n'th row of Pascal's triangle."""
    if n == 0:
        return [1]
    prev_row = pascals_triangle(n-1)
    return [1, *map(sum, sliding_window(2, prev_row)), 1]

pascals_triangle(300)在macbook pro（2.9 GHz Intel Core i5）上大约需要15 ms。请注意，在不增加默认递归深度限制的情况下，您可以更高。

答案 6 :(得分：0)

# combining the insights from Aaron Hall and Hai Vu,
# we get:

def pastri(n):
    rows = [[1]]
    for _ in range(1, n+1):
        rows.append([1] +
                    [sum(pair) for pair in zip(rows[-1], rows[-1][1:])] +
                    [1])
    return rows

# thanks! learnt that "shape shifting" data,
# can yield/generate elegant solutions.

答案 7 :(得分：0)

# call the function ! Indent properly , everything should be inside the function
def triangle():

          matrix=[[0 for i in range(0,20)]for e in range(0,10)]         # This method assigns 0's to all Rows and Columns , the range is mentioned
          div=20/2           # it give us the most middle columns 
          matrix[0][div]=1        # assigning 1 to the middle of first row 
          for i in range(1,len(matrix)-1): # it goes column by column
               for j in range(1,20-1):  #  this loop goes row by row
                   matrix[i][j]=matrix[i-1][j-1]+matrix[i-1][j+1]               # this is the formula , first element of the matrix gets , addition of i index (which is 0 at first ) with third value on the the related row
    # replacing 0s with spaces :) 
          for i in range(0,len(matrix)):
              for j in range(0,20):
                   if matrix[i][j]==0:       #  Replacing 0's with spaces
                        matrix[i][j]=" "

          for i in range(0,len(matrix)-1):           # using spaces , the triangle will printed beautifully 
                for j in range(0,20):
                    print 1*" ",matrix[i][j],1*" ", # giving some spaces in two sides of the printing numbers
triangle() # calling the function

会打印出类似这样的内容

                       1
                1              1
           1           2            1
      1         3            3            1
  1        4        6            4               1

答案 8 :(得分：0)

我在流行的fibonacci sequence解决方案中作弊。对我来说，Pascal三角形的实现与斐波那契的概念相同。在斐波那契，我们一次使用一个数字，并将其添加到前一个数字。在pascal的三角形中，一次使用一行并将其添加到前一行。

以下是完整的代码示例：

>>> def pascal(n):
...     r1, r2 = [1], [1, 1]
...     degree = 1
...     while degree <= n:
...         print(r1)
...         r1, r2 = r2, [1] + [sum(pair) for pair in zip(r2, r2[1:]) ] + [1]
...         degree += 1

<强>测试 

>>> pascal(3)
[1]
[1, 1]
[1, 2, 1]
>>> pascal(4)
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
>>> pascal(6)
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]

注意：要将结果作为生成器，请将print(r1)更改为yield r1。

答案 9 :(得分：0)

初学Python学生。这是我的尝试，一种非常直接的方法，使用两个For循环：

pascal = [[1]]
num = int(input("Number of iterations: "))
print(pascal[0]) # the very first row
for i in range(1,num+1):
    pascal.append([1]) # start off with 1
    for j in range(len(pascal[i-1])-1):
    # the number of times we need to run this loop is (# of elements in the row above)-1
        pascal[i].append(pascal[i-1][j]+pascal[i-1][j+1])
        # add two adjacent numbers of the row above together
    pascal[i].append(1) # and cap it with 1
    print(pascal[i])

答案 10 :(得分：0)

这是实现Pascal三角形的简单方法：

def pascal_triangle(n):
    myList = []
    trow = [1]
    y = [0]
    for x in range(max(n,0)):
        myList.append(trow)
        trow=[l+r for l,r in zip(trow+y, y+trow)]

    for item in myList:
        print(item)

pascal_triangle(5)

Python zip（）函数返回zip对象，它是元组的迭代器，其中每个传递的迭代器中的第一项配对在一起，然后每个传递的迭代器中的第二项配对在一起。 Python zip是在其中保存真实数据的容器。

Python zip（）函数接受可迭代项（可以为零或多个），使其成为一个迭代器，该迭代器根据传递的可迭代项聚合项目，并返回元组的迭代器。

答案 11 :(得分：0)

当我和我的儿子一起制作 Python 介绍片时，我就这样做了。当我们瞄准时，它开始是一个相当简单的作品 -

1
1 2
1 2 3
1 2 3 4

然而，一旦我们达到实际算法，复杂性就超出了我们的预期。无论如何，我们确实构建了这个 -

             1    
           1   1    
         1   2   1    
       1   3   3   1    
     1   4   6   4   1    
   1   5   10   10   5   1    
 1   6   15   20   15   6  1

使用了一些递归 -

def genRow(row:list) :  
  # print(f"generatig new row below {row}")
  # printRow(row)
  l = len(row) #2
  newRow : list = []
  i = 0
  # go through the incoming list
  while i <= l:
    # print(f"working with i = {i}")
    # append an element in the new list
    newRow.append(1)
    # set first element of the new row to 1
    if i ==0:
      newRow[i] = 1
      # print(f"1:: newRow = {newRow}")

    # if the element is in the middle somewhere, add the surroundng two elements in
    # previous row to get the new element
    # e.g. row 3[2] = row2[1] + row2[2]
    elif i <= l-1:
      # print(f"2:: newRow = {newRow}")      
      newRow[i] = row[i-1] + row[i]
    else:
      # print(f"3 :: newRow = {newRow}")      
      newRow[i] = 1
    
    i+=1
  # print(newRow)
  return newRow

def printRow(mx : int, row:list):
  n = len(row)
  spaces = ' ' *((mx - n)*2)
  print(spaces,end=' ')
  for i in row:
    print(str(i) + '  ',end = ' ')
  print(' ')


r = [1,1]
mx = 7
printRow(mx,[1])
printRow(mx,r)

for a in range(1,mx-1):
  # print(f"working for Row = {a}")
  if len(r) <= 2:
    a1 = genRow(r)
    r=a1
  else:
    a2 = genRow(a1)
    a1 = a2
  printRow(mx,a1)

希望它有所帮助。

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