Python中的帕斯卡三角形

Question

作为Python学习经验，我正在尝试编写自己版本的帕斯卡三角形代码。由于我刚开始学习，所以花了几个小时时间来编写以下代码：

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)

这将返回

[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]

然而，我不知道该怎么做。我已经花费了数小时在这个问题上。我想强调的是，我并不希望你替我完成它;只需指导我正确的方向即可。就列表而言，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]]

谢谢。

编辑：我听取了一些好建议，完全重写了我的代码，但现在遇到了另一个问题。以下是我的代码。

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)

然而，我发现输出结果有些不理想：

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

我该如何修复这个问题？

- louie mcconnell

Answer 1

好的代码审查：

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三角形的解释：

这是"n choose k"的公式（即从一个有序列表中选择k个项目（不考虑顺序）的不同方式数）：

from math import factorial

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

有评论者问这是否与itertools.combinations有关 - 的确如此。 "n选k"可以通过从组合中取元素的列表的长度来计算：

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

让我们看一下这个演示：

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]]

我们可以使用嵌套列表推导式避免重复:

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它（这是一种不太高效但可能更数学优美的定义）：

 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，你可以看到每一行的执行时间逐渐变长，因为每一行都必须重新计算前一行中几乎每个元素两次：

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 ...

当你看腻了，按下Ctrl-C退出，它会变得非常慢...

- Russia Must Remove Putin

Answer 2

我知道你想自己实现，但是最好的方式是通过演示一个实现过程来解释。这是我会怎么做，这个实现依赖于我对Python函数如何工作的相当完整的了解，所以你可能不想使用这个代码本身，但它可以让你找到正确的方向。

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

使用方法：

>>> 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]

- Russia Must Remove Putin

Answer 3

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

例子：

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]]

绘制三角形

为了绘制一个美丽的三角形（仅适用于 n < 7，超过这个数字会变形。对于 n>7，请参考 draw_beautiful）

n < 7

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

例子：

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，所以不能使用生成器。

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')

示例（2）： 适用于任何数字：

draw_beautiful(100)

- MortalViews

Answer 4

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)

- Abdel

Answer 5

这里是我的尝试：

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() 函数基本上将相邻的两个数字配对在一起
• 我们仍然需要在开头添加1，在结尾再添加1，因为 zip() 函数只生成下一行的中间部分

- Hai Vu

Answer 6

我从流行的斐波那契数列解决方案中作弊。对我来说，帕斯卡三角形的实现与斐波那契的概念相同。在斐波那契中，我们一次使用一个数字并将其添加到前一个数字上。在帕斯卡三角形中，一次使用一行，并将其添加到前一行中。

这是完整的代码示例：

>>> 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

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：如果要将结果作为生成器返回，请将print(r1)更改为yield r1。

- Aziz Alto

Answer 7

这是一个优雅高效的递归解决方案。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 英特尔 Core i5) 上大约需要 15 毫秒。请注意，如果不增加默认递归深度限制，您无法再提高更多。

- fredcallaway

---

惊人的速度！ - ChoF

Answer 8

# 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.

- ssgam

Answer 9

我是一名初学者Python学生。这是我的尝试，采用了非常字面的方法，使用了两个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])

- user3932000

Answer 10

这里是实现帕斯卡三角形的简单方法：

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的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个项目被配对在一起，然后每个传递的迭代器中的第二个项目被配对在一起。Python zip是容纳实际数据的容器。

Python的zip()函数接受可迭代对象（可以是零个或多个），创建一个基于传递的可迭代对象聚合项的迭代器，并返回元组的迭代器。

- Colm Fitzpatrick