Pascal的Python三角形

Question

作为Python的学习经历,我试图编写自己的Pascal三角形版本.它花了我几个小时(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]]

谢谢.

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]

我怎样才能解决这个问题？

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选择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."n选择k"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,但可能更加数学上优雅的定义):

 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,Ctrl-C退出,速度非常快......

Answer 2

我知道你想要实现自己,但我解释的最好方法是完成一个实现.这是我会怎么做,这实现依赖于我的Python的功能是如何工作的相当完整的知识,所以你可能不希望使用自己的代码,但它可以让你在正确的方向.

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]

Answer 3

不使用zip，而是使用generator：

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：

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

将其绘制为漂亮的三角形（仅适用于n <7，除此之外，它会变形。ref draw_beautiful适用于n> 7）

对于n <7

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

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   

任何尺寸

因为我们需要知道最大宽度，所以我们不能使用发电机

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)