python中的帕斯卡三角形

Question

所以我正在制作一个帕斯卡三角形，但我无法弄清楚为什么这段代码不起作用。它打印出这样的东西

[]
[1]
[1, 2]
[1, 3, 3]
[1, 4, 6, 4]
[1, 5, 10, 10, 5]
[1, 6, 15, 20, 15, 6]
[1, 7, 21, 35, 35, 21, 7]
[1, 8, 28, 56, 70, 56, 28, 8]
[1, 9, 36, 84, 126, 126, 84, 36, 9]

这几乎是正确的，但是似乎这些值似乎是行号太高，所以 1,2 ，如果将 2 计为第一行，将 1 计为第 0 行，则 2 是 1 值太高，因为它在第一行，应该是 1,1。下一行应该是 1,2,1，第一个值是正确的，但是下一行是 1 的值太高，后面是 2 的值太高。

我尝试执行 append(a-i) 之类的操作，但它似乎不起作用，您如何才能正确打印？

def triangle(rows):

    for rownum in range (rows):
        newValue=1
        PrintingList = list()
        for iteration in range (rownum):
            newValue = newValue * ( rownum-iteration ) * 1 / ( iteration + 1 )
            PrintingList.append(int(newValue))
        print(PrintingList)
    print()

提前致谢。

Answer 1

我会将PrintingList = list() 更改为PrintingList = [newValue]。

triangle(10) 然后为您提供以下信息：

[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]

这是一个有效的帕斯卡三角形。

Answer 2

我下面的代码有一些打印语句，有助于显示它是如何工作的。我在 IPython 笔记本中运行它。我的输出如下所示：

                   [1]                        
                  [1, 1]                      
                [1, 2, 1]                     
               [1, 3, 3, 1]                   
             [1, 4, 6, 4, 1]        

除了输出看起来应该的事实之外，计算方法实际上遵循帕斯卡三角形的原始定义，其中每一行都是从上面的一个构建的。 （有些方法使用组合数/阶乘，这可能会给出正确的答案，但不遵循帕斯卡三角的原始精神，afaik。）

代码如下：

def MakeTriangle(numberOfRows):
    """the print statements are not required. They are just to help see how it works."""
    triangle=[[1]] #base case (i.e., 0th row)
    print 'triangle[-1] is ', triangle[-1]
    for _x in range(numberOfRows - 1): #_x isn't used
        zipperd = zip([0] + triangle[-1], triangle[-1] + [0]) 
        print "zipperd: ", zipperd
        newRow = map(sum, zipperd) 
        triangle.append(list(newRow))
        print 'triangle[-1] is ', triangle[-1] #to help see how tri is built
    print "\n\n"
    return triangle

for row in MakeTriangle(5):
    print('{0:^50}'.format(row))

要开始了解它的工作原理，请注意上述函数zip 和map 的以下简单案例：

x = [1, 2, 3]
y = [4, 5, 6]
zipped = zip(x, y)
print zipped #output is [(1, 4), (2, 5), (3, 6)]
print map(sum, zipped) #output is [5, 7, 9]

使用 MakeTriangle 函数中的打印语句，您将获得此输出以帮助您进一步了解代码的工作原理：

triangle[-1] is  [1]
zipperd:  [(0, 1), (1, 0)]
triangle[-1] is  [1, 1]
zipperd:  [(0, 1), (1, 1), (1, 0)]
triangle[-1] is  [1, 2, 1]
zipperd:  [(0, 1), (1, 2), (2, 1), (1, 0)]
triangle[-1] is  [1, 3, 3, 1]
zipperd:  [(0, 1), (1, 3), (3, 3), (3, 1), (1, 0)]
triangle[-1] is  [1, 4, 6, 4, 1]

在查看此输出的同时阅读下面的文档，它将开始有意义：

• http://docs.python.org/2/library/functions.html#zip
• http://docs.python.org/2/library/functions.html#map

如果仍然没有意义，请将这些打印语句粘贴到 MakeTriangle 的 for 循环中：

    print "[0] + triangle[-1] is ", [0] + triangle[-1]
    print "triangle[-1] + [0] is ", triangle[-1] + [0]

请注意，对齐（居中）打印是使用http://docs.python.org/2/library/string.html#format-examples 中的示例完成的。

另见https://stackoverflow.com/a/19895685/463994

Answer 3

我收到一个错误，但您应该尝试以下方法：

row1 = whatever

pascal = [[row/(colum*(row-colum) for row in range(0, colum)] for row in range(1, row1)]]

Answer 4

import math

coef = 1

rows = int(input("Enter the number of rows :"))

stringVal = ""

for i in range(rows):

    for space in range(1,rows - i + 1):

        stringVal = stringVal + "  "
    for j in range(0,i + 1):
        if(i == 0 or j == 0):
            coef = 1
        else:
            coef = coef * (i - j + 1) / j

        temp = coef
        TotalSpace=0
        while(temp != 0):
            TotalSpace = TotalSpace + 1
            temp = int(math.floor(temp / 10))
        p=0
        while((p+TotalSpace)!=4):
            stringVal = stringVal + " "
            p=p+1
        stringVal = stringVal + str(int(math.floor(coef)))
    stringVal = stringVal + "\n"
print(stringVal)