Question

这是我的代码，用于查找第三行pascals三角形

def pascaline(n):
     line = [1]       
     for k in range(max(n,0)):             
         line.append(line[k]*(n-k)/(k+1))             
     return line

我想问两件事。首先，输出整数以.0结尾，总是像

一样

pascaline(2) = [1, 2.0, 1.0]

如何删除那些.0？另外，我怎么能从$ n = 1 $而不是$ 0 $开始呢？例如，在这种情况下，pascaline（2）应为[1,1]而不是[1,2,1]

Answer 1

可以使用//而不是使用/的浮点除法来删除.0，因此您的代码将为line.append(line[k]*(n-k)//(k+1))。要让它重新开始，只需使用n -= 1减少一个。

def pascaline(n):
     n -= 1
     line = [1]
     for k in range(max(n,0)):             
         line.append(line[k]*(n-k)//(k+1))             
     return line


pascaline(2) >>> [1,1]

Answer 2

从n和typecast中减去1。基本上改变你的方法：

def pascaline(n):
    n = n - 1
    line = [1]

    for k in range(max(n ,0)):

        line.append(int(line[k]*(n-k)/(k+1)))

    return line
print(pascaline(5));

Answer 3

我知道你已经得到了答案。问题是你正在处理浮点数，不是整数。这是编程，而不是数学。数字具体表示。我只是想比较这两个实现，第一个实现让你通过使用对称来节省一些计算时间。但两者仍然是O（n）：

def pascal_line(n):
    line = [1]
    mid, even = divmod(n, 2)
    for k in range(1, mid + 1):
        num = int(line[k-1]*(n + 1 - k)/(k))
        line.append(num)
    reverse_it = reversed(line)
    if not even:
        next(reverse_it)
    for n in reverse_it:
        line.append(n)
    return line

def pascal_line2(n):
    line = [1]
    for k in range(1, n + 1):
        num = int(line[k-1]*(n + 1 - k)/(k))
        line.append(num)
    return line

现在，在行动中：

>>> for i in range(21):
...   print(pascal_line(i))
... 
[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, 153, 18, 1]
[1, 19, 171, 969, 3876, 11628, 27132, 50388, 75582, 92378, 92378, 75582, 50388, 27132, 11628, 3876, 969, 171, 19, 1]
[1, 20, 190, 1140, 4845, 15504, 38760, 77520, 125970, 167960, 184756, 167960, 125970, 77520, 38760, 15504, 4845, 1140, 190, 20, 1]

现在，快速而肮脏的时间测试：

>>> def time_me(f, n):
...   start = time.time()
...   f(n)
...   stop = time.time()
... 
>>> times = [time_me(pascal_line,n) for n in range(10, 1001,10)]
>>> times2 = [time_me(pascal_line2,n) for n in range(10, 1001,10)]
>>> import pandas as pd
>>> import matplotlib.pyplot as plt
>>> n = range(10, 1001,10)
>>> df = pd.DataFrame({'pascal_lines':times, 'pascal_lines2':times2},index=list(n))
>>> df.plot()
<matplotlib.axes._subplots.AxesSubplot object at 0x7f6c1c3d1c18>
>>> plt.savefig('pascal.png')

结果：

不确定是否值得，因为我在n = 1000之后很快就会出现OverFlow错误。

修改

正如其他人所指出的那样，使用整数除法而不是从float转换为int更有效。它还有一个额外好处，即在n = 1000左右后不抛出OverflowError: integer division result too large for a float。

def pascal_line0(n):
    line = [1]
    mid, even = divmod(n, 2)
    for k in range(1, mid + 1):
        num = line[k-1]*(n + 1 - k)//(k)
        line.append(num)
    reverse_it = reversed(line)
    if not even:
        next(reverse_it)
    for n in reverse_it:
        line.append(n)
    return line

如何获得pascal三角形的第n行

3 个答案:

修改