确定代码查看：
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)
印刷品：
[1]
[1, 1]
[1, 2, 1]
这是“ n选择k”的公式（即，从n项的有序列表中有多少种不同的方式（不考虑顺序），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)))
一个评论者问这是否与itertools.combinations有关-确实如此。“ n select k”可以通过组合中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]]
我们可以避免使用嵌套列表理解来重复自己：
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]]
我们可以使用三角形所示的关系来递归定义（效率较低，但数学上可能更优雅的定义）：
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)
有趣的是，您可以看到每一行的执行时间逐渐变长，因为每一行必须重新计算上一行中的几乎每个元素两次：
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 ...
厌倦了Ctrl-C退出时，它会变得非常慢非常快…
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