我需要创建一个不使用itertools
的函数,这将创建一个具有给定任何内容的元组排列列表。
例:
perm({1,2,3}, 2)
应该返回[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]
这就是我得到的:
def permutacion(conjunto, k):
a, b = list(), list()
for i in conjunto:
if len(b) < k and i not in b:
b.append(i)
b = tuple(b)
a.append(b)
return a
我知道这不会做任何事情,它会添加第一个组合,没有别的。
正如@John在注释中提到的,itertools.permutations
的代码是:
def permutations(iterable, r=None):
# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
# permutations(range(3)) --> 012 021 102 120 201 210
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
if r > n:
return
indices = list(range(n))
cycles = list(range(n, n-r, -1))
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[i:] = indices[i+1:] + indices[i:i+1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return
这适用于不使用外部导入或递归调用的示例:
for x in permutations([1,2,3],2):
print (x)
(1, 2)
(1, 3)
(2, 1)
(2, 3)
(3, 1)
(3, 2)
我对@Hooked的答案有问题...
首先,在py方面,我是一个完全的新手,但我正在寻找类似于上述代码的东西。 我目前正在 Repl.it 输入它
我的第一个问题是争论
for x in permutations([1,2,3],2):
print x
返回以下错误
line 26
print x
^
SyntaxError: Missing parentheses in call to 'print'
我这样修复了这个问题
for x in permutations([1,2,3],2):
print (x)
但是现在得到错误:
line 25, in <module>
for x in permutations([1,2,3],2):
File "main.py", line 14, in permutations
cycles[i] -= 1
TypeError: 'range' object does not support item assignment
现在在这一点上,我不知道去哪里调试代码。 但是,我看到很多人指出迭代工具在其文档中包含代码。 我复制了它并且它有效。 这是代码:
def permutations(iterable, r=None):
# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
# permutations(range(3)) --> 012 021 102 120 201 210
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
if r > n:
return
indices = list(range(n))
cycles = list(range(n, n-r, -1))
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[i:] = indices[i+1:] + indices[i:i+1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return