Python 具有唯一值的置换
itertools.permutations生成的元素根据其位置而不是价值被视为唯一的元素。所以基本上我想避免像这样的重复:Python 具有唯一值的置换,python,permutation,itertools,Python,Permutation,Itertools,itertools.permutations生成的元素根据其位置而不是价值被视为唯一的元素。所以基本上我想避免像这样的重复: >>> list(itertools.permutations([1, 1, 1])) [(1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1)] 之后过滤是不可能的,因为在我的例子中排列的数量太大了 有人知道一个合适的算法吗 多谢各位 编辑: 我基本上想要的是: x = ite
>>> list(itertools.permutations([1, 1, 1]))
[(1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1)]
之后过滤是不可能的,因为在我的例子中排列的数量太大了
有人知道一个合适的算法吗
多谢各位
编辑:
我基本上想要的是:
x = itertools.product((0, 1, 'x'), repeat=X)
x = sorted(x, key=functools.partial(count_elements, elem='x'))
[''.join(perm) for perm in unique_permutations('abunchofletters')]
from itertools import combinations,chain
l = ["A","B","C","D"]
combs = (combinations(l, r) for r in range(1, len(l) + 1))
list_combinations = list(chain.from_iterable(combs))
这是不可能的,因为sorted
创建了一个列表,并且itertools.product的输出太大
对不起,我应该已经描述了实际问题。您可以尝试使用set:
>>> list(itertools.permutations(set([1,1,2,2])))
[(1, 2), (2, 1)]
调用set删除的重复项您可以尝试使用set:
>>> list(itertools.permutations(set([1,1,2,2])))
[(1, 2), (2, 1)]
调用set removed duplicates听起来像是在寻找itertools.combines()
听起来您正在寻找itertools.compositions() 编辑(如何工作): 我重写了上面的程序,使其更长,但更具可读性 我通常很难解释某些东西是如何工作的,但让我试试。 为了理解这是如何工作的,您必须了解一个类似但更简单的程序,该程序将产生所有重复排列
def permutations_with_replacement(elements,n):
return permutations_helper(elements,[0]*n,n-1)#this is generator
def permutations_helper(elements,result_list,d):
if d<0:
yield tuple(result_list)
else:
for i in elements:
result_list[d]=i
all_permutations = permutations_helper(elements,result_list,d-1)#this is generator
for g in all_permutations:
yield g
def置换与替换(元素,n):
返回置换_helper(元素,[0]*n,n-1)#这是生成器
def置换辅助程序(元素、结果列表、d):
如果d
编辑(如何工作):
我重写了上面的程序,使其更长,但更具可读性
我通常很难解释某些东西是如何工作的,但让我试试。
为了理解这是如何工作的,您必须了解一个类似但更简单的程序,该程序将产生所有重复排列
def permutations_with_replacement(elements,n):
return permutations_helper(elements,[0]*n,n-1)#this is generator
def permutations_helper(elements,result_list,d):
if d<0:
yield tuple(result_list)
else:
for i in elements:
result_list[d]=i
all_permutations = permutations_helper(elements,result_list,d-1)#this is generator
for g in all_permutations:
yield g
def置换与替换(元素,n):
返回置换_helper(元素,[0]*n,n-1)#这是生成器
def置换辅助程序(元素、结果列表、d):
如果d这取决于实现细节,即排序的iterable的任何排列都是按排序顺序排列的,除非它们是先前排列的重复
from itertools import permutations
def unique_permutations(iterable, r=None):
previous = tuple()
for p in permutations(sorted(iterable), r):
if p > previous:
previous = p
yield p
for p in unique_permutations('cabcab', 2):
print p
给予
这取决于实现细节,即排序的iterable的任何排列都是按排序顺序排列的,除非它们是先前排列的副本
from itertools import permutations
def unique_permutations(iterable, r=None):
previous = tuple()
for p in permutations(sorted(iterable), r):
if p > previous:
previous = p
yield p
for p in unique_permutations('cabcab', 2):
print p
给予
那怎么办
np.unique(itertools.permutations([1, 1, 1]))
问题是排列现在是Numpy数组的行,因此使用了更多内存,但您可以像以前一样循环使用它们
perms = np.unique(itertools.permutations([1, 1, 1]))
for p in perms:
print p
那怎么办
np.unique(itertools.permutations([1, 1, 1]))
问题是排列现在是Numpy数组的行,因此使用了更多内存,但您可以像以前一样循环使用它们
perms = np.unique(itertools.permutations([1, 1, 1]))
for p in perms:
print p
前几天我在处理自己的问题时遇到了这个问题。我喜欢Luka Rahne的方法,但我认为在collections library中使用Counter类似乎是一个适度的改进。这是我的密码:
def unique_permutations(elements):
"Returns a list of lists; each sublist is a unique permutations of elements."
ctr = collections.Counter(elements)
# Base case with one element: just return the element
if len(ctr.keys())==1 and ctr[ctr.keys()[0]] == 1:
return [[ctr.keys()[0]]]
perms = []
# For each counter key, find the unique permutations of the set with
# one member of that key removed, and append the key to the front of
# each of those permutations.
for k in ctr.keys():
ctr_k = ctr.copy()
ctr_k[k] -= 1
if ctr_k[k]==0:
ctr_k.pop(k)
perms_k = [[k] + p for p in unique_permutations(ctr_k)]
perms.extend(perms_k)
return perms
这段代码以列表的形式返回每个排列。如果你给它一个字符串,它会给你一个排列列表,其中每个排列都是一个字符列表。如果您希望以字符串列表的形式输出(例如,如果您是一个糟糕的人,您想滥用我的代码来帮助您在拼字游戏中作弊),只需执行以下操作:
x = itertools.product((0, 1, 'x'), repeat=X)
x = sorted(x, key=functools.partial(count_elements, elem='x'))
[''.join(perm) for perm in unique_permutations('abunchofletters')]
from itertools import combinations,chain
l = ["A","B","C","D"]
combs = (combinations(l, r) for r in range(1, len(l) + 1))
list_combinations = list(chain.from_iterable(combs))
前几天我在处理自己的问题时遇到了这个问题。我喜欢Luka Rahne的方法,但我认为在collections library中使用Counter类似乎是一个适度的改进。这是我的密码:
def unique_permutations(elements):
"Returns a list of lists; each sublist is a unique permutations of elements."
ctr = collections.Counter(elements)
# Base case with one element: just return the element
if len(ctr.keys())==1 and ctr[ctr.keys()[0]] == 1:
return [[ctr.keys()[0]]]
perms = []
# For each counter key, find the unique permutations of the set with
# one member of that key removed, and append the key to the front of
# each of those permutations.
for k in ctr.keys():
ctr_k = ctr.copy()
ctr_k[k] -= 1
if ctr_k[k]==0:
ctr_k.pop(k)
perms_k = [[k] + p for p in unique_permutations(ctr_k)]
perms.extend(perms_k)
return perms
这段代码以列表的形式返回每个排列。如果你给它一个字符串,它会给你一个排列列表,其中每个排列都是一个字符列表。如果您希望以字符串列表的形式输出(例如,如果您是一个糟糕的人,您想滥用我的代码来帮助您在拼字游戏中作弊),只需执行以下操作:
x = itertools.product((0, 1, 'x'), repeat=X)
x = sorted(x, key=functools.partial(count_elements, elem='x'))
[''.join(perm) for perm in unique_permutations('abunchofletters')]
from itertools import combinations,chain
l = ["A","B","C","D"]
combs = (combinations(l, r) for r in range(1, len(l) + 1))
list_combinations = list(chain.from_iterable(combs))
大概和卢卡·拉恩的答案一样快,但更简短,更简单
def unique_permutations(elements):
if len(elements) == 1:
yield (elements[0],)
else:
unique_elements = set(elements)
for first_element in unique_elements:
remaining_elements = list(elements)
remaining_elements.remove(first_element)
for sub_permutation in unique_permutations(remaining_elements):
yield (first_element,) + sub_permutation
>>> list(unique_permutations((1,2,3,1)))
[(1, 1, 2, 3), (1, 1, 3, 2), (1, 2, 1, 3), ... , (3, 1, 2, 1), (3, 2, 1, 1)]
它通过设置第一个元素(遍历所有唯一的元素)并遍历所有剩余元素的排列来递归工作
让我们通过(1,2,3,1)的唯一排列来了解它是如何工作的:
唯一元素是1,2,3
- 让我们遍历它们:
第一个元素
以1开头。
剩余的_元素是[2,3,1](即1,2,3,1减去前1)
- 我们(递归地)遍历剩余元素的排列:(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1),(3,2,1)
- 对于每个
子排列
,我们插入第一个元素
:(1,1,2,3),(1,1,3,2)。。。并得出结果
- 现在我们迭代到
第一个\u元素
=2,并执行与上面相同的操作。
剩余的_元素是[1,3,1](即1,2,3,1减去前2)
- 我们迭代剩余元素的排列:(1,1,3),(1,3,1),(3,1,1)
- 对于每个
子排列
,我们插入第一个元素
:(2,1,1,3),(2,1,3,1),(2,3,1,1)。。。并得出结果
- 最后,我们对
第一个\u元素
=3执行相同的操作
大概和卢卡·拉恩的答案一样快,但更简短,更简单
def unique_permutations(elements):
if len(elements) == 1:
yield (elements[0],)
else:
unique_elements = set(elements)
for first_element in unique_elements:
remaining_elements = list(elements)
remaining_elements.remove(first_element)
for sub_permutation in unique_permutations(remaining_elements):
yield (first_element,) + sub_permutation
>>> list(unique_permutations((1,2,3,1)))
[(1, 1, 2, 3), (1, 1, 3, 2), (1, 2, 1, 3), ... , (3, 1, 2, 1), (3, 2, 1, 1)]
它通过设置第一个元素(遍历所有唯一的元素)并遍历所有剩余元素的排列来递归工作
让我们通过(1,2,3,1)的唯一排列来了解它是如何工作的:
唯一元素是1,2,3
- 让我们遍历它们:
第一个元素
以1开头。
剩余的_元素是[2,3,1](即1,2,3,1减去前1)
- 我们(递归地)遍历剩余元素的排列:(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1),(3,2,1)
- 对于每个
子排列
,我们插入第一个元素
:(1,1,2,3),
['moon', 'mono', 'mnoo', 'omon', 'omno', 'nmoo', 'oomn', 'onmo', 'nomo', 'oonm', 'onom', 'noom']
from itertools import combinations,chain
l = ["A","B","C","D"]
combs = (combinations(l, r) for r in range(1, len(l) + 1))
list_combinations = list(chain.from_iterable(combs))
[('A',),
('B',),
('C',),
('D',),
('A', 'B'),
('A', 'C'),
('A', 'D'),
('B', 'C'),
('B', 'D'),
('C', 'D'),
('A', 'B', 'C'),
('A', 'B', 'D'),
('A', 'C', 'D'),
('B', 'C', 'D'),
('A', 'B', 'C', 'D')]
import statsmodels.formula.api as smf
import pandas as pd
# create some data
pd_dataframe = pd.Dataframe(somedata)
response_column = "Y"
# generate combinations of column/variable names
l = [col for col in pd_dataframe.columns if col!=response_column]
combs = (combinations(l, r) for r in range(1, len(l) + 1))
list_combinations = list(chain.from_iterable(combs))
# generate OLS input string
formula_base = '{} ~ '.format(response_column)
list_for_ols = [":".join(list(item)) for item in list_combinations]
string_for_ols = formula_base + ' + '.join(list_for_ols)
Y ~ A + B + C + D + A:B + A:C + A:D + B:C + B:D + C:D + A:B:C + A:B:D + A:C:D + B:C:D + A:B:C:D'
model = smf.ols(string_for_ols, pd_dataframe).fit()
model.summary()
import numba
@numba.njit
def perm_unique_fast(elements): #memory usage too high for large permutations
eset = set(elements)
dictunique = dict()
for i in eset: dictunique[i] = elements.count(i)
result_list = numba.typed.List()
u = len(elements)
for _ in range(u): result_list.append(0)
s = numba.typed.List()
results = numba.typed.List()
d = u
while True:
if d > 0:
for i in dictunique:
if dictunique[i] > 0: s.append((i, d - 1))
i, d = s.pop()
if d == -1:
dictunique[i] += 1
if len(s) == 0: break
continue
result_list[d] = i
if d == 0: results.append(result_list[:])
dictunique[i] -= 1
s.append((i, -1))
return results
import timeit
l = [2, 2, 3, 3, 4, 4, 5, 5, 6, 6]
%timeit list(perm_unique(l))
#377 ms ± 26 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
ltyp = numba.typed.List()
for x in l: ltyp.append(x)
%timeit perm_unique_fast(ltyp)
#293 ms ± 3.37 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
assert list(sorted(perm_unique(l))) == list(sorted([tuple(x) for x in perm_unique_fast(ltyp)]))
def perm_unique_fast_gen(elements):
eset = set(elements)
dictunique = dict()
for i in eset: dictunique[i] = elements.count(i)
result_list = list() #numba.typed.List()
u = len(elements)
for _ in range(u): result_list.append(0)
s = list()
d = u
while True:
if d > 0:
for i in dictunique:
if dictunique[i] > 0: s.append((i, d - 1))
i, d = s.pop()
if d == -1:
dictunique[i] += 1
if len(s) == 0: break
continue
result_list[d] = i
if d == 0: yield result_list
dictunique[i] -= 1
s.append((i, -1))
def perm_helper(head: str, tail: str):
if len(tail) == 0:
yield head
else:
last_c = None
for index, c in enumerate(tail):
if last_c != c:
last_c = c
yield from perm_helper(
head + c, tail[:index] + tail[index + 1:]
)
def perm_generator(word):
yield from perm_helper("", sorted(word))
from itertools import takewhile
word = "POOL"
list(takewhile(lambda w: w != word, (x for x in perm_generator(word))))
# output
# ['LOOP', 'LOPO', 'LPOO', 'OLOP', 'OLPO', 'OOLP', 'OOPL', 'OPLO', 'OPOL', 'PLOO', 'POLO']
ans=[]
def fn(a, size):
if (size == 1):
if a.copy() not in ans:
ans.append(a.copy())
return
for i in range(size):
fn(a,size-1);
if size&1:
a[0], a[size-1] = a[size-1],a[0]
else:
a[i], a[size-1] = a[size-1],a[i]